7.5: Logical Analysis using Truth Tables
Truth tables, which we introduced in the last section, are primarily useful in a different way. Instead of just using them to tell us what the different logical operators mean, we can use them to do some in depth analysis of the logical form of a statement, a set of statements, or an inference.
What’s a logical analysis? A logical analysis is a set of things we can do to learn something about a particular logical structure. The statement “I’ll go if you don’t go or if we can get a babysitter and it’s not too expensive for both of us to go” has a particular logical form—something like
[(B \(\wedge\) ~E) \(\rightarrow\) I] . If we do a logical analysis of this logical form, we’ll find out certain things. For instance: when is this statement true? That is, what must the world be like for this statement to turn out to be true? Is this statement always true? Is it always false? What is its relationship with other similar logical forms like [~(B \(\wedge\) E) \(\rightarrow\) I] ? [~(B \(\wedge\) ~E) \(\rightarrow\) ~I] ? [(B \(\wedge\) E) \(\rightarrow\) ~I] ? Is it possible that all of these statements could be true at the same time? That is, are they consistent? We can find out the answers to all of these questions using Truth Tables.
Conceptually speaking, we’re doing the following when we build a truth table:
1. Collecting all of the possible combinations of truth values and listing them out.
- That is, we’re finding out all of the different ways that T and F can be combined for our atomic sentence letters.
2. Using each set of possible truth values to calculate the output truth value for a complex formula (or a set of complex formulas).
- That is, we’re doing what we did in Computing Truth Values : we’re taking the truth values of atomic sentence letters as an input and calculating the single truth value output.
3. Analyzing the results.
- That is, we’re looking at the resulting output and trying to figure out what it tells us about the logical formulas in question.
The goal is to see what possible conditions of the world (what combinations of true and false for the atomic propositions, each of which either describes the world correctly or incorrectly) give what sorts of truth values for the complex formulas we’re analyzing and then to look for patterns in the outputs to tell us something about the individual statement, the set of statements, or the argument we’re analyzing.
That’s not terribly illuminating, though, about how to actually go about building a truth table. Let’s look more concretely at what to do.
First, I must point out a new notation that will be unfamiliar to you. Well, you will have seen it before lots of times, but not here in logic. The forward slash!
If you see this:
[\(\neg\)[D \(\leftrightarrow\) \(\neg\)(X\(\rightarrow\) (Z \(\bullet\) Q))] \(\vee\) P] / (X\(\rightarrow\) (Z \(\bullet\) Q))]
That means there are two formulas here:
[\(\neg\)[D \(\leftrightarrow\) \(\neg\)(X\(\rightarrow\) (Z \(\bullet\) Q))] \(\vee\) P]
And
(X\(\rightarrow\) (Z \(\bullet\) Q))]
The forward slash (/) has no logical meaning. It simply separates formulas from one another so we can list them on a single line without it seeming like they are parts of the same formula.
Building a Truth Table
Building a truth table is very straightforward, but that doesn’t mean it’s not going to take some getting used to. Let’s divide it into a series of steps.
Step One: Figure out what size you need
How many unique sentence letters are in the formula or set of formulas? Just count ‘em up, counting each unique letter only once (so two B’s just count as one). Here are some examples:
- [~(B\(\wedge\)E)\(\rightarrow\)I] has 3 unique letters: B, E, and I
- [~(B\(\wedge\)E)\(\rightarrow\)B] has 2 unique letter: B and E
- [~(I\(\wedge\)I)\(\rightarrow\)I] has 1 unique letter: I
- [~(B\(\wedge\)E)\(\rightarrow\)I] / [(B\(\wedge\)E)\(\rightarrow\)~I] have 3 unique letters all together: B, E, and I
- [\(\neg\)[D \(\leftrightarrow\) \(\neg\)(X\(\rightarrow\) (Z \(\bullet\) Q))] \(\vee\) P] / Z / (P \(\bullet\) Q) has 5 unique letters: D, X, Z, Q, and P
Once you’ve figure out this magic number, you plug it into a magic formula: 2 n , where n is the magic number: the number of unique sentence letters in the set of propositions the truth table is for. The result of this mathematical formula is the number of rows you’ll need in your truth table.
Here are a set of truth table sizes:
|
# of unique letters |
# of rows |
|---|---|
|
1 |
2 |
|
2 |
4 |
|
3 |
8 |
|
4 |
16 |
|
5 |
32 |
|
6 |
64 |
|
16 |
65,536 |
Notice the pattern? The nice thing is that chances are, your instructor will only assign at most a 4-letter truth table, so 16 rows is the absolute most you’ll usually need to work with. Most instructors stick to 1, 2, and 3-letter truth tables.
The downside of truth tables is that it doesn’t take long before you have to start making tables with 32, 64, 128, 256, 512, 65,536 rows! That’s too much to really make making a truth table worthwhile. This is called the problem of Combinatorial Explosion because all of the combinations that are possible “explode” to astronomical numbers . Nevertheless, they are quite useful for relatively simple problems.
Step Two: Make a truth table
A truth table is a table with a column for each unique sentence letter (usually in the order in which they show up in the formulas you are analyzing) and then a column for each formula you are analyzing. Here are some examples:
|
R |
R \(\equiv\) (R \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) R |
|---|---|---|
|
Q |
R |
R \(\equiv\) (Q \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) Q |
|---|---|---|---|
Notice how each column gets its own label and there’s a bigger border separating the individual sentence letters (the inputs) from the complex formulas (the outputs). So far so good.
Step Three: Fill in the input side
This step is always the same no matter what the columns labels are. There are basically 3 different tables you’ll make, and for bigger tables you can extrapolate the same basic pattern. Remember that the goal when filling in the input side is to make a list of all of the possible combinations of the two truth values T and F.
So when we only have one unique sentence letter. The two possibilities are that the letter is True and the letter is False:
|
R |
R \(\equiv\) (R \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) R |
|---|---|---|
|
T |
||
|
F |
It’s a bit more complicated if we have two unique sentence letters:
|
Q |
R |
R \(\equiv\) (Q \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) Q |
|---|---|---|---|
|
T |
T |
||
|
T |
F |
||
|
F |
T |
||
|
F |
F |
Each row in the table is a possible set of truth values. The possible combos are TT, TF, FT, and FF. See how that works?
Before moving onto an eight row truth table, let’s think about the pattern here. We’ve started on the right nearest the thick line separating the inputs from the outputs and alternated going down: TFTF. For the first table, we alternated too, but just had to do it once: TF. On the second table, we alternated TF going down the R column, and then just repeated so the downward pattern would be TFTF.
Then we moved left and alternated every 2, so the Q column reads TTFF. Why would we do this? Well, we have R is true and R is false, then we need to test for when Q is True with both of these possibilities, and then test again for when Q is false. The result is something like the following:
|
Q is True |
R is True |
|
R is False |
|
|
Q is False |
R is True |
|
R is False |
It may even be helpful to think of it in these terms:
This way, we’re getting all the possible combinations of Q:true, Q:false, R:true, and R:false. If this little explanation is confusing for you, it’s probably best to move on. Perhaps it will make more sense later, and even if it doesn’t, it’s okay since this is sort of conceptual background work rather than something that is vital to understanding the truth table. What you need to understand at minimum is simply that by following this procedure, you’re creating all of the possible combinations of T and F for the atomic sentence letters in the formulas you are trying to analyze using the truth table.
Now let’s look at an eight row table. The first thing we do is start on the rightmost input (sentence letter) column, and alternate T and F every one row. Like so:
|
P |
Q |
R |
P \(\equiv\) (Q \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) Q |
|---|---|---|---|---|
|
T |
||||
|
F |
||||
|
T |
||||
|
F |
||||
|
T |
||||
|
F |
||||
|
T |
||||
|
F |
Then we move one column to the left and alternate every two . Like this:
|
P |
Q |
R |
P \(\equiv\) (Q \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) Q |
|---|---|---|---|---|
|
T |
T |
|||
|
T |
F |
|||
|
F |
T |
|||
|
F |
F |
|||
|
T |
T |
|||
|
T |
F |
|||
|
F |
T |
|||
|
F |
F |
Finally, we move to the left again and alternate every four rows so that we double the amount we are alternating by each time we move to the left. Like this:
|
P |
Q |
R |
P \(\equiv\) (Q \(\rightarrow\) R) |
R \(\bullet\)\(\neg\) Q |
|---|---|---|---|---|
|
T |
T |
T |
||
|
T |
T |
F |
||
|
T |
F |
T |
||
|
T |
F |
F |
||
|
F |
T |
T |
||
|
F |
T |
F |
||
|
F |
F |
T |
||
|
F |
F |
F |
The result is a truth table that’s totally ready to solve: we have our columns labeled, our rows easily distinguishable, and most importantly we have all of the input side filled in in the standard way. This input side will be basically the same for every truth table. That is, you always follow this pattern:
Start below the rightmost atomic sentence letter and alternate every one row. Then move to the left one column and alternate going down every two rows. Finally, move to the left one last column and alternate going down that column every four rows. Extrapolate for bigger tables.
Solving a Truth Table
There are two methods to solving a truth table—filling in the right side or output side of the truth table. On the Brute Force method, you simply calculate each cell in the table by plugging the truth values of the sentence letters and working your way from the inside out. On the Intuitive method, you use your intuitive understanding of the operators to save yourself some work.
Brute Force Method
The Brute Force method to solving a truth table is simply to plug in the truth value for each individual letter by carrying them over from the input side and then calculating the truth value of each complex proposition given the truth values of the input side. This method is safer, so if you feel lost or lose confidence, then just revert back to the brute force method. It does have two downsides, though: a) It requires a lot more work and so takes more time, and b) it involves more individual steps and so the probably of making a simple mistake increases a bit. The second problem is probably balanced out by the riskiness of the Intuitive Method.
Write the truth value of each * letter * as it appears on the left side of the table under each letter as it appears on the right side
1. Starting with the “inner most” operators (inside the most parentheses) calculate the truth value of the whole relationship.
2. Then work your way out until you’ve calculated the truth value of the main operator.
So, step 0 is to make the truth table, as discussed in Basic Symbolization :
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
||
|
T |
F |
||
|
F |
T |
||
|
F |
F |
Step 1 is to fill in the truth values on the right side for the sentence letters
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
F |
T |
T T T |
T T T |
|
F |
F |
||
|
T |
T |
||
|
T |
F |
Step 2 is to find the inner most operators (the ones with the most amount of parentheses outside of them.
|
P |
Q |
(P \(\rightarrow\) \(\require{enclose} \enclose{circle}{(Q\leftrightarrow P)}\)) |
(Q \(\leftrightarrow\) \(\neg\)\(\enclose{circle}{(P\vee Q)}\)) |
|---|---|---|---|
|
F |
T |
T T T |
T T T |
|
F |
F |
||
|
T |
T |
||
|
T |
F |
And then solve for those values using the truth tables that we use to define each operator.
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
T T |
T T |
|
T |
F |
||
|
F |
T |
||
|
F |
F |
Repeat the steps in 2 (working from “in” to “out”) until you’ve solved the truth value of the main operators (the operators that are only inside the outermost parentheses. Keep in mind that if there are no outermost parentheses, then they are implied).
So Q \(\leftrightarrow\) (P\(\vee\)Q) is actually supposed to read: (Q \(\leftrightarrow\) (P\(\vee\)Q))
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
T T |
T F |
|
T |
F |
||
|
F |
T |
||
|
F |
F |
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
T |
F |
|
T |
F |
||
|
F |
T |
||
|
F |
F |
Repeat in each row:
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
\(\downarrow\) T \(\downarrow\) \(\downarrow\) |
\(\downarrow\) F \(\downarrow\) \(\downarrow\) |
|
T |
F |
T F T |
F T F |
|
F |
T |
F T F |
T F T |
|
F |
F |
F F F |
F F F |
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
\(\downarrow\) T \(\downarrow\) |
\(\downarrow\) F \(\downarrow\) |
|
T |
F |
T F |
F T |
|
F |
T |
F F |
T T |
|
F |
F |
F T |
F F |
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
\(\downarrow\) T \(\downarrow\) |
\(\downarrow\) F \(\downarrow\) |
|
T |
F |
T F |
F F |
|
F |
T |
F F |
T F |
|
F |
F |
F T |
F T |
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
T |
F |
|
T |
F |
F |
T |
|
F |
T |
T |
F |
|
F |
F |
T |
F |
Then you analyze the results by looking at the right side of the truth table!
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\leftrightarrow\) \(\neg\)(P\(\vee\)Q)) |
|---|---|---|---|
|
T |
T |
T |
F |
|
T |
F |
F |
T |
|
F |
T |
T |
F |
|
F |
F |
T |
F |
Intuitive Method
The Intuitive Method to solving a truth table uses our intuitive understanding of the logical operators and their individual truth tables to save as much work as possible. We can often eliminate half of our work for one column in a single swoop.
It is a faster method, but also increases the chances that we’ll make a mistake by moving too quickly and missing something or being overconfident and ignoring important details.
How does it work? Let’s start with an example and then we’ll look at how it works a bit more.
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
|||
|
T |
F |
|||
|
F |
T |
|||
|
F |
F |
Starting with the first column, we ask ourselves what we know about the main operator (\(\rightarrow\)). I notice that there is only one atomic letter as the antecedent to this conditional. I know that if the antecedent to a conditional is false the whole conditional is true (since it’s only false when T\(\rightarrow\)F and if it’s F\(\rightarrow\)?, then it’s certainly not T\(\rightarrow\)F!). So I just need to find the rows on which P is false and I know the whole first formula will be True!
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
|||
|
T |
F |
|||
|
F |
T |
T |
||
|
F |
F |
T |
Then we solve the rest more-or-less using the Brute Force method. When is (Q\(\leftrightarrow\)P) true? When they’re the same! It’s false if they’re different.
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
T |
||
|
T |
F |
F |
||
|
F |
T |
T |
||
|
F |
F |
T |
Now we look at the second column. When is a conjunction true? It’s only true in one case: when both conjuncts are true. So if Q is false...? Then the whole second formula (Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) is false.
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
T |
||
|
T |
F |
F |
F |
|
|
F |
T |
T |
||
|
F |
F |
T |
F |
Now that we know Q is true in the remaining cells of the second output column, we need to ask ourselves when \(\neg\)(P\(\vee\)Q) is true. It says “neither P nor Q are true”. When would that be true? Only when P and Q are both false (remember it would be equivalent to (\(\neg\)P\(\wedge\)\(\neg\)Q)). Are both false in rows 1 or 3? Nopers. So that means that \(\neg\)(P\(\vee\)Q) is false in both of our remaining rows. If just one conjunct is false, the whole conjunction is false. So:
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
T |
F |
|
|
T |
F |
F |
F |
|
|
F |
T |
T |
F |
|
|
F |
F |
T |
F |
Okay, final column. Another way of doing the intuitive method is simply to understand what the formula says in a more intuitive way than the logical formula. (Q \(\wedge\) \(\neg\)P) says something like “Q is true and P is false.” When we understand it this way, it’s easier to figure out on which row(s) it will be true: we’re looking for the row(s) where Q is true and P is false. Which row is that?
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
T |
F |
|
|
T |
F |
F |
F |
|
|
F |
T |
T |
F |
T |
|
F |
F |
T |
F |
Now we just fill in the rest as false since we know that on those rows P isn’t true while Q is false.
|
P |
Q |
(P \(\rightarrow\) (Q\(\leftrightarrow\)P)) |
(Q \(\wedge\) \(\neg\)(P\(\vee\)Q)) |
(Q \(\wedge\) \(\neg\)P) |
|---|---|---|---|---|
|
T |
T |
T |
F |
F |
|
T |
F |
F |
F |
F |
|
F |
T |
T |
F |
T |
|
F |
F |
T |
F |
F |
Okay, now that we’ve gone through the intuitive method, let’s take a look at some of the rules which we can use while doing the intuitive method. Here are the rules I use:
- Antecedent F or Consequent T \(\rightarrow\) Whole Implication T
- One disjunct T \(\rightarrow\) Whole Disjunction T
- One conjunct F \(\rightarrow\) Whole Conjunction F
-
“\(\neg\)Q” means “Q is
false
”.
- So “(\(\neg\)P\(\bullet\)Q)” means “P is false and Q is true ”
These four rules can save you loads of time. Just identify which is simplest in a formula (surrounding the main operator ): an antecedent? A consequent? A disjunct? A conjunct? Then you identify when that element fits the intuitive rule and eliminate lots of work!
Analyzing a Truth Table
Classification
If you’re being asked to analyze a single proposition using a truth table, then you automatically know that the answer will be one of three options. This is called a Classification problem because you’re classifying a single proposition.
1) Tautology/Tautologous : the column under the proposition is filled only with Ts.
A “Logical Truth” or Tautology is a statement that, regardless of how the world turns out to be, will be true. Think of “we’ll either have a democratic president or we won’t have a democratic president.” Even if the world ends and we have no president at all, that disjunctive statement is still true!
| T |
| T |
| T |
| T |
2) Self-Contradiction/Self-Contradictory : the column under the proposition is filled only with F’s.
A self-contradiction is always false no matter how the world ends up being. Think of the example “Kamala Harris is going to be our next president, but luckily we won’t have to have Kamala Harris as our next president.” It doesn’t matter what actually happens in the world, this statement will always be false. It can’t possibly be true because it contradicts itself.
| F |
| F |
| F |
| F |
3) Contingent : the column under the proposition is filled with a mixture of “T”s and “F”s.
Contingent propositions are true or false depending on how the world is. The simplest contingent propositions are atomic propositions, which either describe the world accurately or don’t describe the world accurately.
| F |
| F |
| T |
| F |
Comparison
If you’re being asked to analyze a set of propositions using a truth table, then you know that the answer will be one of the following four options. This is called a Comparison problem because you’re comparing multiple propositions with one another in order to determine what logical relation holds between them. This is the most complex type of truth table analysis problem.
1) Logically Equivalent : each row of the output side of the truth table is the same on each column. So each row is a homogeneous set of either all T’s or all F’s. Logically equivalent propositions are true and false in exactly the same states of the world: they give the same output every time.
| T | T |
| F | F |
| T | T |
| F | F |
2) Contradictory : the truth values are the opposite on each row of the output side of the truth table. Notice that, since we only have two truth values (T and F), that means that we could only ever have contradictory pairs of statements. A set of three or more couldn’t possibly be contradictory.
| T | F |
| F | T |
| F | T |
| T | F |
3) Consistent : Once you’ve determined that a set of statements is neither contradictory nor logically equivalent, you should check to see whether it is consistent. A consistent set of propositions is one where the logical form or structure of those propositions allows them to all be true at the same time. If the world is a certain way, then all of the statements will end up being true. So when testing for consistency, you’re simply looking at the output side of the truth table for a row completely filled with T’s. If you find it, then that set is consistent: they can all be true at the same time.
| T | T |
| F | T |
| F | F |
| T | F |
4) Inconsistent : The last option is to call a set of propositions inconsistent. If you never find that row filled completely with T’s, then the propositions you are analyzing are inconsistent: they cannot, as a matter of logical structure, be true at the same time, no matter the state of the world.
| T | F |
| F | F |
| F | F |
| T | F |
Note that a set of logically equivalent propositions is likely to be consistent since it’s likely to have one line on which all propositions are true. Furthermore, every contradictory pair of propositions is inconsistent. For the sake of the class, instructors will often require that you choose only one o f the four options on a multiple-choice quiz or exam, though, so how do you decide?
Easy: just test for these four relations in order. Start by asking “are they logically equivalent?” Then, when you’ve found a line on which they have different truth values, as yourself “could they be contradictory?” Next, if you decide that they aren’t contradictory (or couldn’t be because it’s a set of 3 or more), go hunting for one row on which all of the columns have a T. If you find it, then the set is consistent. If you never find such a row, then the set is inconsistent.
In short : always answer with the strongest answer available. If a set of statements is both logically equivalent and consistent, then the correct answer on a multiple-choice test is “Logically Equivalent” since that’s a stronger claim (it’s a claim about every row rather than just one row).
Testing for Validity
If you’re being asked to analyze an argument or inference , then you know that there are only two possible answers: Valid and Invalid. It’s easier to start by discussing an invalid inference:
1) Invalid : an inference or argument is invalid if you find a row of the output side of the truth table where all of the premises are true, but the conclusion is false. So this, depending on how many propositions make up the argument, will typically be a row that looks like TTTTF, TTF, TF, TTTF, etc. If you find even just one row where the output side looks like this, then you’ve proven that the argument is invalid. I like to think of it as searching for a radioactive row. If you find one with *all* true premises and a false conclusion, then you’ve found a radioactive row and therefore you’ve found out that the argument is invalid.
I like to think of the counterexample line—the line that tells you the inference is invalid— as a sort of “radioactive” line you’re searching for. Think of it like this: you’ve got your Geiger counter and you’re scanning through the truth table for radiation. If you find a radioactive line, then the argument is bad (invalid). If you don’t find a radioactive line, then the argument is clean (valid).
2) Valid : If you look through the whole output side of a truth table and never see a row where *every* premise is true and the conclusion is false, then you’ve found a valid argument. Remember that rows where one premise is false don’t count and rows where it’s all false premises and a true conclusion don’t count. Everything is safe except rows somewhat similar to TTF or TTTTF or the like, depending on how many premises and conclusions there are.
Validity means it’s impossible that the premises would be true while the conclusion is false. So if you find a row (even just one!) in a truth table telling you that if the world is like this (what you see on the input side of the row) then the premises will be true while the conclusion is false. This would never happen with a valid argument, so it follows that the argument you’ve found is an invalid argument.
The Reverse Truth Table Method
What happens if we symbolize or translate an argument and we end up with 5, 6, or more atomic sentence letters? Are we doomed to create a truth table with 32, 64, or more rows? That would be a fate worse than many things!
Fear not, dear student. We have a method for directly testing the validity of an inference without having to build a complete truth table. It’s called the indirect or reverse truth table method. The basic idea is to assign “invalid” truth values to the premises and conclusions, figure out what we need the atomic sentence letters to be in order for those truth values to obtain, and then see if we can consistently apply truth values to the atomic sentence letters in order to create a counterexample (a line with all true premises and a false conclusion). So, basically, we’re looking for that “radioactive” row (TF, TTF, TTTF, etc.) from the complete truth table from the previous section, but instead of building a whole truth table, we’re simply going right to the end and testing if there could be such a row. We’re testing to see if a radioactive row is even possible.
So, let’s walk through an example, and then we’ll come up with a set of steps for doing the reverse truth table method. Here’s the English argument:
If you don’t pass the driver’s license written test, then you won’t have a driver's license (until you are able to pass it). But I don’t have a driver’s license, so that means I didn’t pass the driver’s license written test?
Consider how confusing this would be to someone who has passed the written portion of the test, but hasn’t yet completed the driving test. They did pass the written test, but still don’t have a driver’s license! This is confusing, because this argument is what is called “Affirming the Consequent”. It’s a formal fallacy, or an invalid argument that might appear to be valid. Let’s symbolize it (ignoring the parenthetical):
~W \(\rightarrow\) ~L
~L
\(\therefore\) ~W
Remember that “\(\therefore\)” means “therefore"
In order to run the reverse truth table method of analysis on this argument, we’ll first want to set up as if we are doing the output side of a truth table. We’ll want to give ourselves lots of room to work with:
(~W \(\rightarrow\) ~L) / ~L // ~W
Now, the next thing to do is to assign truth values to each whole proposition so that we have a radioactive row or counterexample row. In this case, there are two premises and a conclusion, so the counterexample row will be TTF:
(~W \(\underset{\bf{T}}{\rightarrow}\) ~L) / ~\(\underset{\bf{T}}{L}\) // \(\underset{\bf{F}}{\sim W}\)
The setup part is done. Now we need to do the hard work: actually work out if we can consistently apply truth values to the atomic sentence letters. We’ll go one step at a time, starting with the conclusion. Why start with the conclusion? It's typically easier to make a sentence false given the truth tables for disjunction, negation, and conditional; so the conclusion generally is the easier place to start. Can you figure out why the rules for these operators make establishing falsehood easier?
(~W \(\underset{\text{T}}{\rightarrow}\) ~L) / ~\(\underset{\text{T}}{L}\) // \(\underset{\text{F}}{\sim \overset{\bf{T}}{W}}\)
The conclusion is false, so W will need to be true since the conclusion is ~W. So then we assign T to every W that appears in the argument:
(~\(\overset{\bf{T}}{W}\) \(\underset{\text{T}}{\rightarrow}\) ~L) / ~\(\underset{\text{T}}{L}\) // \(\underset{\text{F}}{\overset{\bf{F}}{\sim }\overset{\bf{T}}{W}}\)
And then work out how that affects the formulas containing the atomic letters I just assigned truth values to. In this case, if W is true, then ~W is false, and if an antecedent is false, then the whole conditional is true. The main operator is the conditional, and so no problems with the first premise:
(\(\overset{\bf{F}}{\sim }\overset{\bf{T}}{W}\overset{\bf{T}}{\underset{\text{T}}{\rightarrow}}\) ~L) / ~\(\underset{\text{T}}{L}\) // \(\underset{\text{F}}{\overset{\bf{F}}{\sim }\overset{\bf{T}}{W}}\)
What about the second premise? Well, since we haven’t yet been forced to assign a truth value to L, we can assign whatever we want to it, so we’ll make it false! The result will be that ~L is true.
(\(\overset{\bf{F}}{\sim }\overset{\bf{T}}{W}\overset{\bf{T}}{\underset{\text{T}}{\rightarrow}}\) ~L) / \(\overset{\bf{T}}{\sim }\underset{\text{T}}{\overset{\bf{F}}{L}}\) // \(\underset{\text{F}}{\overset{\bf{F}}{\sim }\overset{\bf{T}}{W}}\)
At this point, we’ve symbolized the argument, put it in a row as if we were going to make a truth table out of it, and then tried assigning truth values to the atomic sentence letters to make a radioactive row (in this case TTF). Since we were able to do so without coming across a contradiction, we know that the inference is invalid .
Here’s a sort of algorithm for the reverse truth table method of analysis:
1. Symbolize the inference and write out in a single row using slashes between formulas.
2. Assign truth values to each complete formula such that all premises are true and the conclusion is false.
3. Then calculate the truth value of each atomic sentence letter given the truth value of the whole formula. Start with the most restrictive formulas. If you run into a case where multiple truth values would work, then start a new line for each possibility and test each going forward.
4. Then transfer the atomic sentence letter truth value(s) to the other instances throughout the whole inference (transfer the truth values of, for example, ‘A’ to all other A’s throughout the formula).
5. Then calculate whether it is possible to continue assigning truth values to atomic sentence letters and transferring those values to other instances without running into a contradiction (that is, where a single letter must be both T and F) .
6. If you run into no contradiction , then the inference is invalid because the radioactive row is possible . The process is over.
If you run into a contradiction , then possibly the inference is valid. Complete all lines you’ve started to ensure that there is no possible consistent assignment of truth values. You only need one possible row where there is no contradiction to prove that the inference is invalid; whereas you need to eliminate every possible row that could have true premises and a false conclusion before you can know it’s valid.
Okay, let’s try a more complex one:
Lila: We’re either going home or I’m going home alone unless you both assure me you will drive us home later and will phone the babysitter.
Diego: I can’t assure you that I’ll be sober enough to drive us home later.
Lila: Well, I’m not going home alone and you’re not staying the night here.
Diego: Well, then, I guess either we’re both going home now or we’re getting a motel room.
If we conceive of the whole exchange as one big inference, we can symbolize it the following way:
(H \(\vee\) (I \(\vee\) (A \(\wedge\) P))) / ~A / (~I \(\wedge\) ~S) // (H \(\vee\) M)
Now, we assign truth values to the premises and conclusion so that they’ll be radioactive.
Here, I’ve gone ahead and put them under the main operators:
(H \(\underset{\bf{T}}{\vee}\) (I \(\vee\) (A \(\wedge\) P))) / \(\underset{\bf{T}}{\sim }\)A / (~I \(\underset{\bf{T}}{\wedge}\) ~S) // (H \(\underset{\bf{F}}{\vee}\) M)
Next, I’m going to start looking for some simpler formulas to assign truth values to. I’m eyeing the conclusion and the ~A. The conclusion will only admit of one set of truth values: both H and M must be false for the \(\vee\) to be false. A must be false for ~A to be true.
(H \(\underset{\text{T}}{\vee}\) (I \(\vee\) (A \(\wedge\) P))) / \(\underset{\bf{T}}{\sim }\underset{\text{F}}{A}\) / (~I \(\underset{\text{T}}{\wedge}\) ~S) // (\(\underset{\bf{F}}{H}\underset{\text{F}}{\vee} \underset{\bf{F}}{M}\))
Next, you transfer truth values from the ones you just assigned to all identical letters.
Notice how I’d need the right disjunct (I \(\vee\) (A \(\wedge\) P)) to be true for the first premise to turn out true.
But we already know that A is false from premise 2. So that means the conjunction (A \(\wedge\) P) must be false. Now that we know this, we must conclude that I is true for the first premise to turn out true. So we can conclude that I is true.
(\(\underset{\text{F}}{H} \underset{\text{T}}{\vee} (\underset{\bf{T}}{I} \vee (\underset{\text{F}}{A} \underset{\bf{F}}{\wedge}\) P))) / \(\underset{\text{T}}{\sim }\underset{\text{F}}{A}\) / (~I \(\underset{\text{T}}{\wedge}\) ~S) // (\(\underset{\text{F}}{H}\underset{\text{F}}{\vee} \underset{\text{F}}{M}\))
Now we transfer the new atomic truth value:
Now let’s try to work out that third premise. First we can process the negation on I:
(\(\underset{\text{F}}{H} \underset{\text{T}}{\vee} (\underset{\text{T}}{I} \vee (\underset{\text{F}}{A} \underset{\text{F}}{\wedge}\) P))) / \(\underset{\text{T}}{\sim }\underset{\text{F}}{A}\) / (\(\underset{\bf{F}}{\sim }\underset{\text{T}}{I} \underset{\text{T}}{\wedge}\) ~S) // (\(\underset{\text{F}}{H}\underset{\text{F}}{\vee} \underset{\text{F}}{M}\))
We’ve already got a contradiction!!! Ouch! ~I would have had to be true for (~I \(\wedge\) ~S) to turn out true. Both conjuncts need to be true. But ~I is false according to our assignment of false to I. Bummer dude!
What now? We’ve reached a contradiction! At this point we ask ourselves: “was there any step I made that I wasn’t forced to make?” In this case, no, we didn’t arbitrarily choose true or false for any letter, so every step we took was forced by logic. That means there aren’t any alternative assignments to consider and therefore the radioactive counterexample is impossible . Our inference is valid .
You could have also assigned values to I and S given the third premise, but I chose to finishing the first premise. Either way would’ve resulted in a contradiction. This is not an example of an alternative assignment . We’ll cover alternative assignments below:
Here's a handy flow chart for you:
Let’s try a quick example with alternative assignments possible:
(H \(\underset{\text{T}}{\vee}\) (I \(\vee\) (A \(\wedge\) P))) / \(\underset{\text{T}}{\sim }\)A / (H \(\underset{\text{T}}{\vee}\) M) // (~I \(\underset{\text{F}}{\wedge}\) ~S)
There are many ways for Premise 1 to be true, many ways for Premise 3 to be true, and many ways for the Conclusion to be false (an exception to the generalization I made earlier). So we’re not being forced as much as we were in the previous example. Arguments where the conclusion has more than one way of being false are typically the hardest arguments to do using the reverse truth table method. Hardest, that is, in terms of how much work is involved. Remember that this isn’t difficult in the sense of it being a complicated procedure. It’s not like playing chess. We could program a computer to do this whole method in an afternoon. It does, though, sometimes take a bit of work to work out the answer. Let’s start with what we are forced to do:
(H \(\underset{\text{T}}{\vee}\) (I \(\vee\) (\(\underset{\bf{F}}{A}\) \(\wedge\) P))) / \(\underset{\bf{T}}{\sim }\underset{\text{F}}{A}\) / (H \(\underset{\text{T}}{\vee}\) M) // (~I \(\underset{\text{F}}{\wedge}\) ~S)
A must be false because ~A is true. At this point, we aren’t forced to do anything for Premise 1 since there are still many ways for it to come out true. Nothing has changed for Premise 3 and the Conclusion. At this point we need to split our line into all possible successful assignments. I’m going to start with the conclusion (I chose this more or less arbitrarily). Here’s what I do:
\[\begin{array}{} &(H &\vee (I \vee (&A& \wedge P))) / &\sim &A / (H &\vee M) // (&\sim I& \wedge& \sim S) &&\\ \text{Possibility 1} & \rightarrow &T &F &F &T &F &T &\textbf{T} &F &\textbf{F}&&\\\text{Possibility 2} & \rightarrow &T &F &F &T &F &T &\textbf{F} &F &\textbf{T}&&\\\text{Possibility 3} & \rightarrow &T &F &F &T &F &T &\textbf{F} &F &\textbf{F} && \end{array} \nonumber\]
If you focus on the conclusion, it looks sort of like a truth table now, doesn’t it? Now we complete the process for each possible line:
\[\begin{array}{} (H &\vee (I \vee (&A& \wedge P))) / &\sim &A / (H &\vee M) // (&\sim &I& \wedge& \sim &S) \\& T &F &F &T &F &T &T&\textbf{F} &F &F&\textbf{T}\\ &T &F &F &T &F &T &F&\textbf{T} &F &T&\textbf{F}\\ &T &F &F &T &F &T &F&\textbf{T} &F &F&\textbf{T} \end{array} \nonumber\]
Now I’m transferring truth values:
And then calculate what needs to happen given the changes you’ve made. I’ve changed the first premise, so now I notice that (A \(\wedge\) P) is false and now I is false in my first row, so that means H must be true. In the other rows, nothing has yet forced me to assign anything to H.
\[\begin{array}{} (H &\vee (&I \vee (&A& \wedge P))) / &\sim &A / (H &\vee M) // (&\sim &I& \wedge& \sim &S) \\T& T &F &F &F &T &F &T&T &F &F&F &T\\ &T &T &F &F &T &F &F&F &T &F&T &F\\ &T &T &F &F &T &F &F&F &T &F&F &T \end{array} \nonumber\]
Then transfer that H:
At this point, nothing is forcing my hand. If you look carefully, you’ll see that there is no single letter on any single line that must be a given truth value for the truth values we’ve assigned the complete formulas to obtain. So now what???
If we are free to assign any truth values we want, then we aren’t going to run into any contradictions. It follows that there are no contradictions in these three rows and there is therefore at least one row where there are no contradictions. This inference is invalid .
All we need is one row where there are no contradictions to prove that the inference is invalid.
That’s all! That’s how we do the reverse truth table method. One can imagine how to use this to test even super complex sets of sentences for consistency: just assign “true” to each individual sentence and then look for a contradiction. One line with no contradiction? You’ve got a consistent set of sentences.


21: The Birth of Stars and the Discovery of Planets outside the Solar System
“There are countless suns and countless earths all rotating round their suns in exactly the same way as the planets of our system. We see only the suns because they are the largest bodies and are luminous, but their planets remain invisible to us because they are smaller and non-luminous. . . . The unnumbered worlds in the universe are all similar in form and rank and subject to the same forces and the same laws.” —Giordano Bruno in On the Infinite Universe and Worlds (1584). Bruno was tried for heresy by the Roman Inquisition and burned at the stake in 1600.
We’ve discussed stars as nuclear furnaces that convert light elements into heavier ones. A star’s nuclear evolution begins when hydrogen is fused into helium, but that can only occur when the core temperature exceeds 10 to 12 million K. Since stars form from cold interstellar material, we must understand how they collapse and eventually reach this “ignition temperature” to explain the birth of stars. Star formation is a continuous process, from the birth of our Galaxy right up to today. We estimate that every year in our Galaxy, on average, three solar masses of interstellar matter are converted into stars. This may sound like a small amount of mass for an object as large as a galaxy, but only three new stars (out of billions in the G alaxy) are formed each year.
Do planets orbit other stars or is ours the only planetary system? In the past few decades, new technology has enabled us to answer that question by revealing nearly 3500 exoplanets in over 2600 planetary systems. Even before planets were detected, astronomers had predicted that planetary systems were likely to be byproducts of the star-formation process. In this chapter, we look at how interstellar matter is transformed into stars and planets.
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- 21.2: Star Formation
- Most stars form in giant molecular clouds with masses as large as \(3 × 10^6\) solar masses. The most well-studied molecular cloud is Orion, where star formation is currently taking place. Molecular clouds typically contain regions of higher density called clumps, which in turn contain several even-denser cores of gas and dust, each of which may become a star. A star can form inside a core if its density is high enough that gravity can overwhelm the internal pressure and cause the gas and dust t
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- 21.3: The H-R and the Study of Stellar Evolution
- The evolution of a star can be described in terms of changes in its temperature and luminosity, which can best be followed by plotting them on an H–R diagram. Protostars generate energy (and internal heat) through gravitational contraction that typically continues for millions of years, until the star reaches the main sequence.
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- 21.4: Evidence That Planets Form around Other Stars
- Observational evidence shows that most protostars are surrounded by disks with large-enough diameters and enough mass (as much as 10% that of the Sun) to form planets. After a few million years, the inner part of the disk is cleared of dust, and the disk is then shaped like a donut with the protostar centered in the hole—something that can be explained by the formation of planets in that inner zone.
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- 21.5: Planets beyond the Solar System- Search and Discovery
- Several observational techniques have successfully detected planets orbiting other stars. These techniques fall into two general categories—direct and indirect detection. The Doppler and transit techniques are our most powerful indirect tools for finding exoplanets. Some planets are also being found by direct imaging.
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- 21.6: Exoplanets Everywhere - What we are Learning
- Although the Kepler mission is finding thousands of new exoplanets, these are limited to orbital periods of less than 400 days and sizes larger than Mars. Still, we can use the Kepler discoveries to extrapolate the distribution of planets in our Galaxy. The data so far imply that planets like Earth are the most common type of planet, and that there may be 100 billion Earth-size planets around Sun-like stars in the Galaxy. About 2600 planetary systems have been discovered around other stars.
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- 21.7: New Perspectives on Planet Formation
- The ensemble of exoplanets is incredibly diverse and has led to a revision in our understanding of planet formation that includes the possibility of vigorous, chaotic interactions, with planet migration and scattering. It is possible that the solar system is unusual (and not representative) in how its planets are arranged. Many systems seem to have rocky planets farther inward than we do, for example, and some even have “hot Jupiters” very close to their star.
Thumbnail: We see a close-up of part of the Carina Nebula taken with the Hubble Space Telescope. This image reveals jets powered by newly forming stars embedded in a great cloud of gas and dust. Parts of the clouds are glowing from the energy of very young stars recently formed within them. (credit: modification of work by NASA, ESA, and M. Livio and the Hubble 20th Anniversary Team (STScI))


4.6: Summary
4.1 Fertility and Conception
Fertility is the ability to conceive a pregnancy. Many intricate processes must successfully occur to achieve conception. Couples with infertility may seek care from a reproductive endocrinologist who specializes in the treatment of infertility. Infertility can be caused by factors in either or both partners, though in many cases, a specific cause cannot be identified. Assessing a couple who presents with infertility involves looking at both partners and the three main factors that are essential for conception: the release of an egg, the presence of sperm, and a healthy environment for implantation. Though conception and pregnancy are largely physiologic events, the nurse must not neglect the important psychosocial factors that are associated with fertility and conception, including stress and societal pressures. Patient teaching is an essential function for the nurse caring for all patients considering conception.
4.2 Genetics
Humans typically have a total of 46 chromosomes organized in pairs: 22 pairs of autosomes (body chromosomes) and 1 pair of sex chromosomes (XX in people who are genetically female or XY in people who are genetically male). One set of chromosomes (22 autosomes and 1 sex chromosome) is inherited from each parent. Sometimes, an error in replication occurs, and an abnormal number of chromosomes results.
Genetic transmission between generations can lead to autosomal dominant inheritance disorders and autosomal recessive disorders. X-linked disorders disproportionately affect more males than females because males must inherit only one affected X chromosome to be affected.
Patients have many options for genetic screening, both prenatally and postnatally. Prenatal screening is done during pregnancy to determine the risk that the fetus has for aneuploidy, a neural tube defect, or another abnormality suggestive of a genetic disorder. Prenatal diagnostic testing includes chromosomal analysis. Postnatal testing, or the newborn screen, is done after the baby is born and consists of a capillary blood test. The hope is that early detection and treatment of these conditions improves the quality of life for newborns.
4.3 Causes of Infertility
Causes of infertility are multifactorial. Many causes of infertility in both persons assigned female at birth and those assigned male at birth are similar, such as medications, environmental toxins, hormones, and medical conditions. Treatments are available for many causes of infertility; yet not all causes of infertility are diagnosable or treatable. The nurse can be an excellent support to couples experiencing infertility by providing education, referrals, and emotional support.
4.4 Treating Infertility
When a patient or couple presents with difficulty in conceiving, the first step is a comprehensive assessment of both partners. Once the initial assessment has been completed, the provider will create a plan of care. Pharmacologic management of infertility, also known as ovulation induction or ovarian stimulation, is used in conjunction with many of the treatment options available for infertility. Medications are used to control the menstrual cycle and ovarian response to increase the odds of conception. IUI and IVF are procedures available for treatment of infertility. Like any medical procedure, infertility treatments do pose some risks. Educating patients about what to expect, what is normal, and what should be reported is a crucial function of all nurses practicing in this field. In addition, it is important to recognize signs and symptoms of potentially serious complications and how to manage them.


Section 09: Soundness and completeness
This toolkit is incredibly convenient. It is also intuitive, because it seems natural that provability and semantic entailment should agree. Yet, do not be fooled by the similarity of the symbols ‘⊨’ and ‘⊢ .’ The fact that these two are really interchangeable is not a simple thing to prove.
Why should we think that an argument that can be proven is necessarily a valid argument? That is, why think that \(\mathcal{A}\) ⊢ \(\mathcal{B}\) implies \(\mathcal{A}\) ⊨ \(\mathcal{B}\)?
This is the problem of soundness . A proof system is sound if there are no proofs of invalid arguments. Demonstrating that the proof system is sound would require showing that any possible proof is the proof of a valid argument. It would not be enough simply to succeed when trying to prove many valid arguments and to fail when trying to prove invalid ones.
Fortunately, there is a way of approaching this in a step-wise fashion. If using the &E rule on the last line of a proof could never change a valid argument into an invalid one, then using the rule many times could not make an argument invalid. Similarly, if using the &E and ∨E rules individually on the last line of a proof could never change a valid argument into an invalid one, then using them in combination could not either.
The strategy is to show for every rule of inference that it alone could not make a valid argument into an invalid one. It follows that the rules used in combination would not make a valid argument invalid. Since a proof is just a series of lines, each justified by a rule of inference, this would show that every provable argument is valid.
Consider, for example, the &I rule. Suppose we use it to add \(\mathcal{A}\)&\(\mathcal{B}\) to a valid argument. In order for the rule to apply, \(\mathcal{A}\) and \(\mathcal{B}\) must already be available in the proof. Since the argument so far is valid, \(\mathcal{A}\) and \(\mathcal{B}\) are either premises of the argument or valid consequences of the premises. As such, any model in which the premises are true must be a model in which \(\mathcal{A}\) and \(\mathcal{B}\) are true. According to the definition of truth in ql , this means that \(\mathcal{A}\)&\(\mathcal{B}\) is also true in such a model. Therefore, \(\mathcal{A}\)&\(\mathcal{B}\) validly follows from the premises. This means that using the &E rule to extend a valid proof produces another valid proof.
In order to show that the proof system is sound, we would need to show this for the other inference rules. Since the derived rules are consequences of the basic rules, it would suffice to provide similar arguments for the 16 other basic rules. This tedious exercise falls beyond the scope of this book.
Given a proof that the proof system is sound, it follows that every theorem is a tautology.
It is still possible to ask: Why think that every valid argument is an argument that can be proven? That is, why think that \(\mathcal{A}\) ⊨ \(\mathcal{B}\) implies \(\mathcal{A}\) ⊢ \(\mathcal{B}\)?
This is the problem of completeness .  Completeness for a language like QL was first proven by Kurt Gödel in 1929. The proof is beyond the scope of this book.
The important point is that, happily, the proof system for QL is both sound and complete. This is not the case for all proof systems and all formal languages. Because it is true of QL, we can choose to give proofs or construct models— whichever is easier for the task at hand.
Summary of definitions
- A sentence A is a theorem if and only if ⊢\(\mathcal{A}\).
- Two sentences \(\mathcal{A}\) and \(\mathcal{B}\) are provably equivalent if and only if \(\mathcal{A}\)⊢\(\mathcal{B}\) and \(\mathcal{B}\)⊢\(\mathcal{A}\).
- {\(\mathcal{A}\) 1 ,\(\mathcal{A}\) 2 ,...} is provably inconsistent if and only if, for some sentence \(\mathcal{B}\), {\(\mathcal{A}\) 1 ,\(\mathcal{A}\) 2 ,...}⊢(\(\mathcal{B}\) &¬\(\mathcal{B}\)).


5.5: Ohm's Law
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Learning Objectives
By the end of this section, you will be able to:
- Describe Ohm’s law
- Recognize when Ohm’s law applies and when it does not
We have been discussing three electrical properties so far in this chapter: current, voltage, and resistance. It turns out that many materials exhibit a simple relationship among the values for these properties, known as Ohm’s law. Many other materials do not show this relationship, so despite being called Ohm’s law, it is not considered a law of nature, like Newton’s laws or the laws of thermodynamics. But it is very useful for calculations involving materials that do obey Ohm’s law.
Description of Ohm’s Law
The current that flows through most substances is directly proportional to the voltage \(V\) applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied :
\[I \propto V. \nonumber \]
This important relationship is the basis for Ohm’s law . It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law, which is to say that it is an experimentally observed phenomenon, like friction. Such a linear relationship doesn’t always occur. Any material, component, or device that obeys Ohm’s law, where the current through the device is proportional to the voltage applied, is known as an ohmic material or ohmic component. Any material or component that does not obey Ohm’s law is known as a nonohmic material or nonohmic component.
Ohm’s Experiment
In a paper published in 1827, Georg Ohm described an experiment in which he measured voltage across and current through various simple electrical circuits containing various lengths of wire. A similar experiment is shown in Figure \(\PageIndex{1}\). This experiment is used to observe the current through a resistor that results from an applied voltage. In this simple circuit, a resistor is connected in series with a battery. The voltage is measured with a voltmeter, which must be placed across the resistor (in parallel with the resistor). The current is measured with an ammeter, which must be in line with the resistor (in series with the resistor).
In this updated version of Ohm’s original experiment, several measurements of the current were made for several different voltages. When the battery was hooked up as in Figure \(\PageIndex{1a}\), the current flowed in the clockwise direction and the readings of the voltmeter and ammeter were positive. Does the behavior of the current change if the current flowed in the opposite direction? To get the current to flow in the opposite direction, the leads of the battery can be switched. When the leads of the battery were switched, the readings of the voltmeter and ammeter readings were negative because the current flowed in the opposite direction, in this case, counterclockwise. Results of a similar experiment are shown in Figure \(\PageIndex{2}\).
In this experiment, the voltage applied across the resistor varies from −10.00 to +10.00 V, by increments of 1.00 V. The current through the resistor and the voltage across the resistor are measured. A plot is made of the voltage versus the current, and the result is approximately linear. The slope of the line is the resistance, or the voltage divided by the current. This result is known as Ohm’s law :
\[V = IR \label{Ohms} \]
where \(V\) is the voltage measured in volts across the object in question, \(I\) is the current measured through the object in amps, and \(R\) is the resistance in units of ohms. As stated previously, any device that shows a linear relationship between the voltage and the current is known as an ohmic device. A resistor is therefore an ohmic device.
Example \(\PageIndex{1}\): Measuring Resistance
A carbon resistor at room temperature \((20^oC)\) is attached to a 9.00-V battery and the current measured through the resistor is 3.00 mA.
- What is the resistance of the resistor measured in ohms?
- If the temperature of the resistor is increased to \(60^oC\) by heating the resistor, what is the current through the resistor?
Strategy
- The resistance can be found using Ohm’s law. Ohm’s law states that \(V = IR\), so the resistance can be found using \(R = V/I\).
- First, the resistance is temperature dependent so the new resistance after the resistor has been heated can be found using \(R = R_0 (1 + \alpha \Delta T)\). The current can be found using Ohm’s law in the form \(I = V/R\).
Solution
- Using Ohm’s law and solving for the resistance yields the resistance at room temperature: \[R = \dfrac{V}{I} = \dfrac{9.00 \, V}{3.00 \times 10^{-3} A} = 3.00 \times 10^3 \, \Omega = 3.00 k\Omega \nonumber \]
- The resistance at \(60^oC\) can be found using \(R = R_0 (1 + \alpha \Delta T)\) where the temperature coefficient for carbon is \(\alpha = -0.0005\). \[R = R_0 (1 + \alpha \Delta T) = 3.00 \times 10^3 (1 - 0.0005 (60^oC - 20^oC)) = 2.94 \, k\Omega. \nonumber \] The current through the heated resistor is \[I = \dfrac{V}{R} = \dfrac{9.00 \, V}{2.94 \times 10^3 \, \Omega} = 3.06 \times 10^{-3} A = 3.06 \, mA. \nonumber \]
Significance
A change in temperature of \(40^oC\) resulted in a 2.00% change in current. This may not seem like a very great change, but changing electrical characteristics can have a strong effect on the circuits. For this reason, many electronic appliances, such as computers, contain fans to remove the heat dissipated by components in the electric circuits.
Exercise \(\PageIndex{1}\)
The voltage supplied to your house varies as \(V(t) = V_{max} sin \, (2\pi \, ft)\). If a resistor is connected across this voltage, will Ohm’s law \(V = IR\) still be valid?
- Answer
-
Yes, Ohm’s law is still valid. At every point in time the current is equal to \(I(t) = V(t) /R\), so the current is also a function of time, \(I(t) = \dfrac{V_{max}}{R} \, sin \, (2\pi \, ft)\).
Simulation: PhET
See how Ohm’s law (Equation \ref{Ohms}) relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm’s law. The sizes of the symbols in the equation change to match the circuit diagram.
Nonohmic devices do not exhibit a linear relationship between the voltage and the current. One such device is the semiconducting circuit element known as a diode . A diode is a circuit device that allows current flow in only one direction. A diagram of a simple circuit consisting of a battery, a diode, and a resistor is shown in Figure \(\PageIndex{3}\). Although we do not cover the theory of the diode in this section, the diode can be tested to see if it is an ohmic or a nonohmic device.
A plot of current versus voltage is shown in Figure \(\PageIndex{4}\). Note that the behavior of the diode is shown as current versus voltage, whereas the resistor operation was shown as voltage versus current. A diode consists of an anode and a cathode. When the anode is at a negative potential and the cathode is at a positive potential, as shown in part (a), the diode is said to have reverse bias. With reverse bias, the diode has an extremely large resistance and there is very little current flow—essentially zero current—through the diode and the resistor. As the voltage applied to the circuit increases, the current remains essentially zero, until the voltage reaches the breakdown voltage and the diode conducts current. When the battery and the potential across the diode are reversed, making the anode positive and the cathode negative, the diode conducts and current flows through the diode if the voltage is greater than 0.7 V. The resistance of the diode is close to zero. (This is the reason for the resistor in the circuit; if it were not there, the current would become very large.) You can see from the graph in Figure \(\PageIndex{4}\) that the voltage and the current do not have a linear relationship. Thus, the diode is an example of a nonohmic device.
Ohm’s law is commonly stated as \(V = IR\), but originally it was stated as a microscopic view, in terms of the current density, the conductivity, and the electrical field. This microscopic view suggests the proportionality \(V \propto I\) comes from the drift velocity of the free electrons in the metal that results from an applied electrical field. As stated earlier, the current density is proportional to the applied electrical field. The reformulation of Ohm’s law is credited to Gustav Kirchhoff, whose name we will see again in the next chapter.


According to the U.S. Census Bureau in 2011, over sixty percent of families with children under five relied on regular child care arrangements. Around a quarter of those families used organized child care facilities as their primary arrangement.24
Child care involves supervising a child or children, usually from infancy to age thirteen, and typically refers to work done by somebody outside the child's immediate family. Child care is a broad topic covering a wide spectrum of contexts, activities, social and cultural conventions, and institutions. The majority of child care institutions that are available require that child care providers have extensive training in first aid and are CPR certified. In addition, background checks, drug testing, and reference verification are normally required.
It is traditional in Western society for children to be cared for by their parents or their legal guardians. In families where children live with one or both of their parents, the child care role may also be taken on by the child's extended family. If a parent or extended family is unable to care for the children, orphanages and foster homes are a way of providing for children's care, housing, and schooling.
Child Care in the United States
Formal child care options include
center-based care
and
family child care homes
. Each state has different regulations for licensing child care centers, including teacher requirements. In some states, teaching in a child care center requires an associate’s degree in child development. States with quality standards built into their licensing programs may have higher requirements for support staff, such as teacher assistants.
Head Start
(a federally funded child care program for income qualified families) lead teachers must have a bachelor’s degree in Early Childhood Education. States vary in other standards set for daycare providers, such as teacher to child ratios.
State legislation may regulate the number and ages of children allowed before the home is considered an official family child care program and subject to licensing regulations. Often the nationally recognized Child Development Associate credential is the minimum standard for the individual leading this home care program.
In addition to these licensed options, parents may also choose to find their own caregiver or arrange childcare exchanges/swaps with another family. This care is typically provided by nannies, au pairs, or friends and family. The child is watched inside their own home or the caregiver's home, reducing exposure to outside children and illnesses. Depending on the number of children in the home, the children utilizing in-home care can enjoy the greatest amount of interaction with their caregiver and form a close bond.
There are no required licensing or background checks for this type of in-home care, making parental vigilance essential in choosing an appropriate caregiver. The cost of in-home care is the highest of childcare options per child, though a household with many children may find this the most convenient and affordable option. 27
Play-Based Programs in Early Childhood Education
In 1989, UNICEF ( United Nations International Children's Emergency Fund) recognizes play’s significance. Children have a right to play. According to
The UN Convention on the Rights of the Child’s
Article 31 states the following:
“States Parties recognize the right of the child to rest and leisure, to engage in play and recreational activities appropriate to the age of the child and to participate freely in cultural life and the arts. States Parties shall respect and promote the right of the child to participate fully in cultural and artistic life and shall encourage the provision of appropriate and equal opportunities for cultural, artistic, recreational and leisure activity.”
(UNICEF)
28
Children are hard-wired for play. Play is the context in which children optimally learn. Play based programs are child-led (based on their interests), more than teacher directed. An important element in nurturing a happy, well-formed child—but which often gets overlooked in the search for a preschool—is the idea of playful discovery, and creativity. Playful learning is as important to intellectual and personal growth as academics. Play based programs can be center-based or home-based. Through playful discoveries children find joy in learning that they may, or may not, experience in traditional educational settings. Every program is unique and different, even those under the same name (e.g. Reggio Inspired, Montessori, Head Start, etc.) because these programs are staffed with educated humans, who have different teaching styles, different ways of motivating children, different play experiences themselves which can provide various ways to work and play with children. A few things that stand out to help identify play based programs are the following:
One that incorporates the children's natural interest
Outdoor and indoor opportunities to learn
A number of play spaces available, like blocks, water, sensory, art, light areas, to name a few.
Children get to decide where thy want to play (experience and materials to use). 29


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Summary
5.1
Creating Different Document Types in Microsoft Word
-
Being able to create different types of documents in Word is an important skill. You can either use a template or create them from scratch.
-
Although many people use email, printed business memos are still common, and the way they are composed is important to know.
-
It is also important to know how to create and format documents on letterhead and printed envelopes. Letterheads display your name and contact information, while printed envelopes help add a look of professionalism to your communication.
-
Business cards can be designed in Word and printed by yourself, or by a printing company using cardstock paper. They display your contact information.
-
Brochures and flyers are forms of print advertising that can also be used digitally.
-
Invoices always contain what you or your company is billing a client for. Microsoft has a large number of templates you can create an invoice from.
-
A business plan usually has a standard layout, which a template can help you with. But it can be helpful to look at samples of business plans online so you can understand what kind of language to use in the plan.
-
When looking for a job, it is important to have a polished résumé and cover letter. The professional look and feel of your documents can go a long way in helping you start your career.
5.2
Mail Merge in Microsoft Word
-
Mail merge is a tool in Word that can personalize dozens or hundreds of form letters. It allows the user to designate merge fields for any type of customized client data, such as names, addresses, or other details.
-
To do a mail merge, you first need to create your main document, or the document that contains the letter or information going to your recipients. You can temporarily designate where the custom fields will be by highlighting them.
-
Next, you need to create or access a source document. A source document is the place where you have all of your recipients’ information. The information you include here will go into the custom fields in the main document.
-
You complete a mail merge by combining the main document and the source document. The resulting item will be the merged document. Word will create one letter for each recipient you designated in a single document.
5.3
Creating Forms in Microsoft Word
-
The Developer tab is part of Word’s set of command groups, but it does not appear by default, and needs to be manually activated. It is used for making forms, macros, handling add-ons/templates, and other advanced processes.
-
Fillable forms are built using the tools in the Controls command group of the Developer tab. This command group contains options for content control fields, which are used for creating different types of questions for forms. The respondent uses the content control fields to answer the questions in the form.
-
Creating a fillable form in Word is a multistep process. You should first create your questions, then choose the question types, add in the content control fields, then finally edit the properties for each question.
5.4
Creating Different Document Types in Google Docs
-
Docs lets you create and publish your own templates. It does not have as many templates as Word in its default Template Gallery, but it does allow users to search through user-generated templates on the internet.
-
There are no default templates in the Template Gallery for business memos, envelopes, business cards, flyers, brochures, or invoices, but these can all be made from scratch in Google Docs.
5.5
Creating Forms in Google Docs
-
Forms is part of the Google workspace of software programs that allows the writer to send a form electronically, and automatically tabulates their answers. Google provides some templates that users can base their forms on.
-
Forms also allows for creating forms from scratch. Users can choose from a wide array of question types, add multimedia questions, and apply different themes.
-
The results from responses collected through a form can easily be downloaded into Sheets for data analysis.
5.6
Advanced Collaboration in Google Docs
-
You can share a document using an email or a shareable link. There are three types of shareable links: public links, anyone with the link, and restricted.
-
Docs allows many users to edit and type in a document concurrently. This can be done with layers of permissions, from editor, to commenter, to reader.
-
Comment history collects all comments into a sidebar. This can save time for the user because this way, they do not have to scroll through the entire document.


4.6: Inelastic Collisions in One Dimension
Learning Objectives
- Define inelastic collision.
- Explain perfectly inelastic collision.
- Apply an understanding of collisions to everyday situations.
- Determine recoil velocity and loss in kinetic energy given mass and initial velocity.
We have seen that in an elastic collision, total kinetic energy is conserved. An inelastic collision is one in which the total kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add total kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some kinetic energy into thermal energy. Or it may convert stored energy into total kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle.
Definition: INELASTIC COLLISION
An inelastic collision is one in which the total kinetic energy changes (it is not conserved).
Figure \(\PageIndex{1}\) shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total kinetic energy is initially \(\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}=m v^{2}\). The two objects come to rest after sticking together, conserving total momentum. And the total kinetic energy is zero after this collision. A collision in which the objects stick together is called a perfectly inelastic collision because it reduces the total kinetic energy as much as possible, while conserving total momentum.
Definition: PERFECTLY INELASTIC COLLISION
A collision in which the objects stick together is sometimes called “perfectly inelastic.”
Example \(\PageIndex{1}\): Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie
(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure \(\PageIndex{2}\))
Strategy
Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.
Solution for (a)
Momentum is conserved because the net external force on the puck-goalie system is zero.
Conservation of momentum is
\[p_{1}+p_{2}=p_{1}^{\prime}+p_{2}^{\prime} \nonumber\]
or
\[m_{1} v_{1}+m_{2} v_{2}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime}. \nonumber\]
Because the goalie is initially at rest, we know \(v_{2}=0\). Because the goalie catches the puck, the final velocities are equal, or \(v_{1}^{\prime}=v_{2}^{\prime}=v^{\prime}\). Thus, the conservation of momentum equation simplifies to
\[m_{1} v_{1}=\left(m_{1}+m_{2}\right) v^{\prime}. \nonumber\]
Solving for \(v^{\prime}\) yields
\[v^{\prime}=\frac{m_{1}}{m_{1}+m_{2}} v_{1}. \nonumber\]
Entering known values in this equation, we get
\[v^{\prime}=\left(\frac{0.150 \mathrm{~kg}}{0.150 \mathrm{~kg}+70.0 \mathrm{~kg}}\right)(35.0 \mathrm{~m} / \mathrm{s})=7.48 \times 10^{-2} \mathrm{~m} / \mathrm{s}. \nonumber\]
Discussion for (a)
This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect.
Solution for (b)
Before the collision, the total kinetic energy \(\mathrm{KE}_{\text {int }}\) of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, \(\mathrm{KE}_{\text {int }}\) is initially
\[\begin{aligned}
\mathrm{KE}_{\mathrm{int}} &=\frac{1}{2} m v^{2}=\frac{1}{2}(0.150 \mathrm{~kg})(35.0 \mathrm{~m} / \mathrm{s})^{2} \\
&=91.9 \mathrm{~J}
\end{aligned}. \nonumber\]
After the collision, the total kinetic energy is
\[\begin{aligned}
\mathrm{KE}_{\text {int }}^{\prime} &=\frac{1}{2}(m+M) v^{2}=\frac{1}{2}(70.15 \mathrm{~kg})\left(7.48 \times 10^{-2} \mathrm{~m} / \mathrm{s}\right)^{2} \\
&=0.196 \mathrm{~J}
\end{aligned}. \nonumber\]
The change in total kinetic energy is thus
\[\begin{aligned}
\mathrm{KE}_{\mathrm{int}}^{\prime}-\mathrm{KE}_{\mathrm{int}} &=0.196 \mathrm{~J}-91.9 \mathrm{~J} \\
&=-91.7 \mathrm{~J}
\end{aligned} \nonumber\]
where the minus sign indicates that the energy was lost.
Discussion for (b)
Nearly all of the initial total kinetic energy is lost in this perfectly inelastic collision. \(\mathrm{KE}_{\text {int }}\) is mostly converted to thermal energy and sound. Note that in this case, the total kinetic energy does not decrease all the way down to zero, because doing so would violate conservation of momentum, because initial total momentum is not zero. This is the most amount of energy that can be converted from the initial total kinetic energy without violating conservation of momentum.
Example \(\PageIndex{2}\): Calculating Final Velocity and Energy Release: Two Carts Collide
During some collisions stored energy may be converted into kinetic energy during a collision. Figure \(\PageIndex{3}\) shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring.
In the collision pictured in Figure \(\PageIndex{3}\), two carts collide inelastically. Cart 1 (denoted \(m_{1}\) carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to total kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of \(2.00 \mathrm{~m} / \mathrm{s}\). Cart 2 (denoted \(m_{2}\) in Figure \(\PageIndex{3}\)) has a mass of 0.500 kg and an initial velocity of \(-0.500 \mathrm{~m} / \mathrm{s}\). After the collision, cart 1 is observed to recoil with a velocity of \(-4.00 \mathrm{~m} / \mathrm{s}\). (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?
Strategy
We can use conservation of momentum to find the final velocity of cart 2, because \(F_{\text {net }}=0\) (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the total kinetic energy before and after the collision to see how much energy was released by the spring.
Solution for (a)
As before, the equation for conservation of momentum in a two-object system is
\[m_{1} v_{1}+m_{2} v_{2}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime}. \nonumber\]
The only unknown in this equation is \(v_{2}^{\prime}\). Solving for \(v_{2}^{\prime}\) and substituting known values into the previous equation yields
\[\begin{aligned}
v_{2}^{\prime} &=\frac{m_{1} v_{1}+m_{2} v_{2}-m_{1} v_{1}^{\prime}}{m_{2}} \\
&=\frac{(0.350 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})+(0.500 \mathrm{~kg})(-0.500 \mathrm{~m} / \mathrm{s})}{0.500 \mathrm{~kg}}-\frac{(0.350 \mathrm{~kg})(-4.00 \mathrm{~m} / \mathrm{s})}{0.500 \mathrm{~kg}} \\
&=3.70 \mathrm{~m} / \mathrm{s}.
\end{aligned} \nonumber\]
Solution for (b)
The total kinetic energy before the collision is
\[\begin{aligned}
\mathrm{KE}_{\mathrm{int}} &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\
&=\frac{1}{2}(0.350 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2}(0.500 \mathrm{~kg})(-0.500 \mathrm{~m} / \mathrm{s})^{2} \\
&=0.763 \mathrm{~J}
\end{aligned}. \nonumber\]
After the collision, the total kinetic energy is
\[\begin{aligned}
\mathrm{KE}_{\text {int }}^{\prime} &=\frac{1}{2} m_{1} v_{1}^{\prime 2}+\frac{1}{2} m_{2} v_{2}^{\prime 2} \\
&=\frac{1}{2}(0.350 \mathrm{~kg})(-4.00 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2}(0.500 \mathrm{~kg})(3.70 \mathrm{~m} / \mathrm{s})^{2} \\
&=6.22 \mathrm{~J}.
\end{aligned} \nonumber\]
The change in total kinetic energy is thus
\[\begin{aligned}
\mathrm{KE}_{\text {int }}^{\prime}-\mathrm{KE}_{\text {int }} &=6.22 \mathrm{~J}-0.763 \mathrm{~J} \\
&=5.46 \mathrm{~J}.
\end{aligned} \nonumber\]
Discussion
The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The total kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.
Section Summary
- An inelastic collision is one in which the total kinetic energy changes (it is not conserved).
- A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces total kinetic energy as much as possible while conserving total momentum.
Glossary
- inelastic collision
- a collision in which total kinetic energy is not conserved
- perfectly inelastic collision
- a collision in which the colliding objects stick together


1.2.17: Simplifying Expressions with Roots
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Simplifying Expressions with Roots
Remember that when a real number \(n\) is multiplied by itself, we write \(n^{2}\) and read it '\(n^{2}\) squared’. This number is called the square of \(n\), and \(n\) is called the square root . For example,
\(13^{2}\) is read "\(13\) squared"
\(169\) is called the square of \(13\), since \(13^{2}=169\)
\(13\) is a square root of \(169\)
Square and Square Root of a Number
Square
If \(n^{2}=m\), then \(m\) is the square of \(n\).
Square Root
If \(n^{2}=m\), then \(n\) is a square root of \(m\).
Notice \((−13)^{2} = 169\) also, so \(−13\) is also a square root of \(169\). Therefore, both \(13\) and \(−13\) are square roots of \(169\).
So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? We use a radical sign , and write, \(\sqrt{m}\), which denotes the positive square root of \(m\). The positive square root is also called the principal square root .
We also use the radical sign for the square root of zero. Because \(0^{2}=0, \sqrt{0}=0\). Notice that zero has only one square root.
Square Root Notation
\(\sqrt{m}\) is read "the square root of \(m\)."
If \(n^{2}=m\), then \(n=\sqrt{m}\), for \(n\geq 0\).
We know that every positive number has two square roots and the radical sign indicates the positive one. We write \(\sqrt{169}=13\). If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, \(-\sqrt{169}=-13\).
Example \(\PageIndex{1}\)
Simplify:
- \(\sqrt{144}\)
- \(-\sqrt{289}\)
Solution :
- Since \(12^{2}=144\),
\(\sqrt{144}=12\)
- Since \(17^{2}=289\) and the negative is in front of the radical sign,
\(-\sqrt{289}=-17\)
You Try \(\PageIndex{1}\)
Simplify:
- \(-\sqrt{64}\)
- \(\sqrt{225}\)
- Answer
-
- \(-8\)
- \(15\)
Can we simplify \(-\sqrt{49}\)? Is there a number whose square is \(-49\)?
\((\)___\( )^{2}=-49\)
Any positive number squared is positive. Any negative number squared is positive. There is no real number equal to \(\sqrt{-49}\). The square root of a negative number is not a real number.
Example \(\PageIndex{2}\)
Simplify:
- \(\sqrt{-196}\)
- \(-\sqrt{64}\)
Solution :
- There is no real number whose square is \(-196\).
\(\sqrt{-196}\) is not a real number.
- The negative is in front of the radical and \(8^2=64).
\(-\sqrt{64}=-8\)
You Try \(\PageIndex{2}\)
Simplify:
- \(\sqrt{-169}\)
- \(-\sqrt{81}\)
- Answer
-
- not a real number
- \(-9\)
So far we have only talked about squares and square roots. Let’s now extend our work to include higher powers and higher roots.
Let’s review some vocabulary first.
\(\begin{array}{ll}{\text { We write: }} & {\text { We say: }} \\ {n^{2}} & {n \text { squared }} \\ {n^{3}} & {n \text { cubed }} \\ {n^{4}} & {n \text { to the fourth power }} \\ {n^{5}} & {n \text { to the fifth power }}\end{array}\)
The terms ‘squared’ and ‘cubed’ come from the formulas for area of a square and volume of a cube.
It will be helpful to have a table of the powers of the integers from \(−5\) to \(5\). See Figure \(\PageIndex{2}\)
Notice the signs in the table. All powers of positive numbers are positive, of course. But when we have a negative number, the even powers are positive and the odd powers are negative. We’ll copy the row with the powers of \(−2\) to help you see this.
We will now extend the square root definition to higher roots.
N th Root of a Number
If \(b^{n}=a\), then \(b\) is an \(n^{th}\) root of \(a\).
The principal \(n^{th}\) root of \(a\) is written \(\sqrt[n]{a}\).
The \(n\) is called the index of the radical.
Just like we use the word ‘cubed’ for \(b^{3}\), we use the term ‘cube root’ for \(\sqrt[3]{a}\).
We can refer to Figure \(\PageIndex{2}\) to help find higher roots.
\(\begin{aligned} 4^{3} &=64 & \sqrt[3]{64}&=4 \\ 3^{4} &=81 & \sqrt[4]{81}&=3 \\(-2)^{5} &=-32 & \sqrt[5]{-32}&=-2 \end{aligned}\)
Could we have an even root of a negative number? We know that the square root of a negative number is not a real number. The same is true for any even root. Even roots of negative numbers are not real numbers. Odd roots of negative numbers are real numbers.
Properties of \(\sqrt[n]{a}\)
When \(n\) is an even number and
- \(a \geq 0\), then \(\sqrt[n]{a}\) is a real number.
- \(a<0\), then \(\sqrt[n]{a}\) is not a real number.
When \(n\) is an odd number, \(\sqrt[n]{a}\) is a real number for all the values of \(a\).
We will apply these properties in the next two examples.
Example \(\PageIndex{3}\)
Simplify:
- \(\sqrt[3]{64}\)
- \(\sqrt[4]{81}\)
- \(\sqrt[5]{32}\)
Solution :
- Since \(4^{3}=64\),
\(\sqrt[3]{64}=4\)
- Since \((3)^{4}=81\),
\(\sqrt[4]{81}=3\)
- Since \((2)^{5}=32\),
\(\sqrt[5]{32}=2\)
You Try \(\PageIndex{3}\)
Simplify:
- \(\sqrt[3]{27}\)
- \(\sqrt[4]{256}\)
- \(\sqrt[5]{243}\)
- Answer
-
- \(3\)
- \(4\)
- \(3\)
In this example be alert for the negative signs as well as even and odd powers.
Example \(\PageIndex{4}\)
Simplify:
- \(\sqrt[3]{-125}\)
- \(\sqrt[4]{16}\)
- \(\sqrt[5]{-243}\)
Solution :
- Since \((-5)^{3}=-125\),
\(\sqrt[3]{-125}=-5\)
- Think, \((?)^{4}=-16\). No real number raised to the fourth power is negative.
\(\sqrt[4]{16}\) is not a real number.
- Since \((-3)^{5}=-243\),
\(\sqrt[5]{-243}=-3\)
You Try \(\PageIndex{4}\)
Simplify:
- \(\sqrt[3]{-27}\)
- \(\sqrt[4]{-256}\)
- \(\sqrt[5]{-32}\)
- Answer
-
- \(-3\)
- not real
- \(-2\)
Estimating and Approximating Roots
When we see a number with a radical sign, we often don’t think about its numerical value. While we probably know that the \(\sqrt{4}=2\), what is the value of \(\sqrt{21}\) or \(\sqrt[3]{50}\)? In some situations a quick estimate is meaningful and in others it is convenient to have a decimal approximation.
To get a numerical estimate of a square root, we look for perfect square numbers closest to the radicand. To find an estimate of \(\sqrt{11}\), we see \(11\) is between perfect square numbers \(9\) and \(16\), closer to \(9\). Its square root then will be between \(3\) and \(4\), but closer to \(3\).
Similarly, to estimate \(\sqrt[3]{91}\), we see \(91\) is between perfect cube numbers \(64\) and \(125\). The cube root then will be between \(4\) and \(5\).
Example \(\PageIndex{5}\)
Estimate each root between two consecutive whole numbers:
- \(\sqrt{105}\)
- \(\sqrt[3]{43}\)
Solution :
- Think of the perfect square numbers closest to \(105\). Make a small table of these perfect squares and their squares roots.
| \(\sqrt{105}\) | |
| Locate \(105\) between two consecutive perfect squares. | \(100<\color{red}105 \color{black} <121\) |
| \(\sqrt{105}\) is between their square roots. | \(10< \color{red}\sqrt{105}< \color{black}11\) |
- Similarly we locate \(43\) between two perfect cube numbers.
| \(\sqrt[3]{43}\) | |
| Locate \(43\) between two consecutive perfect cubes. | |
| \(\sqrt[3]{43}\) is between their cube roots. |
You Try \(\PageIndex{5}\)
Estimate each root between two consecutive whole numbers:
- \(\sqrt{38}\)
- \(\sqrt[3]{93}\)
- Answer
-
- \(6<\sqrt{38}<7\)
- \(4<\sqrt[3]{93}<5\)
There are mathematical methods to approximate square roots, but nowadays most people use a calculator to approximate square roots. To find a square root you will use the \(\sqrt{x}\) key on your calculator. To find a cube root, or any root with higher index, you will use the \(\sqrt[y]{x}\) key.
When you use these keys, you get an approximate value. It is an approximation, accurate to the number of digits shown on your calculator’s display. The symbol for an approximation is \(≈\) and it is read ‘approximately’.
Suppose your calculator has a \(10\) digit display. You would see that
\(\sqrt{5} \approx 2.236067978\) rounded to two decimal places is \(\sqrt{5} \approx 2.24\)
\(\sqrt[4]{93} \approx 3.105422799\) rounded to two decimal places is \(\sqrt[4]{93} \approx 3.11\)
How do we know these values are approximations and not the exact values? Look at what happens when we square them:
\(\begin{aligned}(2.236067978)^{2} &=5.000000002 &(3.105422799)^{4}&=92.999999991 \\(2.24)^{2} &=5.0176 & (3.11)^{4}&=93.54951841 \end{aligned}\)
Their squares are close to \(5\), but are not exactly equal to \(5\). The fourth powers are close to \(93\), but not equal to \(93\).
Example \(\PageIndex{6}\)
Round to two decimal places:
- \(\sqrt{17}\)
- \(\sqrt[3]{49}\)
- \(\sqrt[4]{51}\)
Solution :
- Use the calculator square root key, \(\sqrt{17} \approx 4.123105626 \dots\)
Round to two decimal places, \(\sqrt{17} \approx 4.12\)
- Use the calculator \(\sqrt[y]{x}\) key, \(\sqrt[3]{49} \approx 3.659305710 \dots\)
Round to two decimal places, \(\sqrt[3]{49} \approx 3.66\)
- Use the calculator \(\sqrt[y]{x}\) key, \(\sqrt[4]{51} \approx 2.6723451177 \dots\)
Round to two decimal places, \(\sqrt[4]{51} \approx 2.67\)
You Try \(\PageIndex{6}\)
Round to two decimal places:
- \(\sqrt{11}\)
- \(\sqrt[3]{71}\)
- \(\sqrt[4]{127}\)
- Answer
-
- \(\approx 3.32\)
- \(\approx 4.14\)
- \(\approx 3.36\)
Simplifying Variable Expressions with Roots
The odd root of a number can be either positive or negative. For example,
But what about an even root? We want the principal root, so \(\sqrt[4]{625}=5\).
But notice,
How can we make sure the fourth root of \(−5\) raised to the fourth power is \(5\)? We can use the absolute value. \(|−5|=5\). So we say that when \(n\) is even \(\sqrt[n]{a^{n}}=|a|\). This guarantees the principal root is positive.
Simplifying Odd and Even Roots
For any integer \(n\geq 2\),
when the index \(n\) is odd \(\sqrt[n]{a^{n}}=a\)
when the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\)
We must use the absolute value signs when we take an even root of an expression with a variable in the radical.
Example \(\PageIndex{7}\)
Simplify:
- \(\sqrt{x^{2}}\)
- \(\sqrt[3]{n^{3}}\)
- \(\sqrt[4]{p^{4}}\)
- \(\sqrt[5]{y^{5}}\)
Solution :
- We use the absolute value to be sure to get the positive root. Since the index \(n\) is even, \(\sqrt[n]{a^{n}}=|a|\),
\(\sqrt{x^{2}}=|x|\)
- This is an odd indexed root so there is no need for an absolute value sign. Since the index is \(n\) is odd, \(\sqrt[n]{a^{n}}=a\),
\(\sqrt[3]{m^{3}}=m\)
- Since the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\),
\(\sqrt[4]{p^{4}}=|p|\)
- Since the index \(n\) is odd, \(\sqrt[n]{a^{n}}=a\),
\(\sqrt[5]{y^{5}}=y\)
You Try \(\PageIndex{7}\)
Simplify:
- \(\sqrt{b^{2}}\)
- \(\sqrt[3]{w^{3}}\)
- \(\sqrt[4]{m^{4}}\)
- \(\sqrt[5]{q^{5}}\)
- Answer
-
- \(|b|\)
- \(w\)
- \(|m|\)
- \(q\)
What about square roots of higher powers of variables? The Power Property of Exponents says \(\left(a^{m}\right)^{n}=a^{m \cdot n}\). So if we square \(a^{m}\), the exponent will become \(2m\).
\(\left(a^{m}\right)^{2}=a^{2 m}\)
Looking now at the square root.
\(\sqrt{a^{2 m}}\)
Since \(\left(a^{m}\right)^{2}=a^{2 m}\).
\(\sqrt{\left(a^{m}\right)^{2}}\)
Since \(n\) is even \(\sqrt[n]{a^{n}}=|a|\).
\(\left|a^{m}\right|\)
So \(\sqrt{a^{2 m}}=\left|a^{m}\right|\).
We apply this concept in the next example.
Example \(\PageIndex{8}\)
Simplify:
- \(\sqrt{x^{6}}\)
- \(\sqrt{y^{16}}\)
Solution :
- Since \(\left(x^{3}\right)^{2}=x^{6}\) and the index \(n\) is even \(\sqrt{a^{n}}=|a|\), \[\begin{align*}\sqrt{x^{6}}&=\sqrt{\left(x^{3}\right)^{2}}\\&=\left|x^{3}\right|\end{align*}\]
Hence, \(\sqrt{x^{6}}=\left|x^{3}\right|\)
- Since \(\left(y^{8}\right)^{2}=y^{16}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}\sqrt{y^{16}}&=\sqrt{\left(y^{8}\right)^{2}}\\&=y^{8}\end{align*}\]
Hence, \(\sqrt{y^{16}}=y^{8}\)
In this case the absolute value sign is not needed as \(y^{8}\) is positive.
You Try \(\PageIndex{8}\)
Simplify:
- \(\sqrt{y^{18}}\)
- \(\sqrt{z^{12}}\)
- Answer
-
- \(|y^{9}|\)
- \(z^{6}\)
The next example uses the same idea for higher roots.
Example \(\PageIndex{9}\)
Simplify:
- \(\sqrt[3]{y^{18}}\)
- \(\sqrt[4]{z^{8}}\)
Solution :
- Since \(\left(y^{6}\right)^{3}=y^{18}\) and \(n\) is odd, \(\sqrt[n]{a^{n}}=a\), \[\begin{align*}\sqrt[3]{y^{18}}=\sqrt[3]{\left(y^{6}\right)^{3}}=y^{6}\end{align*}\]
Hence, \(\sqrt[3]{y^{18}}=y^{6}\)
- Since \(\left(z^{2}\right)^{4}=z^{8}\) and \(z^{2}\) is positive, we do not need an absolute value sign, \[\begin{align*}\sqrt[4]{z^{8}}&=\sqrt[4]{\left(z^{2}\right)^{4}}\\&=z^{2}\end{align*}\]
Hence, \(\sqrt[4]{z^{8}}=z^{2}\)
You Try \(\PageIndex{9}\)
Simplify:
- \(\sqrt[4]{u^{12}}\)
- \(\sqrt[3]{v^{15}}\)
- Answer
-
- \(|u^{3}|\)
- \(v^{5}\)
In the next example, we now have a coefficient in front of the variable. The concept \(\sqrt{a^{2 m}}=\left|a^{m}\right|\) works in much the same way.
\(\sqrt{16 r^{22}}=4\left|r^{11}\right|\) because \(\left(4 r^{11}\right)^{2}=16 r^{22}\).
But notice \(\sqrt{25 u^{8}}=5 u^{4}\) and no absolute value sign is needed as \(u^{4}\) is always positive.
Example \(\PageIndex{10}\)
Simplify:
- \(\sqrt{16 n^{2}}\)
- \(-\sqrt{81 c^{2}}\)
Solution :
- Since \((4 n)^{2}=16 n^{2}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}\sqrt{16 n^{2}}&=\sqrt{(4 n)^{2}}\\&=4|n|\end{align*}\]
Hence, \(\sqrt{16 n^{2}}=4|n|\)
- Since \((9 c)^{2}=81 c^{2}\) and the index \(n\) is even \(\sqrt[n]{a^{n}}=|a|\), \[\begin{align*}-\sqrt{81 c^{2}}&=-\sqrt{(9 c)^{2}}\\&=-9|c| \end{align*}\]
Hence, \(-\sqrt{81 c^{2}}=-9|c|\)
You Try \(\PageIndex{10}\)
Simplify:
- \(\sqrt{64 x^{2}}\)
- \(-\sqrt{100 p^{2}}\)
- Answer
-
- \(8|x|\)
- \(-10|p|\)
This example just takes the idea farther as it has roots of higher index.
Example \(\PageIndex{11}\)
Simplify:
- \(\sqrt[3]{64 p^{6}}\)
- \(\sqrt[4]{16 q^{12}}\)
Solution :
- Rewrite \(64p^{6}\) as \(\left(4 p^{2}\right)^{3}\), \[\sqrt[3]{64 p^{6}}=\sqrt[3]{\left(4 p^{2}\right)^{3}}\nonumber\]
Take the cube root, \[\sqrt[3]{64 p^{6}}=4p^{2}\nonumber\]
- Rewrite the radicand as a fourth power, \[\sqrt[4]{16 q^{12}}=\sqrt[4]{\left(2 q^{3}\right)^{4}}\nonumber\]
Take the fourth root, \[\sqrt[3]{64 p^{6}}=2|q^{3}|\nonumber\]
You Try \(\PageIndex{11}\)
Simplify:
- \(\sqrt[3]{27 x^{27}}\)
- \(\sqrt[4]{81 q^{28}}\)
- Answer
-
- \(3x^{9}\)
- \(3|q^{7}|\)
The next examples have two variables.
Example \(\PageIndex{12}\)
Simplify:
- \(\sqrt{36 x^{2} y^{2}}\)
- \(\sqrt{121 a^{6} b^{8}}\)
- \(\sqrt[3]{64 p^{63} q^{9}}\)
Solution :
- Since \((6 x y)^{2}=36 x^{2} y^{2}\), \[\sqrt{36 x^{2} y^{2}}=\sqrt{(6 x y)^{2}}\nonumber\]
Take the square root, \(\sqrt{36 x^{2} y^{2}}=6|xy|\)
- Since \(\left(11 a^{3} b^{4}\right)^{2}=121 a^{6} b^{8}\), \[\sqrt{121 a^{6} b^{8}}=\sqrt{\left(11 a^{3} b^{4}\right)^{2}}\nonumber\]
Take the square root, \[\sqrt{121 a^{6} b^{8}}=11\left|a^{3}\right| b^{4}\nonumber\]
- Since \(\left(4 p^{21} q^{3}\right)^{3}=64 p^{63} q^{9}\), \[\sqrt[3]{64 p^{63} q^{9}}=\sqrt[3]{\left(4 p^{21} q^{3}\right)^{3}}\nonumber\]
Take the cube root, \[\sqrt[3]{64 p^{63} q^{9}}=4p^{21}q^{3}\nonumber\]
You Try \(\PageIndex{12}\)
Simplify:
- \(\sqrt{100 a^{2} b^{2}}\)
- \(\sqrt{144 p^{12} q^{20}}\)
- \(\sqrt[3]{8 x^{30} y^{12}}\)
- Answer
-
- \(10|ab|\)
- \(12p^{6}q^{10}\)
- \(2x^{10}y^{4}\)


8: Further Applications of Trigonometry
In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully.
-
- 8.0: Prelude to Further Applications of Trigonometry
- The world’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California. Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure.
-
- 8.1: Non-right Triangles - Law of Sines
- In this section, we will find out how to solve problems involving non-right triangles. The Law of Sines can be used to solve oblique triangles. According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution.
-
- 8.2: Non-right Triangles - Law of Cosines
- Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.
-
- 8.3: Polar Coordinates
- When we think about plotting points in the plane, we usually think of rectangular coordinates (x,y) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled (r,θ) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.
-
- 8.4: Polar Coordinates - Graphs
- polar equation describes a relationship between rr and θ on a polar grid. It is easier to graph polar equations if we can test the equations for symmetry. There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry.
-
- 8.5: Polar Form of Complex Numbers
- In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.
-
- 8.6: Parametric Equations
- We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.
-
- 8.8: Vectors
- Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s ground speed and bearing, while investigating another approach to problems of this type.


16: Solubility equilibrium
We previously learned about aqueous solutions and their importance, as well as about solubility rules. While this gives us a picture of solubility, that picture is not complete if we look at the rules alone. Solubility equilibrium, which we will explore in this chapter, is a more complex topic that allows us to determine the extent to which a slightly soluble ionic solid will dissolve, and the conditions under which precipitation.
Contributors
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).