Section 6: Formal languages
Here is a famous valid argument:
Socrates is a man.
All men are mortal.
.˙. Socrates is mortal.
This is an iron-clad argument. The only way you could challenge the conclusion is by denying one of the premises— the logical form is impeccable. What about this next argument?
Socrates is a man.
All men are carrots.
.˙. Socrates is a carrot.
This argument might be less interesting than the first, because the second premise is obviously false. There is no clear sense in which all men are carrots. Yet the argument is valid. To see this, notice that both arguments have this form:
\(S\) is \(M\).
All \(M\)s are \(C\).
.˙. \(S\) is \(C\).
In both arguments \(S\) stands for Socrates and \(M\) stands for man. In the first argument, \(C\) stands for mortal; in the second, \(C\) stands for carrot. Both arguments have this form, and every argument of this form is valid. So both arguments are valid.
What we did here was replace words like ‘man’ or ‘carrot’ with symbols like ‘M’ or ‘C’ so as to make the logical form explicit. This is the central idea behind formal logic. We want to remove irrelevant or distracting features of the argument to make the logical form more perspicuous.
Starting with an argument in a natural language like English, we translate the argument into a formal language . Parts of the English sentences are replaced with letters and symbols. The goal is to reveal the formal structure of the argument, as we did with these two.
There are formal languages that work like the symbolization we gave for these two arguments. A logic like this was developed by Aristotle, a philosopher who lived in Greece during the 4th century BC. Aristotle was a student of Plato and the tutor of Alexander the Great. Aristotle’s logic, with some revisions, was the dominant logic in the western world for more than two millennia.
In Aristotelean logic, categories are replaced with capital letters. Every sentence of an argument is then represented as having one of four forms, which medieval logicians labeled in this way: (A) All \(A\)s are \(B\)s. (E) No \(A\)s are \(B\)s. (I) Some \(A\) is \(B\). (O) Some \(A\) is not \(B\).
It is then possible to describe valid syllogisms , three-line arguments like the two we considered above. Medieval logicians gave mnemonic names to all of the valid argument forms. The form of our two arguments, for instance, was called Barbara . The vowels in the name, all As, represent the fact that the two premises and the conclusion are all (A) form sentences.
There are many limitations to Aristotelean logic. One is that it makes no distinction between kinds and individuals. So the first premise might just as well be written ‘All \(S\)s are \(M\)s’: All Socrateses are men. Despite its historical importance, Aristotelean logic has been superceded. The remainder of this book will develop two formal languages.
The first is SL, which stands for sentential logic . In SL, the smallest units are sentences themselves. Simple sentences are represented as letters and connected with logical connectives like ‘and’ and ‘not’ to make more complex sentences.
The second is QL, which stands for quantified logic . In QL, the basic units are objects, properties of objects, and relations between objects.
When we translate an argument into a formal language, we hope to make its logical structure clearer. We want to include enough of the structure of the English language argument so that we can judge whether the argument is valid or invalid. If we included every feature of the English language, all of the subtlety and nuance, then there would be no advantage in translating to a formal language. We might as well think about the argument in English.
At the same time, we would like a formal language that allows us to represent many kinds of English language arguments. This is one reason to prefer QL to Aristotelean logic; QL can represent every valid argument of Aristotelean logic and more.
So when deciding on a formal language, there is inevitably a tension between wanting to capture as much structure as possible and wanting a simple formal language— simpler formal languages leave out more. This means that there is no perfect formal language. Some will do a better job than others in translating particular English-language arguments.
In this book, we make the assumption that true and false are the only possible truth-values. Logical languages that make this assumption are called bivalent , which means two-valued . Aristotelean logic, SL, and QL are all bivalent, but there are limits to the power of bivalent logic. For instance, some philosophers have claimed that the future is not yet determined. If they are right, then sentences about what will be the case are not yet true or false. Some formal languages accommodate this by allowing for sentences that are neither true nor false, but something in between. Other formal languages, so-called paraconsistent logics, allow for sentences that are both true and false.
The languages presented in this book are not the only possible formal languages. However, most nonstandard logics extend on the basic formal structure of the bivalent logics discussed in this book. So this is a good place to start.
Summary of logical notions
~An argument is (deductively)
valid
if it is impossible for the premises to be true and the conclusion false; it is
invalid
otherwise.
~A
tautology
is a sentence that must be true, as a matter of logic.
~A
contradiction
is a sentence that must be false, as a matter of logic.
~A
contingent sentence
is neither a tautology nor a contradiction.
~Two sentences are
logically equivalent
if they necessarily have the same truth value.
~A set of sentences is
consistent
if it is logically possible for all the members of the set to be true at the same time; it is
inconsistent
otherwise.


11.1: Voting Methods
- Apply plurality voting to determine a winner.
- Apply runoff voting to determine a winner.
- Apply ranked-choice voting to determine a winner.
- Apply Borda count voting to determine a winner.
- Apply pairwise comparison and Condorcet voting to determine a winner.
- Apply approval voting to determine a winner.
- Compare and contrast voting methods to identify flaws.
Today is the day that you begin your quest to collaborate on the constitution of Imaginaria! Let’s begin by thinking about the selection of a leader who can serve as president. It seems straightforward; if the majority of citizens prefer a particular candidate, that candidate should win. But not all votes are decided by a simple majority. Why not? What are the options?
Majority versus Plurality Voting
When an election involves only two options, a simple majority is a reasonable way to determine a winner. A majority is a number equaling more than half, or greater than 50 percent of the total.
Let’s take a look at the outcomes of U.S. presidential elections to understand more. Table 11.1 displays the results of the 2000 U.S. presidential election. Like most presidential elections, this election involved more than two options. If that is the case, is it possible that no single candidate will receive more than half of the votes cast?
| Candidate (Party Label) | Popular Vote Total |
|---|---|
| Al Gore (Democrat) | 50,999,897 |
| George W. Bush (Republican) | 50,456,002 |
| Ralph Nader (Green) | 2,882,955 |
| Patrick J. Buchanan (Reform/Independent) | 448,895 |
| Harry Browne (Libertarian) | 384,431 |
| Howard Phillips (Constitution) | 98,020 |
| Other | 134,900 |
| Total: | 105,405,100 |
Refer back to Table 11.1. Did any single candidate secure the majority of popular votes?
- Answer
-
Step 1: Calculate 50 percent of 105,405,100 by multiplying the decimal form of 50 percent, which is 0.50, by 158,394,605:
Step 2: Determine the minimum number of votes needed to have a majority. The minimum number of votes required is the lowest counting number that is larger than 50 percent of the votes. To have a majority, an individual candidate must have more than 52,702,550; so, a majority candidate must have 52,702,551 votes or more.
Step 3: Compare the number of votes each candidate received to 52,702,551. According to the data in Table 11.1, none of the candidates secured a majority.
According to the Cook Political Report, a total of 158,394,605 votes were cast in the 2020 U.S. presidential election. Of those, 81,281,502 were cast for Joe Biden, 74,222,593 for Donald Trump, and 2,890,510 for other candidates. Although U.S. presidential elections are not determined by popular vote, did either Joe Biden or Donald Trump secure a majority of the votes?
Unlike in the 2000 U.S. presidential election, a candidate won the majority of votes in the 2020 election (see Table 11.1). It is a common occurrence for no single candidate to receive a majority of the votes in an election with more than two candidates. When this occurs, the candidate with the largest portion of the votes is said to have a plurality .
Refer again to Table 11.1. In the 2000 U.S. presidential election, which candidate had a plurality of popular votes?
- Answer
-
Al Gore secured 50,999,897 votes which was more than any other single candidate. Therefore, he had a plurality of the popular votes.
Your plans for Imaginarian elections will likely include primary elections, or preliminary elections to select candidates for a principal or general election. Table 11.2 displays the results of the 2018 U.S. Senate Republican primary for Maryland.
| Top Four Republican Candidates | Votes | Percentage of Party Votes |
|---|---|---|
| Cambell, Tony | 51,426 | 29.22% |
| Chaffee, Chris | 42,328 | 24.05% |
| Grigorian, Christina J. | 30,756 | 17.48% |
| Graziani, John R. | 15,435 | 8.77% |
| Total Votes | 175,981 | 100% |
Consider the results of the 2018 U.S. Senate primary for Maryland. Determine which candidate won the primary for the Republicans (R). Did the candidate win a majority or a plurality of Republican votes?
Consider how election by plurality, not majority, is the most common method of selecting candidates for public office.
The U.S. Electorial College: Winning the Presidency without a Plurality
In the 2000 U.S. presidential election, Al Gore had a plurality of the popular votes, but he did not win the election. Why? This occurred because the U.S. president and vice president are elected by electors rather than a direct vote by the citizens. The electors are part of the Electoral College, a body of people representing the states. Why was the Electoral College created? The Electoral College was created as a compromise between those authors of the U.S. Constitution who believed Congress should elect the president, and those who believed the citizens should vote directly. The popular vote was not recorded until the presidential election of 1824. Since then, only five presidents have been elected without winning a plurality of the popular vote: John Quincy Adams in 1824, Rutherford B. Hayes in 1876, Grover Cleveland in 1888, George Bush in 2000, and Donald Trump in 2016.
Runoff Voting
Has your family ever debated what to have for dinner? Suppose your family is deciding on a restaurant and exactly half of you want to have pizza but the other half want hamburgers. How do you decide when the result is a tie? You need a tiebreaker!
Will the new democracy of Imaginaria need tiebreakers? When no candidate satisfies the requirements to win the election, a runoff election , or second election, is held to determine a winner.
How would runoff voting work in Imaginaria? There are many types of runoff voting systems , which are voting systems that utilize a runoff election when the first round does not result in a winner. The method for implementing a runoff election can vary widely, particularly in the criteria used to determine whether a candidate will be on the ballot in the second election. For example, a two-round system is a runoff voting system in which only the top candidates advance to the runoff election. In some two-round systems, only the top two candidates are on the second ballot, or it may be any candidate who secures a certain percentage of the vote will advance. The Hare Method is another runoff voting system in which only the candidate(s) with the very least votes are eliminated. This can potentially result in several rounds of runoff elections.
A condominium association elects a new president every two years by a two-round system of voting. If none of the candidates receive a majority, the association charter states that the top two candidates will be eligible to participate in a runoff election. In a particular year, five residents were nominated. The results of the first round are given in the table below.
| Candidate | Votes in First Round |
|---|---|
| Abou | 18 |
| Baiocchi | 10 |
| Campana | 5 |
| Dali | 11 |
| Eugene | 4 |
- Is there a winner based on the first round? Why or why not?
- If there is a winner, who won? If there should be a runoff, who will advance to the second round?
- Answer
-
- A majority of 48 total votes is required to win. Begin by finding 50 percent of 48, which is calculated as follows: . A majority is 25 or more. No candidate has a majority, so there is no winner based on the first round.
- Abou and Dali advance to the second round.
The student government bylaws of a particular college require that a new president is elected annually by plurality voting. In the event of a tie, the bylaws require the candidate(s) with the fewest votes to be removed from the ballot and a runoff election to be held with the remaining candidates. This process is repeated until a single candidate receives a plurality and wins the election. The results of two voting rounds are given in the table below.
| Candidate | Votes in First Round | Votes in Second Round |
|---|---|---|
| Ferguson | 158 | 168 |
| Garcia | 103 | 104 |
| Hearn | 157 | 180 |
| Isaac | 123 | 123 |
| Jackman | 58 | Eliminated |
| Kelly | 72 | 74 |
| Lim | 158 | 180 |
Which two candidates tied in the second round? Does this mean there will be a third election?
If there was a winner, who won? If there should be a runoff, who must be removed from the ballot?
Steps to Determine Winner by Plurality or Majority Election with Runoff
To determine the winner by plurality or when a majority election with runoff occurs, we take these three steps:
Step 1: If a majority is required to win the election, determine the number of votes needed to achieve a majority. This is the least whole number greater than 50 percent of the total votes. If a majority is not required, move to Step 2.
Step 2: Count the number of votes for each candidate in the current round of voting. If a single candidate has enough votes to win a plurality, or a majority as appropriate, then you are done! Otherwise, eliminate a predetermined number of candidates based on the rules of the election. Elimination conditions may vary. For example, the rules may state that the candidate(s) with the fewest votes will be eliminated (as in the Hare method), or that only the candidates meeting a certain threshold will move on (as in a two-round system). Once the appropriate candidates are eliminated, move on to Step 3.
Step 3: Hold a runoff election. If the runoff is simulated using a list of voter’s preferences, renumber the preferences to reflect the remaining number of options in such a way that the original order of preference is retained. Then repeat Step 2.
Note: The second and third steps may be repeated as many times as necessary for voting procedures that allow multiple runoffs.
The five members of the Chionilis family—Annette, Rene, Seema, Titus, and Galen—have decided to get takeout for dinner. They are trying to decide on a restaurant. The options are Rainbow China, Dough Boys Pizza, Taco City, or Caribbean Flavor. They will use majority election with runoffs where the restaurant with the fewest votes is eliminated in each round. The preferences of each family member are listed by first initial in the table below. An entry of 1 represents the person’s first choice; 2, their second; and so on. For example, Annette’s second choice is Dough Boys Pizza.
| Options | A | R | S | T | G |
|---|---|---|---|---|---|
| Rainbow China | 1 | 3 | 3 | 1 | 3 |
| Dough Boys Pizza | 2 | 2 | 1 | 2 | 1 |
| Taco City | 3 | 4 | 2 | 4 | 2 |
| Caribbean Flavor | 4 | 1 | 4 | 3 | 4 |
Use the information in the table to answer the following questions.
- Which common type of runoff voting method is this?
- List the results of each round of voting based on this information and determine which restaurant was ultimately chosen.
- Answer
-
- the Hare Method
-
Step 1:
Determine the number of votes necessary to have a majority. There are five family members, so 50 percent of 5 is
. A majority is three or more votes.
Step 2: Count the number of votes for each restaurant in the first election. In a list of voter preferences, the 1s represent the top choice of each voter, which corresponds to their vote in the first round.
Results of Round 1:- Rainbow China — 2 votes
- Dough Boys — 2 votes
- Taco City — 0 votes
- Caribbean Flavor — 1 vote
No restaurant received a majority. Eliminate Taco City, which has the fewest first place votes:Options A R S T G Rainbow China 1 3 3 1 3 Dough Boys Pizza 2 2 1 2 1 Caribbean Flavor 4 1 4 3 4
Step 3: Hold a runoff election. In other words, hold a second round. Since we have a list of the voters’ preferences with the eliminated option removed, we will renumber the preferences as first, second, and third so that we keep the original order of preference. The result is that we will count the second-place vote of any voter whose first choice was eliminated.Options A R S T G Rainbow China 1 3 2 1 2 Dough Boys Pizza 2 2 1 2 1 Caribbean Flavor 3 1 3 3 3
Step 4: Repeat the process from Step 2. Count the number votes for each restaurant in the first-round election. Since we are using a list of preferences, we need to count the number of 1s received by each restaurant.
Results of Round 2:- Rainbow China — 2 votes
- Dough Boys — 2 votes
- Caribbean Flavor — 1 vote
No single restaurant has three votes. Eliminate Caribbean Flavor, which has the fewest first place votes.OPTIONS A R S T G Rainbow China 1 3 2 1 2 Dough Boys Pizza 2 2 1 2 1
Step 5: Repeat the process from Step 3. Hold another runoff election. This will be Round 3. Renumber the voters' preferences as first and second this time.Options A R S T G Rainbow China 1 2 2 1 2 Dough Boys Pizza 2 1 1 2 1
Step 6: Repeat the process from Step 2 one last time. Count the number of first place votes for each remaining restaurant.Results of Round 3:
- Rainbow China — 2 votes
- Dough Boys — 3 votes
Determine whether any one choice has a majority. Yes! Dough Boys has three votes, so it is the winner!
There are six members on the board of a Parent Teacher Association (PTA) at a local elementary school: the president (P), the vice president (V), the recording secretary (R), the liaison to the administration (L), the treasurer (T), and the chief fundraiser (C). The board must decide which equipment to purchase for the classrooms with moneys from their annual fundraisers. The preferences of the board members are shown in the table below.
| P | V | R | L | T | C | |
|---|---|---|---|---|---|---|
| Option A | 3 | 4 | 1 | 3 | 4 | 1 |
| Option B | 1 | 3 | 2 | 1 | 3 | 2 |
| Option C | 2 | 2 | 3 | 2 | 1 | 4 |
| Option D | 4 | 1 | 4 | 4 | 2 | 3 |
The board uses plurality method and a runoff in the event of a tie, such that the option(s) with the least votes will be eliminated in each round. Which option will be chosen?
Ranked-Choice Voting
In Example preference ranking . A ballot in which a voter is required to give an ordering of their preferences is a ranked ballot , and any voting system in which a voter uses a ranked ballot is referred to as ranked voting .
and Your Turn , you were given a list that ordered each voter’s preferences. This ordering is called aThe vote for the Academy Awards uses a ranked ballot. The table below provides an example of a ranked ballot for the 2020 Academy Award nominees for Best Director.
| Candidate for Best Director | Rank top choice as 1, next choice as 2, and so on. | ||||
|---|---|---|---|---|---|
| Martin Scorsese, The Irishman | 1 | 2 | 3 | 4 | 5 |
| Todd Phillips, Joker | 1 | 2 | 3 | 4 | 5 |
| Sam Mendes, 1917 | 1 | 2 | 3 | 4 | 5 |
| Quentin Tarantino, Once Upon a Time in Hollywood | 1 | 2 | 3 | 4 | 5 |
| Bong Joon-ho, Parasite | 1 | 2 | 3 | 4 | 5 |
As you decide on the voting methods that will be used in your new democracy, budget must be a consideration. You might consider a particular type of ranked voting called ranked-choice voting (RCV) , which simulates a series of runoff elections without the usual time and expense involved when voters must repeatedly return to the polls, like we did in Example .
The method of ranked-choice voting (RCV), also called instant runoff voting (IRV) , is a version of the Hare Method, using preference ranking so that, if no single candidate receives a majority, the least popular selections can be eliminated and the results can be recounted, without the need for more elections.
Ranked voting can be confused with ranked-choice voting, but ranked voting is a more general category which includes ranked-choice voting and several other voting methods.
As we explore examples of ranked voting, we will summarize the voters’ preference rankings using a table in which the top row shows the number of ballots that ranked the options in the same order. Let’s practice interpreting the information in this type of table.
Refer to the table below containing voters’ preference rankings to answer the following questions.
| Number of Ballots | 100 | 200 | 150 | 75 |
|---|---|---|---|---|
| Option A | 1 | 4 | 3 | 4 |
| Option B | 2 | 3 | 4 | 2 |
| Option C | 4 | 2 | 1 | 1 |
| Option D | 3 | 1 | 2 | 3 |
- How many voters ranked the options in the following order: Option A in fourth place, Option B in second place, Option C in first place, and Option D in third place?
- How many ballots in total were collected?
- How many voters indicated that Option C was their first choice?
- Answer
-
- The column farthest to the right displays this ordering. The top entry in this column is 75; so, there were 75 voters who ranked the options in this way.
- The sum of the top row gives the total number of ballots collected: . So, there were 525 ballots collected.
- In the Option C row, there are two entries of 1 which indicate a first choice for that option. These occur in the last two columns. The sum of the top entries in these columns is . So, 225 voters indicated Option C as their first choice.
A kindergarten class votes on their favorite colors using a ranked ballot. Use the results in the following table to answer the questions.
| Number of Ballots | 4 | 6 | 4 | 7 |
|---|---|---|---|---|
| Red | 2 | 6 | 2 | 5 |
| Blue | 1 | 2 | 1 | 4 |
| Green | 6 | 5 | 5 | 2 |
| Yellow | 5 | 4 | 6 | 3 |
| Purple | 4 | 3 | 4 | 1 |
| Pink | 3 | 1 | 3 | 6 |
How many students voted in total?
How many students voted for yellow as their favorite color?
How many students in the class indicated that blue was their favorite color while green was their least favorite color?
Now that we’ve covered how to read a summary of preference rankings, let’s practice using the ranked-choice method to determine the winner of an election. Recall that ranked-choice voting is still the Hare Method where the candidate with the very least number of votes is eliminated each round until a majority is attained. The difference here is that the voters have completed a ranked ballot, so they don't have to visit the polls multiple times. Here are the steps for ranked-choice voting.
Steps to Determine Winner by Ranked-Choice Voting
To determine the winner when ranked-choice voting occurs, we take these three steps:
Step 1: Determine the number of votes needed to achieve a majority. This is the least whole number greater than 50 percent of the total votes.
Step 2: Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election and we are done! Otherwise, eliminate the candidate(s) with the fewest votes and complete Step 3.
Step 3: Reallocate the votes to the remaining candidates, and repeat Step 2.
Be methodical to avoid arithmetic errors. Make sure that each time you count the number of first place votes they sum to the number of ballots.
How Does Ranked-Choice Voting Work?
Let’s review the kindergarten class color preferences again, and this time determine which color would be selected based on these results using the ranked-choice method.
| Number of Ballots | 4 | 6 | 4 | 7 |
|---|---|---|---|---|
| Red | 2 | 6 | 2 | 5 |
| Blue | 1 | 2 | 1 | 4 |
| Green | 6 | 5 | 5 | 2 |
| Yellow | 5 | 4 | 6 | 3 |
| Purple | 4 | 3 | 4 | 1 |
| Pink | 3 | 1 | 3 | 6 |
- Answer
-
Step 1: Determine whether any candidate received a majority. There were 21 ballots. Fifty percent of 21 is . A majority is 11.
Step 2: Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes.
- Red: 0
- Blue:
- Green: 0
- Yellow: 0
- Purple: 7
- Pink: 6
Notice that , which is the total number of ballots. Confirming this helps to catch any arithmetic or counting errors. No candidate has a majority with 11 or more votes. We must eliminate red, green, and yellow which had the fewest votes with 0 each. The remaining votes that must be counted for Round 2 are given in the table below.
Number of Ballots 4 6 4 7 Blue 1 2 1 4 Purple 4 3 4 1 Pink 3 1 3 6 Step 3: Reallocate the votes to the remaining candidates. We can do this by numbering the choices as 1, 2, and 3 in such a way that the order of preference is retained as seen in the table below:
Number of Ballots 4 6 4 7 Blue 1 2 1 2 Purple 3 3 3 1 Pink 2 1 2 3 Step 4: Repeat the process from Step 2. Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes.
- Blue:
- Purple: 7
- Pink: 6
Confirm that . Great! No candidate has 11 or more votes. We must eliminate pink which had the fewest votes with 6. The remaining votes that must be counted for Round 2 are shown in the table below.
Number of Ballots 4 6 4 7 Blue 1 2 1 2 Purple 3 3 3 1 Step 5: Repeat the process from Step 3. Reallocate the votes to the remaining candidates. We can do this by numbering the choices as 1 and 2;
Number of Ballots 4 6 4 7 Blue 1 1 1 2 Purple 2 2 2 1 Step 6: Repeat the process from Step 2 one last time. Count the number of first place votes for each candidate. If a candidate has a majority, that candidate wins the election. Otherwise, eliminate the candidate(s) with the fewest votes.
- Blue:
- Purple: 7
Blue has a majority and wins the election!
Suppose that 58 Star Wars fans were asked to vote for their favorite Star Wars character. They were given a ranked ballot, and the results are shown in the following table. Use ranked-choice voting to determine the winner.
| Number of Ballots | 7 | 6 | 10 | 8 | 4 | 5 | 6 | 7 | 2 | 3 |
|---|---|---|---|---|---|---|---|---|---|---|
| Han Solo | 3 | 5 | 1 | 3 | 2 | 4 | 2 | 2 | 1 | 4 |
| Princess Leia | 2 | 1 | 3 | 6 | 4 | 3 | 3 | 1 | 3 | 2 |
| Luke Skywalker | 6 | 6 | 6 | 1 | 5 | 5 | 5 | 4 | 6 | 1 |
| Chewbacca | 4 | 4 | 5 | 4 | 6 | 1 | 6 | 5 | 4 | 5 |
| Yoda | 1 | 2 | 4 | 2 | 3 | 2 | 1 | 3 | 2 | 3 |
| R2-D2 | 5 | 3 | 2 | 5 | 1 | 6 | 4 | 6 | 5 | 6 |
Borda Count Voting
Ranked-choice voting is one type of ranked voting that simulates multiple runoffs based on ranked ballots. Another type of ranked voting is the Borda count method , which uses ranked ballots that award candidates points corresponding to the number of candidates ranked lower on each ballot.
To understand how this works, let’s review the favorite colors of our kindergarten class from the table below. Let’s focus on the votes represented by the first column of the preference summary.
| Number of Ballots | 4 | 6 | 4 | 7 |
|---|---|---|---|---|
| Red | 2 | 6 | 2 | 5 |
| Blue | 1 | 2 | 1 | 4 |
| Green | 6 | 5 | 5 | 2 |
| Yellow | 5 | 4 | 6 | 3 |
| Purple | 4 | 3 | 4 | 1 |
| Pink | 3 | 1 | 3 | 6 |
Each student had six options. This first column tells us that four students ranked blue as their first choice, red as their second choice, pink as their third choice, purple as their fourth choice, yellow as their fifth choice, and green as their sixth choice. Blue was ranked higher than other colors. For each of the four students who completed their ballot in this way, blue would receive five points. Since there were four ballots with this ordering, blue would receive points from the first column. To determine the total points for each candidate, we have to find the sum of the points they received in each column.
To determine the winner of a contest using the Borda count method, we must compare total number of points earned by each candidate. The candidate with the most points is the winner. Each row of the preference summary corresponds to a single candidate. To find the number of points received by a particular candidate in the preference summary, or their Borda score , we will need to focus on the row in which that candidate appears.
Before we practice determining the winner of a Borda count election, let’s examine how to find the Borda score for a single candidate.
Let’s review the ballots from the kindergarten class again, as shown in the table below. This time, let’s determine the Borda score received by the color purple.
| Number of Ballots | 4 | 6 | 4 | 7 |
|---|---|---|---|---|
| Red | 2 | 6 | 2 | 5 |
| Blue | 1 | 2 | 1 | 4 |
| Green | 6 | 5 | 5 | 2 |
| Yellow | 5 | 4 | 6 | 3 |
| Purple | 4 | 3 | 4 | 1 |
| Pink | 3 | 1 | 3 | 6 |
- Answer
-
Step 1: Find the number of points received by the candidate in each column.
Column 1:
Column 2:
Column 3:
Column 4:
Step 2: Find the sum of the points received in each column. This is the total number of points received by this candidate:
This process can also be combined into one step as shown here.
Purple received 69 points in this election.
When calculating a Borda score in one step, be careful to use the correct order of operations. Perform the subtraction inside each pair of parentheses first, then perform each multiplication, and then perform each addition.
Consider the color preferences of the kindergarten class once more. Using the Borda count method, determine the total number of points the color blue received.
Now let’s determine the winner of an election by comparing the Borda scores for each of the candidates.
Use the table below, which displays a sample preference summary, to answer the questions that follow.
| Number of Ballots | 95 | 90 | 110 | 115 |
|---|---|---|---|---|
| Option A | 4 | 4 | 1 | 1 |
| Option B | 2 | 2 | 2 | 2 |
| Option C | 3 | 1 | 3 | 4 |
| Option D | 1 | 3 | 4 | 3 |
- Use the ranked-choice voting method to determine the winner of the election.
- Use the Borda count method to determine the winner of the election.
- Answer
-
-
Step 1:
Determine the number of votes needed to achieve a majority. The number of ballots is
. Fifty percent of 410 is
. So, 206 votes or more is a majority.
Step 2: Count the number of first place votes for each candidate.
- Option A has
- Option B has 0
- Option C has 90
- Option D has 95
Since Option A has a majority, Option A is the winner by the ranked-choice method.
-
The Borda scores would be:
- Option A:
- Option B:
- Option C:
- Option D:
Since Option B has a Borda score of 820 points, Option B is the winner by the Borda count method.
-
Step 1:
Determine the number of votes needed to achieve a majority. The number of ballots is
. Fifty percent of 410 is
. So, 206 votes or more is a majority.
Answer the following questions using the table below, which summarizes Imaginarian voter preferences.
| Number of Ballots | 12 | 19 | 27 | 29 | 31 | 21 |
|---|---|---|---|---|---|---|
| Candidate A | 1 | 1 | 2 | 2 | 3 | 3 |
| Candidate B | 2 | 3 | 1 | 3 | 1 | 2 |
| Candidate C | 3 | 2 | 3 | 1 | 2 | 1 |
Use the ranked-choice voting method to determine the winner of the election.
Use the Borda count method to determine the winner of the election.
Compare the results of the two methods. Did the same candidate win? What observations can you make about the results?
The Borda count method may seem too complicated to even consider using for Imaginaria, but each voting method has its own pros and cons. The Borda count method, for example, favors compromise candidates over divisive candidates. A compromise candidate is not the first choice of most of the voters, but is more acceptable to the population as a whole than the other candidates.
In Example 11.8, Candidate A was ranked first by 225 voters, but was ranked last by 185 voters. No voters ranked Candidate A as second or third. It appears that, although Candidate A had the majority of first place votes, there was a significant minority who strongly disliked them. Candidate A was a divisive candidate. Candidate B, on the other hand, was the second choice of every voter, making Candidate B a good compromise. The Borda count method chose Candidate B, a compromise candidate, that was more acceptable to the population as a whole. This scenario is cited by both opponents and proponents of the Borda count method.
Pairwise Comparison and Condorcet Voting
We have discussed two kinds of ranked voting methods so far: ranked-choice and Borda count. A third type of ranked voting is the pairwise comparison method , in which the candidates receive a point for each candidate they would beat in a one-on-one election and half a point for each candidate they would tie. If one candidate earns more points than the others, then that candidate wins. This method is one of several Condorcet voting methods , which are methods in which candidates are ranked and then compared pairwise to each other, a candidate having to beat all others in order to win. These methods vary in the way candidates are scored, and there is not always a clear winner. A candidate who wins each possible pairing is known as a Condorcet candidate . These terms are named after the Marquis de Condorcet, a French philosopher and mathematician who preferred the pairwise comparison method to the Hare method and made public arguments in its favor.
Condorcet voting methods are named for the Marquis de Condorcet, a French philosopher and mathematician known for, among other accomplishments, writing “ Sur l'admission des femmes au droit de Cité ” (“On the Admission of Women to the Rights of Citizenship”), in 1789, the first published essay on the political rights of women.
If you include a Condorcet voting method in the constitution of Imaginaria, the election supervisors may want to use a pairwise comparison matrix like the one in Figure 11.3. It’s a tool used to list the number of wins associated with each pairing of two candidates. Each candidate will receive a point for each win and a half a point for each tie. Each pairing is listed twice, once for the number of wins of a candidate over a particular challenger and once for the number of wins of the challenger over that candidate.
Steps to Determine a Winner by Pairwise Comparison Method Using a Matrix
To determine the winner when the pairwise comparison method is used, we take these three steps:
Step 1: On the matrix, indicate a losing matchup by crossing out a box, , and tie match ups by drawing a slash through the box, .
Step 2: Award each candidate 1 point for a win, half a point for a tie, and 0 points for a loss.
Step 3: Identify the winner, which is the candidate with the most points.
Before you decide on the pairwise comparison method for Imaginaria, review what’s involved in constructing a pairwise comparison matrix from a summary of ranked ballots. Then we can use the matrix to determine the winner of the election. Does the winner using the Borda method still win?
Consider the summary of ranked ballots shown in the table below. Determine the winner of an election using the pairwise comparison method.
| Number of Ballots | 95 | 90 | 110 | 115 |
|---|---|---|---|---|
| Option A | 4 | 4 | 1 | 1 |
| Option B | 2 | 2 | 2 | 2 |
| Option C | 3 | 1 | 3 | 4 |
| Option D | 1 | 3 | 4 | 3 |
- Construct a pairwise comparison matrix for the sample summary of ranked ballots in the table above.
- Use the pairwise comparison method to determine a winner.
- Recall that in Example 11.8, Candidate A won by the ranked-ballot method, and Candidate B won by the Hare method. Did the same candidate win using the pairwise comparison method?
- Is the winner a Condorcet candidate?
- Answer
-
There are four candidates on the ballots. We will need a row and a column for each candidate in addition to the headings, so we will draw a five by five matrix.
Notice that the pairwise vote totals are not used to determine the points. Vote totals are only used to determine a win or a loss. Avoid the common error of adding the values in each row to get the points.
According to Variety magazine, there were 8,469 eligible Oscar voters in 2020. To make our matrix easier to work with, we've rounded this number up to 8,700. We'd never do this in a real voting situation. Suppose that the voter preferences for the ballot for Best Director had been as shown in the table below.
| Number of Ballots | 2,400 | 2,100 | 1,900 | 1,200 | 1,100 |
|---|---|---|---|---|---|
| Martin Scorsese, The Irishman | 3 | 1 | 4 | 5 | 2 |
| Todd Phillips, Joker | 1 | 4 | 5 | 2 | 4 |
| Sam Mendes, 1917 | 5 | 5 | 3 | 4 | 5 |
| Quentin Tarantino, Once Upon a Time in Hollywood | 4 | 2 | 1 | 3 | 3 |
| Bong Joon-ho, Parasite | 2 | 3 | 2 | 1 | 1 |
Refer to the table to answer each question.
Construct a pairwise comparison matrix for the Best Director ballots.
Who is the “Best Director” according to the pairwise comparison method?
Is the winner a Condorcet candidate?
Three Key Questions
Before you decide if you want to use the pairwise comparison method for Imaginarian elections, let’s consider three questions that might affect your decision.
- Is there always a winner?
- If there is a winner, is the winner always a Condorcet candidate?
- If there is a Condorcet candidate, does that candidate always win?
Let’s think about why these questions might be important to you if you chose the pairwise comparison method. First, if no candidate meets the criteria to win an election, you will need a backup plan such as a runoff election. Second, if the winner is not a Condorcet candidate, then there is at least one candidate who beat the winner in a pairwise matchup and the supporters of that candidate might question the validity of the election. Finally, if there is a Condorcet candidate who beat every other candidate in a pairwise matchup, it is reasonable to conclude that it would be unfair for anyone else to win. The rest of the examples in this section should illustrate these key concepts.
Suppose that three people are playing the game Rock, Paper, Scissors. On the count of three, each person shows a hand signal for rock, paper, or scissors. Each hand signal beats another hand signal. The group keeps having a tie because Person A always picks rock, Person B always picks paper which beats rock, and Person C always picks scissors which beats paper, and is beaten by rock! This leads to a disagreement about which choice is best. They decide to use the pairwise comparison method determine the winner. Their preference rankings are given in the following table.
| Voters | A | B | C |
|---|---|---|---|
| Rock (R) | 1 | 3 | 2 |
| Paper (P) | 2 | 1 | 3 |
| Scissors (S) | 3 | 2 | 1 |
- Answer
-
Construct the comparison matrix:
Figure There is a tie! There is no winner.
Example
illustrates the answer to the first key question. The pairwise comparison method does not always result in a winner. For example, much like the game of Rock, Paper, Scissors, it is possible for a cyclic pattern to emerge in which each candidate beats the next until the last candidate who beats the first.A pairwise comparison matrix is given. Determine the winner by the pairwise comparison method. If there is not a winner, explain why. If there is a winner, tell whether the winner is a Condorcet candidate.
Now, you have the answer to the second key question. The pairwise comparison matrix in YOUR TURN 11.10 is an example of a scenario where a winner is not a Condorcet candidate.
The answer to the third question is not as clear. If there is a Condorcet candidate, does that candidate win? So far, we have not come across a contradictory example where the Condorcet candidate didn't win, but we cannot know with certainty that it is not possible by looking at examples. Instead, we will need to use some reasoning. Let’s review some particular cases of elections with a certain number of candidates, and then we will try to generalize the scenario to an election with candidates.
- Suppose there is an election with five candidates—A, B, C, D, and E—and that Candidate C is a Condorcet candidate. How many points did Candidate C win?
- What is the greatest number of points that any one of the other candidates could win?
- Is it possible for Candidate C to lose or tie?
- Answer
-
- In any pairwise election with five candidates, each candidate must compete against four other candidates. It follows that the most points a single candidate can win is four points, which would occur if the candidate won every matchup. As a Condorcet candidate, Candidate C won all the pairwise matchups against Candidates A, B, D, and E, earning four points.
- The rest of the candidates lost to Candidate C. The most points a particular candidate could win if they won matchups with each of the other three candidates is three points.
- Since Candidate C has four points and the rest of the candidates have three points or less, Candidate C is the winner. Therefore, it is not possible for Candidate C to tie or lose.
Suppose there is an election with 26 candidates, A through Z, and that Candidate C is a Condorcet candidate.
How many points did Candidate C win?
What is the greatest number of points that any one of the other candidates could win?
Is it possible for Candidate C to lose or tie?
Let’s consider a general case where there are candidates. One of the candidates is a Condorcet candidate. Since the Condorcet candidate wins all matchups, the Condorcet candidate wins points. Since each of the other candidates lost to the Condorcet candidate, the most a single candidate could win is . Since the Condorcet candidate won points and each other candidate won points or fewer, the Condorcet candidate is the winner. You have your answer to the third key question! If there is a Condorcet candidate, that candidate is always the winner.
Approval Voting
The last type of voting system you will consider for your budding democracy is an approval voting system . In this system, each voter may approve any number of candidates without rank or preference for one over another (among the approved candidates), and the candidate approved by the most voters wins. This voting system has aspects in common with plurality voting and Condorcet voting methods, but it has characteristics that distinguish it from both. An approval voting ballot lists the candidates and provides the option to approve or not approve each candidate.
The term “approval voting” was not used until the 1970s Brams, Steven J.; Fishburn, Peter C. (2007), Approval Voting , Springer-Verlag, p. xv, ISBN 978-0-387-49895-9, although its use has been documented as early as the 13th century (Brams, Steven J. (April 1, 2006). The Normative Turn in Public Choice (PDF) (Speech). Presidential Address to Public Choice Society. New Orleans, Louisiana.) Approval voting has the appeal of being simpler than ranked voting methods. It also allows an individual voter to support more than one candidate equally. This has appeal for those who do not want a split vote among a few mainstream candidates to lead to the election of a fringe candidate. It also has appeal for those who want an underdog to have a chance of success because voters will not worry about wasting their vote on a candidate who is not believed likely to win.
Suppose that Person A/B/C were just about to give up on their game of Rock, Paper, Scissors when they were joined by Person D who reminded them that their updated version, Rock, Paper, Scissors, Lizard, Spock was a far superior game with the added rules that Lizard eats Paper, Paper disproves Spock, Spock vaporizes Rock, Rock crushes Lizard, Lizard poisons Spock, Spock smashes Scissors, and Scissors decapitates Lizard.
Person D encourages their friends to hold a new election. This time, for the sake of simplicity, the group decides to use approval voting to determine the best move in the game. The summary of approval ballots for Rock, Paper, Scissors, Lizard, Spock is given in the table below.
| Voters | A | B | C | D |
|---|---|---|---|---|
| Rock | Yes | No | No | No |
| Paper | No | Yes | No | No |
| Scissors | No | No | Yes | No |
| Lizard | No | No | No | Yes |
| Spock | Yes | Yes | Yes | Yes |
- Answer
-
Count the number of approval votes for each candidate by counting the number of “Yes” votes in each row of the table.
- Rock: 1
- Paper: 1
- Scissors: 1
- Lizard: 1
- Spock: 4
Spock is the winning candidate, approved by four voters!
The Chionilis family is trying to decide on a restaurant again, but now they don’t want to deal with multiple runoffs or even ranking. They will use the approval voting method shown in the following table. Each family member will approve their top two choices. Rainbow China won when multiple runoffs were used. Find the winner for tonight’s dinner.
| Options | A | R | S | T | G | E | M | D |
|---|---|---|---|---|---|---|---|---|
| Rainbow China | Yes | No | No | Yes | No | No | Yes | Yes |
| Dough Boys Pizza | Yes | Yes | Yes | Yes | No | Yes | No | Yes |
| Taco City | No | No | Yes | No | Yes | Yes | No | No |
| Caribbean Flavor | No | Yes | No | No | Yes | No | Yes | No |
The eight members of the Chionilis family—Annette, Rene, Seema, Titus, Galen, Elena, Max and Demitri—have another decision to make. Approval voting worked out nicely the last time. They are going to use it again, but this time, Annette, Rene, Seema, and Galen are feeling a little indecisive. They can't narrow their choice down to two. They will approve their three top choices, but the other family members will only approve two. These choices are reflected in the following table. Determine the restaurant that will be chosen.
| Options | A | R | S | T | G | E | M | D |
|---|---|---|---|---|---|---|---|---|
| Rainbow China | Yes | Yes | Yes | Yes | Yes | No | Yes | Yes |
| Dough Boys Pizza | Yes | Yes | Yes | Yes | No | Yes | No | Yes |
| Taco City | Yes | No | Yes | No | Yes | Yes | No | No |
| Caribbean Flavor | No | Yes | No | No | Yes | No | Yes | No |
- Answer
-
Count the number of approval votes for each restaurant by counting the number of “Yes” votes in each row.
- Rainbow China: 7
- Dough Boys Pizza: 6
- Taco City: 4
- Caribbean Flavor: 3
This time, Rainbow China won!
What would be the outcome of the election if every member of the Chionilis family approved their top three choices from the table below?
| Options | A | R | S | T | G |
|---|---|---|---|---|---|
| Rainbow China | 1 | 3 | 3 | 1 | 3 |
| Dough Boys Pizza | 2 | 2 | 1 | 2 | 1 |
| Taco City | 3 | 4 | 2 | 4 | 2 |
| Caribbean Flavor | 4 | 1 | 4 | 3 | 4 |
Compare and Contrast Voting Methods to Identify Flaws
Wow! We have covered a lot of options for the voting methods. Now, you need to decide which one is best for Imaginaria. Imaginarians might consider characteristics of certain voting systems desirable and others undesirable. In some cases, voters may consider these undesirable traits to be flaws in a voting system that are significant enough to motivate them to reject that system. If you are feeling a bit overwhelmed by this decision, maybe it would help to read about the experiences of others who have faced similar questions.
Consider the 2000 U.S. presidential election in which Green Party candidate Ralph Nader and Reform Party candidate Pat Buchanan were on the ballet running against the mainstream candidates, Democrat Al Gore and Republican George W. Bush. The voting results for Florida are given in Table 11.3.
| Candidate | Party | Votes | Percentage |
|---|---|---|---|
| (G) George W. Bush | Republican | 2,912,790 | 48.85% |
| (A) Al Gore | Democrat | 2,912,253 | 48.84% |
| (R) Ralph Nader | Green | 97,488 | 1.63% |
| (P) Pat Buchanan | Reform | 17,484 | 0.29% |
| (H) Harry Brown | Libertarian | 16,415 | 0.28% |
| (O) 7 Other Candidates | Other | 6,680 | 0.11% |
| Total | 5,963,110 |
In more than one state, Buchanan was able to split the Republican vote enough to allow Gore to win that state. Nader split the Democrat vote in Florida and New Hampshire by enough votes to prevent Gore from winning those states. Had Gore won either state, he would have had enough electoral votes to win the election. Instead, Bush won. This is an example of a flaw in the plurality system of voting: the spoiler.
A spoiler is a less popular candidate who takes votes from a more popular candidate with similar positions, swinging the race to another candidate with vastly different views that they would not support. This encourages voters not to vote for the candidate that they perceive to be the best, but instead for the candidate they can live with who they perceive to have a better chance of winning. Some voters may prefer a method such as approval voting, which does not have this trait in common with plurality voting.
Because the vote counts for George W. Bush and Al Gore differed by only 537 votes, many Democrats blamed Ralph Nader and the Green Party for their loss. Let’s consider how the election results might have differed if the approval voting method had been used.
Use Table 11.3 and the following assumptions to extrapolate the results of an approval method election:
- 100 percent of Pat Buchanan supporters would approve George W. Bush.
- 100 percent of Ralph Nader supporters would approve Al Gore.
- 72 percent of Libertarians would approve George W. Bush.
- 28 percent of Libertarians would approve Al Gore (as was roughly the known percentage at the time according to the Cato Institute).
- 50 percent of the supporters of other candidates would approve George Bush while 50 percent would approve Al Gore.
- Answer
-
Step 1: Create a summary of approval ballots based on the given assumptions. For the Libertarian candidate, 72 percent of 16,415 of the votes is and 28 percent is . For the other candidates, 50 percent of the votes is .
Number of Votes 2,912,790
(G)2,912,253
(A)97,488
(R)17,484
(B)11,819
(72% H)4,596
(28% H)3,340
(50% O)3,340
(50% O)(G) George W. Bush Yes No No Yes Yes No No Yes (A) Al Gore No Yes Yes No No Yes Yes No (R) Ralph Nader No No Yes No No No No No (P) Pat Buchanan No No No Yes No No No No (H) Harry Brown No No No No Yes Yes No No (O) 7 Other Candidates No No No No No No Yes Yes Step 2: Count the number of approval votes for each candidate.
- George W. Bush:
- Al Gore:
- Ralph Nader:
- Pat Buchanan:
- Harry Brown:
- Other Candidates:
In this scenario, Al Gore is the winner.
Extrapolate the results of an approval method election using Table 11.3, the assumptions from Example 11.14, and the additional assumptions that the supporters of Al Gore would all approve Ralph Nader and that half of the supporters of George Bush would approve Pat Buchanan while half would approve Harry Brown.
The results in Example 11.14 and Your Turn 11.14 highlight one of the characteristics of approval voting. Ralph Nader moved up from a distant third place finish to a close second place finish when Al Gore’s supporters approved him on their ballots. In this way, fringe candidates have a better chance of winning, which some voters consider a flaw but others consider a benefit.
Another aspect of approval voting systems that is a concern to many voters is that candidates in approval elections might encourage their loyal supporters to approve them and only them to avoid giving support to any other candidate. If this occurred, the election in effect becomes a traditional plurality election. This is a flaw that cannot occur in an instant runoff system since all candidates are ranked.
In the future, humans have explored distant solar systems and found three habitable planets which could be colonized. Since it will take all available resources to colonize one planet, humans must agree on the planet. Planet A has the most comfortable climate and most plentiful resources, but it is the farthest from Earth making travel to the planet a challenge. Planet B is half the distance but will require more resources to make comfortable. Planet C is the least suitable of the three and terraforming will be required, but it is close enough to make travel between Earth and Planet C possible on a more regular basis. The table below provides the voter preferences for the colonization of each planet.
| Percentage of Voters | 45% | 15% | 40% |
|---|---|---|---|
| Planet A | 1 | 3 | 3 |
| Planet B | 2 | 1 | 2 |
| Planet C | 3 | 2 | 1 |
If the entire population were able to vote, determine the winning planet using each of the methods listed below.
- Plurality
- Ranked-choice method
- Borda count
- Answer
-
- The plurality method only considers the top choice of each voter. By this system, Planet A has 45 percent of the vote, Planet B has 15 percent of the vote, and Planet C has 40 percent of the vote. Planet A wins.
- Using either instant runoff or a two-round system, Planet B with only 15 percent of the vote will be eliminated in the first round. In Round 2, the 15 percent that voted for Planet B would vote for their second choice, Planet C. This leaves Planet A with 45 percent and Planet C with 55 percent. Planet C has a majority and wins the election.
-
To find the Borda score for each candidate, imagine there are exactly 100 voters. Then the summary of ranked ballots looks like:
Out of 100 Voters 45 15 40 Planet A 1 3 3 Planet B 2 1 2 Planet C 3 2 1
The Borda score for each candidate is as follows:
Planet A:
Planet B:
Planet C:
Planet B wins.
The juniors at a high school in Central Florida are voting for a theme park to visit for an end of the year field trip. The options are Disney’s four theme parks: Animal Kingdom, Magic Kingdom, EPCOT Center, or Hollywood Studios.
| Percentage of Voters | 35% | 25% | 30% |
|---|---|---|---|
| Animal Kingdom | 1 | 3 | 4 |
| Magic Kingdom | 2 | 1 | 3 |
| Epcot Center | 3 | 4 | 1 |
| Hollywood Studios | 4 | 2 | 2 |
Determine the outcome of the vote by each of the given voting methods.
- Plurality
- Ranked-choice method
- Borda count
The election in Example
involves a scenario in which there are two extreme candidates, Planet A and Planet B, and a moderate candidate, Planet C. The supporters of the extreme candidates prefer the moderate candidate to the other extremist ones. This makes Planet C a compromise candidate. In this case, both the plurality method and ranked-choice voting resulted in the election of one of the extreme candidates, but the Borda count method elected the compromise candidate in this scenario. Depending on a person’s perspective, this may be perceived as a flaw in either ranked-choice and plurality systems, or the Borda count method.In Fairness in Voting Methods, we will analyze the fairness of each voting system in greater detail using objective measures of fairness.
It is possible to create Excel spreadsheets that complete the calculations necessary to determine the winner of an election by various voting methods. In some cases, this work has already been done and posted online. As you practice applying the various voting methods that could be used in Imaginaria, quick Internet search will lead to sites such as Ms. Hearn Math with free specialty calculators.
These sites can be a great way to check your results!
Check Your Understanding
- Name three voting methods that use a ranked ballot.
- Determine whether the following statement is true or false: The same ranked ballots may result in a different winner depending on which voting method is used.
- Determine whether the following statement is true or false. A majority candidate is always a Condorcet candidate.
- The _______________ method is a system of voting using ranked ballots in which each candidate is awarded points corresponding to the number of candidates ranked lower on each ballot.
- The ________________ method is a system of voting using ranked ballots (or multiple elections) in which each candidate receives a point for each candidate they would beat in a one-on-one election and half a point for each candidate they would tie.
- The __________________ method is a runoff voting system in which only the candidate(s) with the very least votes are eliminated.
- Explain the differences between two-round voting and ranked-choice voting.


39.9: Breathing - The Mechanics of Human Breathing
- Describe how the structures of the lungs and thoracic cavity control the mechanics of breathing
The Mechanics of Human Breathing
The relationship between gas pressure and volume helps to explain the mechanics of breathing. Boyle’s Law is the gas law which states that in a closed space, pressure and volume are inversely related. As volume decreases, pressure increases and vice versa. When discussing the detailed mechanics of breathing, it is important to keep this inverse relationship in mind.
Inhalation and Exhalation
The thoracic cavity, or chest cavity, always has a slight, negative pressure which aids in keeping the airways of the lungs open. During the process of inhalation, the lung volume expands as a result of the contraction of the diaphragm and intercostal muscles (the muscles that are connected to the rib cage), thus expanding the thoracic cavity. Due to this increase in volume, the pressure is decreased, based on the principles of Boyle’s Law. This decrease of pressure in the thoracic cavity relative to the environment makes the cavity pressure less than the atmospheric pressure. This pressure gradient between the atmosphere and the thoracic cavity allows air to rush into the lungs; inhalation occurs. The resulting increase in volume is largely attributed to an increase in alveolar space because the bronchioles and bronchi are stiff structures that do not change in size.
During this process, the chest wall expands out and away from the lungs. The lungs are elastic; therefore, when air fills the lungs, the elastic recoil within the tissues of the lung exerts pressure back toward the interior of the lungs. These outward and inward forces compete to inflate and deflate the lung with every breath. Upon exhalation, the lungs recoil to force the air out of the lungs. The intercostal muscles relax, returning the chest wall to its original position. During exhalation, the diaphragm also relaxes, moving higher into the thoracic cavity. This increases the pressure within the thoracic cavity relative to the environment. Air rushes out of the lungs due to the pressure gradient between the thoracic cavity and the atmosphere. This movement of air out of the lungs is classified as a passive event since there are no muscles contracting to expel the air.
Protection of the Lung
Each lung is surrounded by an invaginated sac. The layer of tissue that covers the lung and dips into spaces is called the visceral pleura. A second layer of parietal pleura lines the interior of the thorax. The space between these layers, the intrapleural space, contains a small amount of fluid that protects the tissue by reducing the friction generated from rubbing the tissue layers together as the lungs contract and relax. If these layers of tissues become inflamed, this is categorized as pleurisy: a painful inflammation that increases the pressure within the thoracic cavity, reducing the volume of the lung.
Key Points
- The mechanics of breathing follow Boyle’s Law which states that pressure and volume have an inverse relationship.
- The process of inhalation occurs due to an increase in the lung volume (diaphragm contraction and chest wall expansion) which results in a decrease in lung pressure in comparison to the atmosphere; thus, air rushes in the airway.
- The process of exhalation occurs due to an elastic recoil of the lung tissue which causes a decrease in volume, resulting in increased pressure in comparison to the atmosphere; thus, air rushes out of the airway.
- There is no contraction of muscles during exhalation; it is considered a passive process.
- The lung is protected by layers of tissue referred to as the visceral pleura and parietal pleura; the intrapleural space contains a small amount of fluid that protects the tissue by reducing friction.
Key Terms
- visceral pleura : the portion of protective tissue that is attached directly to the lungs
- parietal pleura : the portion of the protective tissue that lines the inner surface of the chest wall and covers the diaphragm


31 .
Describe one of the ways you can engage your audience in your presentation.
32 .
Describe why it is important to have a strong closing presentation.
33 .
Why would you apply a theme to a presentation, and why might you want to modify it?
34 .
Discuss reasons why the options on the Review tab are useful.
35 .
Discuss how speaker’s notes are added to a slide. How can they be used in a presentation? What information would you include in the speaker’s notes?
36 .
What are some things you need to address regarding the hardware and environmental components of your presentation?
37 .
When preparing for a virtual/hybrid presentation versus a fully in-person presentation, what are some items to consider?
38 .
Discuss how to add transitions to all slides in a Google Slides file and how to change the timing of the transitions.
39 .
Discuss how to use the Q&A feature in a presentation.
40 .
Explain how to insert audio into a presentation.


5.5: Exercises
Exercise \(\PageIndex{1}\)
In Section 5.2, we encountered the complex frequencies \[\omega_\pm = -i\gamma \pm \sqrt{\omega_0^2 - \gamma^2}.\] For fixed \(\omega_0\) and \(\omega_0 > \gamma\) (under-damping), prove that \(\omega_\pm\) lie along a circular arc in the complex plane.
Exercise \(\PageIndex{2}\)
Derive the general solution for the critically damped harmonic oscillator, Eq. (5.3.16), by following these steps:
-
Consider the complex ODE, in the under-damped regime
\(\omega_0 > \gamma\)
. We saw in Section 5.3 that the general solution has the form \[z(t) = \psi_+ \, \exp\left[\left(-\gamma - i \sqrt{\omega_0^2 - \gamma^2}\right)t\right] \; +\; \psi_- \, \exp\left[\left(-\gamma +i\sqrt{\omega_0^2 - \gamma^2}\right)t\right]\] for some complex parameters
\(\psi_+\)
and
\(\psi_-\)
. Define the positive parameter
\(\varepsilon = \sqrt{\omega_0^2 - \gamma^2}\)
. Re-write
\(z(t)\)
in terms of
\(\gamma\)
and
\(\varepsilon\)
(i.e., eliminating
\(\omega_0\)
).
-
The expression for
\(z(t)\)
is presently parameterized by the independent parameters
\(\psi_+\)
,
\(\psi_-\)
,
\(\varepsilon\)
, and
\(\gamma\)
. We are free to re-define the parameters, by taking \[\begin{align} \alpha &= \psi_+ + \psi_- \\ \beta &= -i\varepsilon(\psi_+ - \psi_-). \end{align}\] Using these equations, express
\(z(t)\)
using a new set of independent complex parameters, one of which is
\(\varepsilon\)
. Explicitly identify the other independent parameters, and state whether they are real or complex.
-
Expand the exponentials in
\(z(t)\)
in terms of the parameter
\(\varepsilon\)
. Then show that in the limit
\(\varepsilon \rightarrow 0\)
,
\(z(t)\)
reduces to the critically-damped general solution (5.3.16).
Exercise \(\PageIndex{3}\)
Repeat the above derivation for the critically-damped solution, but starting from the over-damped regime \(\gamma > \omega_0\) .
Exercise \(\PageIndex{4}\)
Let \(z(t)\) be a complex function of a real input \(t\) , which obeys the differential equation \[\frac{dz}{dt} = -i\,(\omega_1 - i \gamma)\; z(t),\] where \(\omega_1\) and \(\gamma\) are real. Find the general solution for \(z(t)\) , and hence show that \(z(t)\) satisfies the damped oscillator equation \[\left[\frac{d^2}{dt^2} + 2\gamma \frac{d}{dt} + \omega_0^2 \right] z(t) = 0\] for some \(\omega_0^2\) . Finally, show that this harmonic oscillator is always under-damped.
- Answer
-
The general solution is \[z(t) = A \exp\left[-i(\omega_1 - i \gamma) t\right].\] It can be verified by direct substitution that this is a solution to the differential equation. It contains one free parameter, and the differential equation is first-order, so it must be a general solution. Next, \[\begin{align} \frac{d^2z}{dt^2} + 2 \gamma \frac{dz}{dt} &= (-i)^2(\omega_1 - i\gamma)^2 z(t) - 2i \gamma (\omega_1 - i \gamma) z(t) \\ &= \left[- \omega_1^2 + \gamma^2 + 2i\gamma\omega_1 - 2i \gamma \omega_1 - 2\gamma^2)\right] z(t) \\ &= -\left(\omega_1^2 + \gamma^2\right)z(t).\end{align}\] Hence, \(z(t)\) obeys a damped harmonic oscillator equation with \(\omega_0^2 = \omega_1^2 + \gamma^2.\) This expression for the natural frequency ensures that \(\omega_0^2 > \gamma^2\) (assuming the parameters \(\gamma\) and \(\omega_1\) are both real); hence, the harmonic oscillator is always under-damped.


Phyllis Nissila
“Books” by Hermann is in the Public Domain, CC0
Most students entering college have not yet dealt with the level of difficulty involved in reading–and comprehending–scholarly textbooks and articles. The challenge may even surprise some who have pretty good reading and comprehension skills so far. Other students for whom reading has mostly consisted of social media, texts, forum chat rooms, and emails, find they are intimidated by the sheer
amount
of reading there is in college classes.
Reading Comprehension Definition
Reading comprehension is defined as the level of understanding of a message. This understanding comes from the interaction between the words that are written and how they trigger knowledge outside the text/message. Comprehension is a “creative, multifaceted process” dependent upon four language skills: phonology, syntax, semantics, and pragmatics. Proficient reading depends on the ability to recognize words quickly and effortlessly. It is also determined by an individual’s cognitive development, which is “the construction of thought processes.” Some people learn through education or instruction and others through direct experiences.
There are specific traits that determine how successfully an individual will comprehend text, including prior knowledge about the subject, well-developed language, and the ability to make inferences. Having the skill to monitor comprehension is a factor: “Why is this important?” and “Do I need to read the entire text?” are examples. Another trait is the ability to be self-correcting, which allows for solutions to comprehension challenges.
 Deep processing involves semantic processing, which happens when we encode the meaning of a word and relate it to similar words. Shallow processing involves structural and phonemic recognition, the processing of sentence and word structure and their associated sounds. This theory was first identified by Fergus I. M. Craik and Robert S. Lockhart.
Brain Region Activation
Comprehension levels can now be observed through the use of a fMRI, functional magnetic resonance imaging. fMRIs’ are used to determine the specific neural pathways of activation across two conditions, narrative-level comprehension and sentence-level comprehension. Images showed that there was less brain region activation during sentence-level comprehension, suggesting a shared reliance with comprehension pathways. The scans also showed an enhanced temporal activation during narrative levels tests indicating this approach activates situation and spatial processing.
History
Initially most comprehension teaching was based on imparting selected techniques that when taken together would allow students to be strategic readers. However, in 40 years of testing these methods never seemed to win support in empirical research. One such strategy for improving reading comprehension is the technique called SQ3R: Survey, Question, Read, Recite, and Review, that was introduced by Francis Pleasant Robinson in his 1946 book
Effective Study
.
Between 1969 and to about 2000 a number of “strategies” were devised for teaching students to employ self-guided methods for improving reading comprehension. In 1969 Anthony Manzo designed and found empirical support for the ReQuest, or Reciprocal Questioning Procedure, it was the first method to convert emerging theories of social and imitation learning into teaching methods through the use of a talk rotation between students and teacher called cognitive modeling.
Since the turn of the 21st century, comprehension lessons usually consist of students answering teachers’ questions, writing responses to questions on their own, or both. The whole group version of this practice also often included “Round-robin reading”, wherein teachers called on individual students to read a portion of the text. In the last quarter of the 20th century, evidence accumulated that the read-test methods were more successful assessing rather than teaching comprehension. Instead of using the prior read-test method, research studies have concluded that there are much more effective ways to teach comprehension. Much work has been done in the area of teaching novice readers a bank of “reading strategies,” or tools to interpret and analyze text.
Instruction in comprehension strategy use often involves the gradual release of responsibility, wherein teachers initially explain and model strategies. Over time, they give students more and more responsibility for using the strategies until they can use them independently. This technique is generally associated with the idea of self-regulation and reflects social cognitive theory, originally conceptualized by Albert Bandura.
Vocabulary
Reading comprehension and vocabulary are inextricably linked. The ability to decode or identify and pronounce words is self-evidently important, but knowing what the words mean has a major and direct effect on knowing what any specific passage means. Students with a smaller vocabulary than other students comprehend less of what they read and it has been suggested that the most impactful way to improve comprehension is to improve vocabulary.
Most words are learned gradually through a wide variety of environments: television, books, and conversations. Some words are more complex and difficult to learn, such as homonyms, words that have multiple meanings and those with figurative meanings, like idioms, similes, and metaphors.
Three Tier Vocabulary Words
Several theories of vocabulary instruction exist, namely, one focused on intensive instruction of a few high value words, one focused on broad instruction of many useful words, and a third focused on strategies for learning low frequency, context specific vocabulary.
Broad Vocabulary Approach
The method of focusing of broad instruction on many words was developed by Andrew Biemiller who argued that more words would benefit students more, even if the instruction was short and teacher-directed. He suggested that teachers teach a large number of words before reading a book to students, by merely giving short definitions, such as synonyms, and then pointing out the words and their meaning while reading the book to students. The method contrasts with the approach by emphasizing quantity versus quality. There is no evidence to suggest the primacy of either approach.
Morphemic Instruction
Another vocabulary technique, strategies for learning new words, can be further subdivided into instruction on using context and instruction on using morphemes, or meaningful units within words to learn their meaning. Morphemic instruction has been shown to produce positive outcomes for students reading and vocabulary knowledge, but context has proved unreliable as a strategy and it is no longer considered a useful strategy to teach students. This conclusion does not disqualify the value in “learning” morphemic analysis – prefixes, suffixes and roots – but rather suggests that it be imparted incidentally and in context. Accordingly, there are methods designed to achieve this, such as Incidental Morpheme Analysis.
Reciprocal Teaching
In the 1980s Annemarie Sullivan Palincsar and Ann L. Brown developed a technique called reciprocal teaching that taught students to predict, summarize, clarify, and ask questions for sections of a text. The use of strategies like summarizing after each paragraph have come to be seen as effective strategies for building students’ comprehension. The idea is that students will develop stronger reading comprehension skills on their own if the teacher gives them explicit mental tools for unpacking text.
Instructional Conversations
“Instructional conversations”, or comprehension through discussion, create higher-level thinking opportunities for students by promoting critical and aesthetic thinking about the text. There are several types of questions that a teacher should focus on: remembering; testing understanding; application or solving; invite synthesis or creating; and evaluation and judging. Teachers should model these types of questions through “think-alouds” before, during, and after reading a text. When a student can relate a passage to an experience, another book, or other facts about the world, they are “making a connection.” Making connections help students understand the author’s purpose and fiction or non-fiction story.
Text Factors
There are factors, that once discerned, make it easier for the reader to understand the written text. One is the genre, like folktales, historical fiction, biographies or poetry. Each genre has its own characteristics for text structure, that once understood help the reader comprehend it. A story is composed of a plot, characters, setting, point of view, and theme. Informational books provide real world knowledge for students and have unique features such as: headings, maps, vocabulary, and an index. Poems are written in different forms and the most commonly used are: rhymed verse, haiku, free verse, and narratives. Poetry uses devices such as: alliteration, repetition, rhyme, metaphors, and similes. “When children are familiar with genres, organizational patterns, and text features in books they’re reading, they’re better able to create those text factors in their own writing.”
The Reading Apprenticeship (RA) Approach to Comprehension
Now to some strategies to help you with some typical college-level comprehension challenges as well as some of your specific challenges identified in the previous exercise.
This lesson focuses on a method called Reading Apprenticeship. It is based on the premise that people who have become expert readers can assist learners by modeling what they have learned to do. As explained in the text,
Reading for Understanding, How Reading Apprenticeship Improves Disciplinary Learning in Secondary and College Classrooms,
“One literacy educator describes the idea of the cognitive apprenticeship in reading by comparing the process of learning to read with that of learning to ride a bike. In both cases, a more proficient other is present to support the beginner, engaging the beginner in the activity and calling attention to often overlooked or hidden strategies.”
This is a strategy that takes a metacognitive approach to comprehension, utilizing various strategies readers may already know they know how to do, then adding more. For example, most readers have learned to make predictions, ask questions concerning meanings (“I wonder about…”), visualize a scene being described, associate the material being read to some other material, and, at the end, summarize the material.
I like to call these our “hard drive” skills. Like a computer hard drive always humming in the background doing its thing behind the scenes, our metacognitive skills have already been assisting us as readers. We just don’t usually talk about what we are doing, for example, “Well, by golly, now I’m predicting what Godzilla will do to the poor villagers in about two scenes from now,” we just automatically predict, especially if we are familiar with characters and plot lines. For another example, “I am now going to visualize this scene in graveyard when Hamlet comes across the deceased court jester’s skull in Act V, Scene 1.” We just
see
it in our mind’s eye.
Now review and affirm important comprehension skills you already possess and complete the exercise below.
EXERCISE 22-1
Go back through the excerpt, above, on reading comprehension and THIS time, write marginal notes where you used any of the comprehension tools listed below:
-
predicting
-
asking questions of the material such as, “I wonder about,” “Could this mean?”
-
visualizing
-
connecting this material to something else you have learned
-
noting where you think you might need to read something over again for comprehension
-
summarizing
Licenses and Attributions:
CC licensed content, previously shared:
How to Learn Like a Pro! Authored by Phyllis Nissila. Located at: openoregon.pressbooks.pub/co...level-reading/ License: CC-BY Attribution.
Adaptions: Changed formatting, removed one exercise, removed some hyperlinks, removed two class discussion sentences and related footnotes.
“Books” image by Hermann is in the Public Domain, CC0