6.10: Comments and solutions
Note: It is important to separate the underlying idea of “induction” from the formal way we have chosen to present proofs. As ever in mathematics, the ideas are what matter most. But the process of struggling with (and slowly coming to understand why we need) the formal structure behind the written proofs is part of the way the ideas are tamed and organised.
Readers should not be intimidated by the physical extent of the solutions to this chapter. As explained in the main text it is important for all readers to review the way they approach induction proofs: so we have erred in favour of completeness – knowing that as each reader becomes more confident, s/he will increasingly compress, or abbreviate, some of the steps.
(a) Yes.
(b) Yes.
(c) No.
and
229
.
Note:
The statement in the problem includes the
quantifier
“
for all
What is to be proved is the compound statement
“
is true for all P ( n ) ”. n ≥ 1
In contrast, each individual statement
It is essential to be clear when you are dealing with the
compound
statement, and when you are referring to some
particular
statement
Let
“
is divisible by 5 2 n + 2 − 24 n − 25 ”. 576
•
“
is divisible by 5 4 − 24 × 1 − 25 ”. 576
That is:
“
is divisible by 625 − 49 = 576 ”, 576
which is evidently true.
-
Now suppose that we know
is true for someP ( k ) . We must show thatk ≥ 1 is then also true.P ( k + 1 )
To prove that
To prove that
“
is divisible by 5 2 ( k + 1 ) + 2 − 24 ( k + 1 ) − 25 ”. 576
So we must start with
The first term on the RHS is a multiple of
Hence
-
is true; andP ( 1 ) -
whenever
is true for someP ( k ) , we have proved thatk ≥ 1 must be true.P ( k + 1 )
230
.
Let
“the angles of any p -gon, for any value of p with
, have sum exactly 3 ≤ p ≤ n radians”. ( p − 2 ) π
-
is the statement:P ( 3 ) “the angles of any triangle have sum
radians”.π This is a known fact: given triangle ΔABC , draw the line
X A Y through A parallel to , with X on the same side ofB C as B and Y on the same side ofA C as C . ThenA B and∠ X A B = ∠ C B A (alternate angles), so∠ Y A C = ∠ B C A ∠ B + ∠ A + ∠ C = ∠ X A B + ∠ A + ∠ Y A C = ∠ X A Y = π . -
Now we suppose that
is known to be true for someP ( k ) . We must show thatk ≥ 3 is then necessarily true.P ( k + 1 )
To prove that
“the angles of a p -gon, for any value of p with
, have sum exactly 3 ≤ p ≤ k + 1 radians” ( p − 2 ) π
This can be reworded by splitting it into two parts:
“the angles of any p -gon for
have sum exactly 3 ≤ p ≤ k radians;” ( p − 2 ) π
and
“the angles of any (k + 1) -gon have sum exactly
radians”. ( ( k + 1 ) − 2 ) π
The first part of this revised version is precisely the statement
Let
[
Note:
The usual move at this point is to say “draw the chord
-
if the triangle
A k A 1 “sticks out” rather than in, andA 0
So what is usually presented as a “proof” does not work in general .
If we want to prove the general result – for polygons of all shapes – we have to get round this unwarranted assumption. Experiment may persuade you that “there is always some vertex that sticks out and which can be safely “cut off”; but it is not at all clear how to prove this fact (we know of no simple proof). So we have to find another way.]
Consider the vertex
Imagine each half-line, which starts at
Consider the locus of all such points
X
as the half line swings from
(a) these points X all belong to a single side of the polygon; or
(b) they don't.
(a) In the first case none of the vertices or sides of the polygon
(b) In the second case, as the half-line
Because of the way the point
X
was chosen, the chord
Hence
231
.
Let
-
LHS of
; RHS ofP ( 1 ) = 1 . Since these two are equal,P ( 1 ) = 1 2 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 .1 + 3 + 5 + ⋯ + ( 2 k − 1 ) = k 2 We wish to prove that
must then be true.P ( k + 1 ) Now
is an equation, so we start with the LHS ofP ( k + 1 ) and try to simplify it in an appropriate way to get the RHS ofP ( k + 1 ) :P ( k + 1 ) LHS of P ( k + 1 ) = 1 + 3 + 5 + … + ( 2 ( k + 1 ) - 1 ) = ( 1 + 3 + 5 + … + ( 2 k - 1 ) ) + ( 2 k + 1 ) . If we now use
, which we are supposing to be true, then the first bracket is equal to k 2 , so this sum is equal to:P ( k ) = k 2 + ( 2 k + 1 ) = ( k + 1 ) 2 = RHS of P ( k + 1 ) . Hence
holds.P ( k + 1 )
Combining these two bullet points then shows that “
(a) The only way to learn is by trying and failing; then trying again and failing slightly better! So don't give up too quickly. It is natural to try to relate each term to the one before. You may then notice that each term is slightly less than 3 times the previous term.
(b)
Note:
The recurrence relation for
Let
“
for all m , u m = 2 m + 3 m ”. 0 ≤ m ≤ n
-
LHS of
; RHS ofP ( 0 ) = u 0 = 2 . Since these two are equal,P ( 0 ) = 2 0 + 3 0 = 1 + 1 is true.P ( 0 ) combinesP ( 1 ) , and the equality ofP ( 0 ) andu 1 = 5 ; since these two are equal,2 1 + 3 1 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 “
for all m ,u m = 2 m + 3 m .”0 ≤ m ≤ k We wish to prove that
must then be true.P ( k + 1 ) Now
requires us to prove thatP ( k + 1 ) “
for all m ,u m = 2 m + 3 m .”0 ≤ m ≤ k + 1 Most of this is guaranteed by
, which we assume to be true. It remains for us to check that the equality holds forP ( k ) . We know thatu k + 1 u k + 1 = 5 u k − 6 u k − 1 . And we may use
), which we are supposing to be true, to conclude that:P ( k ) u k + 1 = 5 ( 2 k + 3 k ) - 6 ( 2 k - 1 + 3 k - 1 ) = ( 10 - 6 ) 2 k - 1 + ( 15 - 6 ) 3 k - 1 = 2 k + 1 + 3 k + 1 . Hence
holds.P ( k + 1 )
Combining these two bullet points then shows that “
233
.
Let
“
for all m , F m = α m − β m 5 ”, 0 ≤ m ≤ n
where
-
LHS of
; RHS ofP ( 0 ) = F 0 = 0 . Since these two are equal,P ( 0 ) = 1 − 1 5 = 0 is true.P ( 0 ) LHS of
; RHS ofP ( 1 ) = F 1 = 1 . Since these two are equal,P ( 1 ) = α − β 5 = 1 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 “
for all m ,F m = α m − β m 5 .”0 ≤ m ≤ k We wish to prove that
must then be true.P ( k + 1 ) Now
requires us to prove thatP ( k + 1 ) “
for all m ,F m = α m − β m 5 .”0 ≤ m ≤ k + 1 Most of this is guaranteed by
, which we assume to be true. It remains to check this forP ( k ) . We know thatF k + 1 F k + 1 = F k + F k − 1 . And we may use
, which we are supposing to be true to conclude that:P ( k ) F k + 1 = α k - β k 5 + α k - 1 - β k - 1 5 = α k + α k - 1 5 - β k + β k - 1 5 = α k + 1 - β k + 1 5 (since
andα are roots of the equationβ ) Hencex 2 − x − 1 = 0 holds.P ( k + 1 )
Combining these two bullet points then shows that “
Note: You may understand the above solution and yet wonder how such a formula could be discovered. The answer is fairly simple. There is a general theory about linear recurrence relations which guarantees that the set of all solutions of a second order recurrence (that is, a recurrence in which each term depends on the two previous terms) is “two dimensional” (that is, it is just like the familiar 2D plane, where every vector (p, q) is a combination of the two “base vectors” (1, 0) and (0, 1) :
Once we know this, it remains:
-
to find two
special
solutions (like the vectors
(1, 0)
and
(0, 1)
in the plane), which we do here by looking for sequences of the form “
” that satisfy the recurrence, which implies that1 , x , x 2 , x 3 , … , so1 + x = x 2 , orx = α ;x = β -
then to choose a linear combination
F m = p α m + q of these two power solutions to give the correct first two terms:β m ,0 = F 0 = p + q , so1 = F 1 = p α + q β ,p = 1 5 .q = − 1 5
234
. Let
“
is divisible by 5 2 n + 1 · 2 n + 2 + 3 n + 2 · 2 2 n + 1 ”. 19
-
is the statement: “P ( 0 ) is divisible by 19”, which is true.5 × 4 + 9 × 2 -
Now suppose that we know that
is true for someP ( k ) . We must show thatk ≥ 0 is then also true.P ( k + 1 ) -
To prove that
is true, we have to show thatP ( k + 1 ) “
is divisible by5 2 k + 3 · 2 k + 3 + 3 k + 3 · 2 2 k + 3 ”.19 5 2 k + 3 ⋅ 2 k + 3 + 3 k + 3 ⋅ 2 2 k + 3 = 5 2 ⋅ 2 ( 5 2 k + 1 ⋅ 2 k + 2 + 3 k + 2 ⋅ 2 2 k + 1 ) - 5 2 ⋅ 2 ⋅ 3 k + 2 ⋅ 2 2 k + 1 + 3 k + 3 ⋅ 2 2 k + 3 = 5 2 ⋅ 2 ⋅ ( 5 2 k + 1 ⋅ 2 k + 2 + 3 k + 2 ⋅ 2 2 k + 1 ) - ( 5 2 - 3 ⋅ 2 ) 3 k + 2 ⋅ 2 2 k + 2 . The bracket in the first term on the RHS is divisible by 19 (by
), and the bracket in the second term is equal to 19. Hence both terms on the RHS are divisible by 19, so the RHS is divisible by 19. Therefore the LHS is also divisible by 19, soP ( k ) is true.P ( k + 1 )
235 .
Note: The proofs of identities such as those in Problem 235 ., which are given in many introductory texts, ignore the lessons of the previous two problems. In particular,
• they often fail to distinguish between
– the single statement
– the “quantified” result to be proved (“for all
and
• they proceed in the `wrong' direction (e.g. starting with the identity
This latter strategy is psychologically and didactically misleading – even though it can be justified logically when proving very simple identities. In these very simple cases, each statement
Readers should try to write each proof in the intended spirit, and to learn from the solutions – since this style has been chosen to highlight what mathematical induction is really about, and it is this approach that is needed in serious applications.
(a) Let
“
”. 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2
-
LHS of
; RHS ofP ( 1 ) = 1 . Since these two are equal,P ( 1 ) = 1 · ( 1 + 1 ) 2 = 1 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 “
”.1 + 2 + 3 + ⋯ + k = k ( k + 1 ) 2 We wish to prove that
must then be true.P ( k + 1 ) Now
is an equation, so we start with the LHS ofP ( k + 1 ) and try to simplify it in an appropriate way to get the RHS ofP ( k + 1 ) ):P ( k + 1 ) LHS of P ( k + 1 ) = 1 + 2 + 3 + ⋯ + k + ( k + 1 ) = ( 1 + 2 + 3 + ⋯ + k ) + ( k + 1 ) . If we now use
, which we are supposing to be true, then the first bracket is equal toP ( k ) , so the complete sum is equal to:k ( k + 1 ) 2 = k ( k + 1 ) 2 + ( k + 1 ) = ( k + 1 ) ( k + 2 ) 2 = RHS of P ( k + 1 ) . Hence
holds.P ( k + 1 ) If we combine these two bullet points, then we have proved that “
holds for allP ( n ) ”.n ≥ 1
(b) Let
“
”. 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + ⋯ + 1 n · ( n + 1 ) = 1 − 1 n + 1
-
LHS of
; RHS ofP ( 1 ) = 1 1 · 2 = 1 2 . Since these two are equal,P ( 1 ) = 1 − 1 2 = 1 2 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 “
”.1 1 · 2 + 1 2 · 3 + 1 3 · 4 + ⋯ + 1 k · ( k + 1 ) = 1 − 1 k + 1 We wish to prove that
must then be true.P ( k + 1 ) Now
is an equation, so we start with the LHS ofP ( k + 1 ) and try to simplify it in an appropriate way to get the RHS ofP ( k + 1 ) :P ( k + 1 ) LHS of P ( k + 1 ) = 1 1 ⋅ 2 + 1 2 ⋅ 3 + 1 3 ⋅ 4 + ⋯ + 1 ( k + 1 ) ( k + 2 ) = [ 1 1 ⋅ 2 + 1 2 ⋅ 3 + 1 3 ⋅ 4 + 1 k ( k + 1 ) ] + 1 ( k + 1 ) ( k + 2 ) . If we now use
, which we assume is true, then the first bracket is equal toP ( k ) , so the complete sum is equal to:1 − 1 k + 1 = [ 1 - 1 k + 1 ] + 1 ( k + 1 ) ( k + 2 ) = 1 - [ 1 k + 1 - 1 ( k + 1 ) ( k + 2 ) ] = 1 - 1 k + 2 = RHS of P ( k + 1 ) . Hence
holds.P ( k + 1 ) If we combine these two bullet points, we have proved that “
holds for allP ( n ) ”.n ≥ 1
(c)
Note:
If
Let
“
”. 1 + q + q 2 + q 3 + ⋯ + q n − 1 = 1 1 − q − q n 1 − q
-
LHS of
; RHS ofP ( 1 ) = 1 . Since these two are equal,P ( 1 ) = 1 1 − q − q 1 − q = 1 is true.P ( 1 ) -
Suppose that
is true for some particular (unspecified)P ( k ) ; that is, we know that, for this particular k ,k ≥ 1 “
”.1 + q + q 2 + q 3 + ⋯ + q k − 1 = 1 1 − q − q k 1 − q We wish to prove that
must then be true.P ( k + 1 )
Now
If we now use
so the complete sum is equal to:
Hence
If we combine these two bullet points, we have proved that “
(d) Note: The statement to be proved starts with a term involving “0!”. The definition
does not immediately tell us how to interpret “
(i) When
(ii) When
(iii) The definition of
Let
“
”. 0 · 0 ! + 1 · 1 ! + 2 · 2 ! + ⋯ + ( n − 1 ) · ( n − 1 ) ! = n ! − 1
• LHS of
• Suppose that
“
”. 0 · 0 ! + 1 · 1 ! + 2 · 2 ! + ⋯ + ( k − 1 ) · ( k − 1 ) ! = k ! − 1
We wish to prove that
Now
If we now use
If we combine these two bullet points, we have proved that “
(e) Let
“
( cos θ + i sin θ ) n = cos n θ + i sin n θ ”
• LHS of
• Suppose that
“
( cos θ + i sin θ ) k = cos k θ + i sin k θ ”.
We wish to prove that
Now
If we combine these two bullet points, we have proved that “
236
.
Let
“
”. ( 1 + 2 + 3 + ⋯ + n ) 2 = 1 3 + 2 3 + 3 3 + ⋯ + n 3
• LHS of
• Suppose that
“
”. ( 1 + 2 + 3 + ⋯ + k ) 2 = 1 3 + 2 3 + 3 3 + ⋯ + k 3
We wish to prove that
Now
If we now use
so the complete RHS is:
If we combine these two bullet points, we have proved that “
Note:
A slightly different way of organizing the proof can sometimes be useful. Denote the two sides of the equation in the statement
•
• simple algebra allows one to check that, for each
It then follows (by induction) that
(a)
and
may eventually lead one to guess that
Note 1: This will certainly be easier to guess if you remember what you found in Problem 17 . and Problem 63 .
Note 2: There is another way to help in such guessing. Suppose you notice that
– adding values for
and
– adding values for
Then you might guess that the sum
will give an answer that is a polynomial of degree 3 in n . Suppose that
We can then use small values of n to set up equations which must be satisfied by A , B , C , D and solve them to find A , B , C , D :
– when
– when
– when
– when
This method assumes that we know that the answer is a polynomial and that we know its degree: it is called “the method of undetermined coefficients”.
There are various ways of improving the basic method (such as writing the polynomial
which tailors it to the values
(b) Let
“
”. T 1 + T 2 + T 3 + ⋯ + T n = n ( n + 1 ) ( n + 2 ) 6
• LHS of
• Suppose that
We wish to prove that
Now
so the complete sum is equal to:
Hence
If we combine these two bullet points, we have proved that “
Note:
The triangular numbers
You might like to interpret Problem 237 . in the language of binomial coefficients, and prove it by repeated use of the basic Pascal triangle relation (Pascal (1623--1662)):
Start by rewriting
(a) We know from Problem 237 .(b) that
Also
Therefore
(b) Guess:
The proof by induction is entirely routine, and is left for the reader.
Therefore
239 .
(a) Let
for some polynomial
Since
(b) Let
“Given any two labelled sets of
real numbers n + 1 , and a 0 , a 1 , a 2 , … , a n , where the b 0 , b 1 , b 2 , … , b n are all distinct (but the a i need not be), there exists a polynomial b i of degree n , such that f n , f n ( a 0 ) = b 0 , f n ( a 1 ) = b 1 , …, f n ( a 2 ) = b 2 .” f n ( a n ) = b n
• When
• Suppose that
“Given any two labelled sets of
real numbers k + 1 , and a 0 , a 1 , a 2 , … , a k , where the b 0 , b 1 , b 2 , … , b k are all distinct (but the a i need not be), there exists a polynomial b i of degree k , such that f k , f k ( a 0 ) = b 0 , f k ( a 1 ) = b 1 .” f k ( a 2 ) = b 2 , … , f k ( a k ) = b k
We wish to prove that
Now
“Given any two labelled sets of
real numbers ( k + 1 ) + 1 , and a 0 , a 1 , … , a k + 1 , where the b 0 , b 1 , b 2 , … , b k + 1 are all distinct (but the a i need not be), there exists a polynomial b i of degree f k + 1 , such that k + 1 , f k + 1 ( a 0 ) = b 0 , f k + 1 ( a 1 ) = b 1 , …, f k + 1 ( a 2 ) = b 2 .” f k + 1 ( a k + 1 ) = b k + 1
So to prove that
any two labelled sets of
real numbers ( k + 1 ) + 1 where the a 0 , a 1 , a 2 , … , a k + 1 , and b 0 , b 1 , b 2 , … , b k + 1 , are all distinct (but the a i need not be). b i
We must then somehow construct a polynomial function
Because we are supposing that
The next step is slightly indirect: we make use of the polynomial
satisfying
Part (a) guarantees the existence of such a polynomial
If we combine these two bullet points, we have proved that “
240 .
(a) 5 cents cannot be made;
Guess: Every amount > N = 5 can be paid directly (without receiving change).
(b) Let
“ n cents can be paid directly (without change) using 3 cent and 4 cent coins”.
–
– Now suppose that we know
To prove
“
cents can be paid directly”. k + 1
We know that
– If a direct method of paying
k
cents involves at least one 3 cent coin, then we can replace one 3 cent coin by a 4 cent coin to produce a way of paying
Hence we only need to worry about a situation in which the only way to pay
k
cents directly involves no 3 cent coins at all – that is, paying
k
cents uses only 4 cent coins. But then there must be at least two 4 cent coins (since
Hence
•
• whenever
(a)
(b)
(c) (d)
As soon as one starts calculating
z
2
and
so whenever
is also an integer.

Note:
In the solution to Problem
241
. we included the condition on
z
as part of the statement
In Problem 242 . the result to be proved has a similar background hypothesis – “Let p be a prime number”. It may make the induction clearer if, as in the statement of the Problem, this hypothesis is stated before starting the induction proof.
Let
p
be any prime number. We let
“
is divisible by p ”. n p − n
•
• Suppose that
“
is divisible by p ”. k p − k
We wish to prove
“
is divisible by p ” ( k + 1 ) p − ( k + 1 )
must then be true. Using the Binomial Theorem again we see that
By
• is an integer, and
• has a factor “ p ” in the numerator and no such factor in the denominator.
Hence each term in the second bracket is a multiple of p . So the RHS (and hence the LHS) is divisible by p .
Hence
If we combine these two bullet points, we have proved that “
243 .
This is a geometric series with first term
244 .
(a) Each person receives in total:
(here
(b) You have
(here
(here
245 . The trains are 120 km apart, and the fly travels at 50 km/h towards Train B , which is initially 120 km away and travelling at 30 km/h.
The relative speed of the fly and Train
B
is 80 km/h, so it takes
The relative speed of the fly and Train
A
is then also 80 km/h, so it takes just
Train
B
is
Continuing in this way, we see that the fly takes
Note: The two trains are approaching each other at 60 km/h, so they crash in exactly 2 hours – during which time the fly travels 100 km.
246 .
(a) (i)
so
(ii)
so
(b) The argument in part (a) gives an upper bound for each bracketed expression in the sum
Replacing each bracket by its upper bound, we see that the sum is
(c) The finite partial sums
• increase steadily as we take more and more terms, and
• every partial sum
It follows that there is some (unknown) number
To see, for example, that
Note 1: The claim that
“an increasing sequence of partial sums
, all less than 2, must converge to some number S n ” S ≤ 2
is a fundamental property of the real numbers – called completeness .
Note 2:
Just as one can obtain better and better lower bounds for
S
– like “
247 .
(a) Let
“
”. 1 1 2 + 1 2 2 + 1 3 2 + ⋯ + 1 n 2 ≤ 2 − 1 n
• Then LHS of
• Suppose we know that
Hence
(b) Let
“
1 1 2 + 1 2 2 + 1 3 2 + ⋯ + 1 n 2 < 1.68 − ”. 1 n
• Then
and RHS of
• Suppose we know that
Hence
(a) (i)
(ii) As we go on adding more and more terms, each finite sum
is bigger than the previous sum. Since every finite sum
the sums increase, but never reach 3, so they accumulate closer and closer to a value “
e
”
(b) (i) Let
“
”. 1 0 ! + 1 1 ! + 1 2 ! + ⋯ + 1 n ! ≤ 3 − 1 n ⋅ n !
•
• Suppose we know that
Hence
(ii) [The reasoning here uses the constant “3” while ignoring the refinement “
is bigger than the previous sum. Since every finite sum
the partial sums increase, but never reach 3; so they accumulate closer and closer to a value “
e
”
(c)
Note:
Examine carefully the role played by the number “3” in the above induction proof in (b)(ii). It is needed precisely to validate the statement
The exact value “
e
” of the infinite series is not really affected by what happens when
The only part of the induction proof where
that is,
(i) Let
“
”. 1 0 ! + 1 1 ! + 1 2 ! + ⋯ + 1 n ! ≤ 2.75 − 1 n ⋅ n !
• LHS of
• Suppose we know that
Hence
(ii) As we add more terms, each finite sum
is bigger than the previous sum.
Since every finite sum
the finite sums increase, but never reach 2.75, so they accumulate closer and closer to a value “
e
”
so
(d) (i) Let
“
”. 1 0 ! + 1 1 ! + 1 2 ! + ⋯ + 1 n ! ≤ 2.7222 ⋯ ( for ever) − 1 n . n !
•
• Suppose we know that
Hence
(ii) As we add more terms, each finite sum
is bigger than the previous sum.
Since every finite sum
the finite sums increase, but never reach
so
Note: This process of refinement can continue indefinitely. But we only have to go one further step to pin down the value of “ e ” with surprising accuracy.
The next step uses the same proof to show that
“
, for all 1 0 ! + 1 1 ! + 1 2 ! + ⋯ + 1 n ! ≤ 2.7185 − 1 n ⋅ n ! ”, n ≥ 4
and to conclude that the endless sum
has a definite value “
e
” that lies somewhere between
We could then repeat the same proof to show that
and use the lower bound
has a definite value “
e
” that lies somewhere between
249
.
Let
“
”. 1 1 + 1 2 + 1 3 + ⋯ + 1 n ≥ n
• LHS of
• Suppose we know that
Hence
250
. Let
a
,
b
be real numbers such that
One of
a
,
b
is then the greater, and we may suppose this is
a
– so that
Let
“
”. a n + b n 2 ≥ a + b 2 n
• LHS of
• Suppose we know that
Hence
251
. Let
x
be any real number
If
Thus we may assume that
Let
• LHS of
• Suppose we know that
Hence
252 . The problem is discussed after the statement of Problem 252 . in the main text.
253 .
(a) (i)
(ii)
(iii) Let
“
Then
•
• Suppose that
The first bracket is
Hence the LHS of
Hence
(iv) At the very first stage (part (i)) we replaced
(b)
(i)
(ii)
(iii) Let
“
1 1 + 1 2 + 1 3 + ⋯ + 1 2 n > 1 + ”. n 2
Then
•
• Suppose that
LHS of
The first bracket is
Hence
254 .
(a) We use induction. Let
“ n identical rectangular strips of length 2 balance exactly on the edge of a table if the successive protrusion distances (first beyond the edge of the table, then beyond the leading edge of the strip immediately below, and so on) are the terms
of the finite harmonic series in reverse order.” 1 n , 1 n − 1 , 1 n − 2 , … , 1 3 , 1 2 , 1 1
• When
• Suppose that we know that
Let
The statement
Let
M
be the mass of each strip; since the leading edge of the bottom strip is distance
The centre of gravity of the bottom strip is distance
Hence
(b) Problem
253
.(b)(iii) now guarantees that a stack of
Note: Ivars Petersen's Mathematical Tourist blog contains an entry in 2009
which explores how one can obtain large overhangs with fewer strips if one is allowed to use strips to counterbalance those that protrude beyond the edge of the table.
255 .
(a) (i) Let
“
s 2 p − 2 < s 2 p < s 2 q + 1 < for all s 2 q − 1 such that p , q ”. 1 ≤ p , q ≤ n
•
• Suppose that
which are true, since
and
Hence
Hence
(ii) The “even sums”
The “odd sums”
The “even sums”
The “odd sums”
Moreover, the difference between successive sums is
(b) The proof from part (a) carries over word for word, with “
256
.
Let
“
has at least k distinct prime factors”. f k
•
• Suppose that
Then
And the second factor
Moreover
Hence
Hence
Note: This problem [suggested by Serkan Dogan] gives a different proof of the result (Problem 76 .(d)) that the list of prime numbers goes on for ever.
257 .
(a) Let
• 0 points leave the line in pristine condition – namely a single interval – so
• Suppose that
Consider an arbitrary straight line divided by
Then the
k
points
The additional point
Hence
Hence
(b) (i) We want a function R satisfying
If we notice that in part (a)
(ii) Let
“ n distinct straight lines in the plane divide the plane into at most
regions”. f ( n ) = 1 + ( n 1 ) + ( n 2 )
• 0 lines leave the plane in pristine condition – namely a single region – so
• Suppose that
Consider the plane divided by
Then the
k
lines
regions (by
The additional line
regions altogether.
Hence
Hence
(c) (i) We want a function S satisfying
After thinking about the differences between successive terms in part (b), we might guess that
(ii) Let
“ n distinct planes in 3-space divide space into at most
S n = ( n 0 ) + ( n 1 ) + ( n 2 ) + ( n 3 ) regions”.
• 0 planes leave our 3D space in pristine condition – namely a single region – so
• Suppose that
Consider 3D divided by
Then the
k
planes
regions (by
The additional plane
regions, and each of these regions on the plane
regions altogether. (There is no need for any algebra here: one only needs to use the Pascal triangle condition.)
Hence
Hence
258
. Notice that, given a dissection of a square into
k
squares, we can always cut one square into four quarters (by lines through the centre, and parallel to the sides), and so create a dissection with
Let
“Any given square can be cut into m (not necessarily congruent) smaller squares, for each m ,
”. 6 ≤ m ≤ n
• Let
Let
Let
• Suppose that
Then
Hence
259
. Let
“Any tree with n vertices has exactly
edges”. n − 1
• A tree with 1 vertex is simply a vertex with 0 edges (since any edge would have to be a loop, and would then create a cycle). Hence
• Suppose that
Consider an arbitrary tree
T
with
[
Idea
: We need to find some way of reducing
T
to a tree
Definition The number of edges incident with a given vertex v is called the valency of v .
Lemma
Let
S
be a finite tree with
Proof
Choose any vertex
If
If
If
Continue in this way.
All of the vertices
in the tree
S
). Since we know that the tree is finite (having precisely
s
vertices), the process must terminate at some stage. The final vertex
If we apply the Lemma to our arbitrary tree
T
with
Hence
260 .
Note: All the polyhedra described in this solution are “spherical” by virtue of having their vertices located on the unit sphere.
(a) (i) A regular tetrahedron.
(ii) A square based pyramid with its apex at the North pole.
(b) If a spherical polyhedron has V vertices, E edges, and F faces, then
Now each edge has exactly two end vertices, so
On the other hand, in a spherical polyhedron, each vertex
v
has valency at least 3; so each vertex
v
occurs in at least 3 pairs
(v, e)
of this kind. Hence
In the same way, each edge
e
lies on the boundary of exactly 2 faces, so
On the other hand, in a spherical polyhedron, each face
f
has at least 3 edges; so each face
f
occurs in at least 3 pairs
(f, e)
of this kind. Hence
If
(c) We show by induction how to construct certain “spherical” polyhedra, with at most one non-triangular face. Let
“There exists a spherical polyhedron with at most one non-triangular face, and with e edges for each
e , ”. 8 ≤ e ≤ n
• We know that there exists a such a spherical polyhedron with
We know there is no such polyhedron with
When
When
When
This provides us with a starting point for the inductive construction. In particular
• Suppose that
Since
Hence
261 . To prove a result that is given in the form of an “if and only if” statement, we have to prove two things: “if”, and “only if”.
We begin by proving the “only if” part:
“a map can be properly coloured with two colours only if every vertex has even valency”.
Let M be a map that can be properly coloured with two colours. Let v be any vertex of the map M .
The edges
We now prove the “if” part:
“a map can be properly coloured with two colours if every vertex has even valency”.
Suppose that we have a map M in which each vertex has even valency. We must prove that any such map M can be properly coloured using just two colours.
Let
“any map with m edges, in which each vertex has even valency, can be properly coloured with two colours whenever m satisfies
,”. 1 ≤ m ≤ n
• If
• Suppose that
Most of the contents of the statement
any map with exactly
edges, in which every vertex has even valency, can be properly coloured using just two colours. k + 1
Consider an arbitrary map
M
with
[
Idea:
We need to find some way of reducing the map
M
to a map
Since
Suppose first that
Hence we may assume that
which is also even.
Hence every vertex of the new map
Hence
262
. Let
“The
sequences of length n consisting of 0s and 1s can be arranged in a cyclic list such that any two neighbouring sequences (including the last and the first) differ in exactly one coordinate position.” 2 n
• When
• The general construction is perhaps best illustrated by first showing how
The above cycle for sequences of length 2 gives rise to two disjoint cycles for sequences of length 3:
– first by adding a third coordinate “0”:
– then by adding a third coordinate “1”:
Now eliminate the final join in each cycle (
In general, suppose that
Take the cycle of the
sequences of length k guaranteed by 2 k , and form two disjoint cycles of length P ( k ) 2 k • first by adding a final coordinate “0“
• then by adding a final coordinate “1”.
 Hence k + 1 is true. P ( k + 1 )
Hence
Note: The significance of what we call Gray codes was highlighted in a 1953 patent by the engineer Frank Gray (1887-1969) – where they were called reflected binary codes (since the crucial step in their construction above involves taking two copies of the previous cycles, reversing one of the cycles, and then producing half of the required cycle by traversing the first copy before returning backwards along the second copy). Their most basic use is to re-encode the usual binary counting sequence
where a single step can lead to the need to change arbitrarily many binary digits (e.g. the step from
263 .
(a) The whole construction is inductive (each label derives from an earlier label). So let
“if
H C F ( r , s ) = 1 and, then the positive rational 2 ≤ r + s ≤ n occurs once and only once as a label, and it occurs in its lowest terms”. r s
• By construction the root is given the label
Notice that the basic construction:
“if
is a `parent' vertex, then we label its `left descendant' as i j , and its `right descendant' i i + j ” i + j j
guarantees that, since we start by labelling the root with the positive rational
Moreover, if any `descendant' were suddenly to appear not “in lowest terms”, then either
–
–
Since we begin by labelling the root
• Suppose that
Most parts of the statement
(i) Suppose that
(ii) Suppose that
Hence
Hence
(b) The fact that the labels are left-right symmetric is also an inductive phenomenon. We note that the one fully “left-right symmetric” label, namely
All other labels occur in reciprocal pairs:
So if we know that the earlier reciprocal pair reciprocal pair
“
: if P ( n ) , and r , s > 0 , then the reciprocal pair 2 ≤ r + s ≤ n , r s occur as labels of vertices at the same level below the root, with the two labelled vertices being mirror images of each other about the vertical mirror through the root vertex.” s r
264
. The intervals in this problem may be of any kind (including finite or infinite). Each interval has two “endpoints”, which are either ordinary real numbers, or
Let
“if a collection of n intervals on the x -axis has the property that any two intervals overlap in an interval (of possibly zero length – i.e. a point), then the intersection of all intervals in the collection is a non-empty interval”.
When
Suppose that
So consider a collection of
Among the
By
We have to show that the intersection
The proof that follows works if the endpoint b is included in the interval I . The slight adjustment needed if b is not included in the interval I is left to the reader.
Since the right hand endpoint of
Moreover, for each point
Hence
Hence
265 . If one experiments a little, it should become clear
• that if tank
• that if, at some stage in the linking sequence, tank
T
contains an interim amount of
a
litres, and is about to link successively to a tank containing
b
litres, and then to the tank containing
c
litres, this ordered pair of changes alters the amount in the tank
T
to
• once the tap linking tank T to another tank has been opened, so that the two levels become equal, there is no benefit from opening the tap linking these two tanks ever again, so tank T should be linked with each other tank at most once.
These three observations essentially determine the answer – namely that tank T should be joined to the other tanks in increasing order of their initial contents.
For a proof by induction, let
“given n tanks containing
litres respectively, where a 1 , a 2 , a 3 , … , a n a 1 < a 2 < a 3 < … < a n , if T is the tank containing the smallest amount
litres, then the optimal sequence for linking the other a 1 tanks to tank T (optimal in the sense that it transfers the maximum amount of water to tank T ) is the sequence that links T successively to the other tanks in increasing order of their initial contents”. n − 1
• When
• Suppose that
and where
T
is the tank which initially contains
Suppose that in the optimal sequence of
k
successive joins to the other
k
tanks (that is, that transfers the largest possible amount of water to tank
T
), the succession of joins is to join
T
first to tank
(i)
(ii)
(i) Suppose tank
(ii) Now suppose that tank
By the very first bullet point, in order to guarantee the optimal overall outcome of the final linking with tank
Hence
Hence
266 .
Note: Like all practical problems, this one requires an element of initial “modelling” in order to make the situation amenable to mathematical analysis.
`Residue' clings to surfaces; so the total amount of `residue' will depend on
(a) the viscosity of the chemical (how `thick', or `sticky' it is), and
(b) the total surface area of the inside of the flask.
Since we are given no information about quantities, we may fix the amount of residue remaining in the `empty' flask at “1 unit”, and the amount of solvent in the other flask as “ s units”.
If we add the solvent, we get a combined amount of
The first modelling challenge arises when we try to make mathematical sense of what remains at each stage after we empty the flask. The internal surface area of the flask, to which any diluted residue may adhere, is fixed. If we make the mistake of thinking of the original chemical as “thick and sticky” and the solvent as “thin”, then the viscosity of the diluted residue will change relative to the original, and will do so in ways that we cannot possibly know. Hence the only reasonable assumption (which may or may not be valid in a particular instance) is to
assume
that the viscosity of the original chemical is roughly the same as the viscosity of the chemical-solvent mixture. This then suggests that, on emptying the diluted mixture, roughly the same amount (1 unit) of diluted mixture will remain adhering to the walls of the flask. So we will be left with 1 unit of residue, with a concentration of
But what if we use only half of the solvent first, empty the flask, and then use the other half? Adding
If we then add the other
Suppose that we use a four-stage strategy – using first one quarter of the solvent, then another quarter, and so on. We then land up with roughly 1 unit of residue with concentration 1 part per
Note: The situation here is similar to that faced by a washing machine designer, who wishes to remove traces of detergent from items that have been washed, without using unlimited amounts of water. The idea of having a “fixed amount of solvent” corresponds to the goal of “water efficient” rinsing. However, the washing machine cycle, or programme, clearly cannot repeat the rinsing indefinitely (as would be required in the limiting case above).
267 .
(i) If
Hence m 2 is even.
It follows that m must be even.
Note:
It is a fact that, if
Claim m cannot be odd.
Proof Suppose m is odd.
But then
would be odd, contrary to “ m 2 must be even”.
Hence m cannot be odd.
(ii) Since
m
is even, we may write
(iii) If
In the same way, it follows that
Continuing in this way then produces an endless decreasing sequence of positive denominators
contrary to the fact that such a sequence can have length at most
268 .
(i) If
If
We now know that
(ii) Suppose
Suppose
Then
So
269 .
(a) Suppose that
S
is not empty. Then by the
Least Element Principle
the set
S
must contain a least element
k
: that is, a smallest integer
k
which is not in the set
T
. Then
Therefore
(b) Suppose that T does not have a smallest element. Clearly 1 does not belong to the set T (or it would be the smallest element of T ). Hence 1 must be an element of the set S .
Now suppose that, for some
“whenever
is an element of S , we can be sure that k ≥ 1 is also an element of S .” k + 1
The
Principle of Mathematical Induction
then guarantees that these two observations (that 1 is an element of
S
, and that whenever
k
is an element of
S
, so is
Hence T must have a smallest element.
270 .
(i) Triangle
(ii) Triangles
(iii) If
and
(iv)
(v) In the square
Zarathustra's last, most vital lesson: “Now do without me.”
George Steiner (1929-)


11.25: Reading- Coastal Hazards
Natural geomorphic processes are considered hazards when human populations are affected by them. For example, storms, hurricanes, and tsunamis are natural driving forces of coastal processes and landforms, but also cause loss of life and property in coastal communities. Storms provide much of the sediment to shallow-marine and estuarine ecosystems. Many wetland environments and barrier islands depend on storm activities for sediment build-up and survival when faced with rising sea levels. When combined with increasing sea levels, these events may have tremendous impacts on coastal environments and beaches.
Rip Currents
Rip currents are strong and swift, moving 3 to 6 feet (1 to 2 m) per second, which is faster than an average person can swim! These currents are created because of “set-up” near the shoreline. Set-up is a slight increase in water levels compared to those found seaward of the surf zone and creates unstable conditions that eventually are relieved through the formation of rip currents. These dangerous currents generally form at a low point or saddle in a sandbar. Because rip currents tend to be narrow, swimmers caught in a rip current should swim parallel to the shore to escape being taken out to sea. Swimmers should be especially cautious during storm events, which may increase the frequency and strength of rip currents.
Storm Surges
Storm surges—extraordinarily high water levels—are generated by the combined effects of low atmospheric pressure and very high wind speeds. The strong onshore winds that accompany tropical storms, hurricanes, and frontal storms of the midlatitudes drag and stack water against coasts, creating a storm surge. In the center of tropical-storm systems, atmospheric pressure may drop as much as 100 millibars below normal, and this can “suck up” sea level below the center of the cyclone by as much as 3.3 feet (1 m) ( Summerfield 1991 ). Winds blowing toward coastal embayments produce the largest storm surges, which are accentuated if they coincide with high tides. By the time the waves generated by the storm surge reach the coast, they may have built up to a height of nearly 10 feet (several meters) above normal high tides.
When low-pressure storm systems approach land, strong winds can affect a coast in a variety of ways. High velocity onshore winds, particularly the kind associated with hurricanes, drive water ashore and elevate the water line well above the predicted tidal variation. The effect of storm surges can be catastrophic because the elevated water surface results in widespread coastal flooding and allows waves to break much further inland than they would normally. In addition, torrential rainfall associated with the storm causes widespread river flooding. The combination of these effects can result in extensive property damage and loss of life ( Pinet 1992 ).
Although comparatively uncommon on a global basis, storm surges occur repeatedly along coasts experiencing tropical cyclones and in midlatitude areas subject to intense storms where the coastal configuration is particularly favorable. Storm surges are most destructive along very low-lying coasts where their effects can extend many miles (kilometers) inland, but their geomorphic significance arises in large part from the way in which they lead to wave attack at much higher levels along a shoreline than is reached by normal waves.
For more information about hurricanes, check out these Web sites:
Tsunamis
Tsunamis are very large seismic ocean waves that are radially generated from volcanic eruptions, earthquakes, or subaqueous slumping. In the open ocean, these waves may travel at speeds in excess of 493 miles per hour (793 kph)! Strangely, sailors on deep ocean vessels may not notice the passing of these waves on account of the waves’ flat, low propagation. In contrast, when tsunamis reach shallow water, they slow down considerably and may reach great heights [up to 33 ft (10 m)]. Tsunamis have caused great destruction and loss of life because of abrupt changes in water levels above the normal high water mark. Numerous areas in the United States have experienced tsunamis including Hawaii, the Pacific Northwest, and Alaska. A new system, the International Tsunami Warning System, is used now to alert the public to impending tsunamis.
Use these Web links for more information about tsunamis: where they occur, why they occur, and what happens in a tsunami.
Contributors and Attributions
- Coastal Processes. Provided by : National Park Service. Located at : http://www.nature.nps.gov/views/KCs/CoastalG/HTML/ET_Processes.htm . Project : Views of the National Park. License : Public Domain: No Known Copyright


6.10: Cash Dividends Calculations
Preferred stockholders are paid a designated dollar amount per share before common stockholders receive any cash dividends. However, it is possible that the dividend declared is not enough to pay the entire amount per preferred share that is guaranteed before common stockholders receive dividends. In that case the amount declared is divided by the number of preferred shares. Common stockholders would then receive no dividend payment.
Preferred stock may be cumulative or non-cumulative. This determines whether preferred shares will receive dividends in arrears, which is payment for dividends missed in the past due to inadequate amount of dividends declared in prior periods. If preferred stock is non-cumulative, preferred shares never receive payments for past dividends that were missed. If preferred stock is cumulative, any past dividends that were missed are paid before any payments are applied to the current period.
25,000 shares of $3 non-cumulative preferred stock and 100,000 shares of common stock. Preferred shares would receive $75,000 in dividends (25,000 * $3) before common shares would receive anything.
| Preferred Stockholders | Common Stockholders | Owed to | ||||
| Year | Total Dividend | Total | Per Share | Total | Per Share | Preferred |
| 1 | $0 | $0 | $0 | $0 | $0 | $0 |
| 2 | $20,000 | $20,000 | $0.80 | $0 | $0 | $0 |
| 3 | $60,000 | $60,000 | $2.40 | $0 | $0 | $0 |
| 4 | $175,000 | $75,000 | $3.00 | $100,000 | $1.00 | $0 |
| 5 | $200,000 | $75,000 | $3.00 | $125,000 | $1.25 | $0 |
| 6 | $375,000 | $75,000 | $3.00 | $300,000 | $3.00 | $0 |
In years 1 through 3, dividends of less than $75,000 were declared. Since preferred stockholders are entitled to receive the first $75,000 in each year, they receive the entire amount of the dividend declared and the common shareholders receive nothing.
In years 4 through 6, dividends of more than $75,000 were declared. In each of those years, the preferred stockholders receive the first $75,000 and the common stockholders receive the remainder.
The preferred stockholders are never “caught up” for the amounts that were less than $75,000 that they missed out on in years 1 through 3.
25,000 shares of $3 cumulative preferred stock and 100,000 shares of common stock. Preferred shares would receive $75,000 in dividends (25,000 * $3) before common shares would receive anything.
| Preferred Stockholders | Common Stockholders | Owed to | ||||
| Year | Total Dividend | Total | Per Share | Total | Per Share | Preferred |
| 1 | $0 | $0 | $0 | $0 | $0 | $75,000 |
| 2 | $20,000 | $20,000 | $0.80 | $0 | $0 | $130,000 |
| 3 | $60,000 | $60,000 | $2.40 | $0 | $0 | $145,000 |
| 4 | $175,000 | $175,000 | $7.00 | $0 | $0 | $45,000 |
| 5 | $200,000 | $120,000 | $4.80 | $80,000 | $0.80 | $0 |
| 6 | $375,000 | $75,000 | $3.00 | $300,000 | $3.00 | $0 |
In years 1 through 3, dividends of less than $75,000 were declared. Since dividends on preferred stock are cumulative, each year that dividends are declared there is a look back to previous years to ensure that preferred shareholders received their full $75,000 for all past years before any dividends are paid to common stockholders in the current year.
In this example, no dividends are paid on either class of stock in year 1.
In year 2, preferred stockholders must receive $150,000 ($75,000 for year 1 and $75,000 for year 2) before common shareholders receive anything. Since only $20,000 is declared, preferred stockholders receive it all and are still “owed” $130,000 at the end of year 2.
In year 3, preferred stockholders must receive $205,000 ($130,000 in arrears and $75,000 for year 3) before common shareholders receive anything. Since only $60,000 is declared, preferred stockholders receive it all and are still “owed” $145,000 at the end of year 3.
In year 4, preferred stockholders must receive $220,000 ($145,000 in arrears and $75,000 for year 4) before common shareholders receive anything. Since only $175,000 is declared, preferred stockholders receive it all and are still “owed” $45,000 at the end of year 4.
In year 5, preferred stockholders must receive $120,000 ($45,000 in arrears and $75,000 for year 5) before common shareholders receive anything. Since $200,000 is declared, preferred stockholders receive $120,000 of it and common shareholders receive the remaining $80,000.
In year 6, preferred stockholders are not owed any dividends in arrears. Of the $375,000 that is declared, they receive the $75,000 due to them in year 6. Common shareholders receive the remaining $300,000.
The accounts that are highlighted in bright yellow are the new accounts you just learned. Those highlighted in pale yellow are the ones you learned previously.
At this point you have learned all of the accounts and calculated amounts that are shown below on the income statement, retained earnings statement, and balance sheet. Understanding what these accounts are and how their balances are determined provides you with a sound foundation for learning managerial accounting concepts. You will move from preparing and reading financial statements to using these results for decision making purposes in a business.


6.3: Skew and Kurtosis
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There are two more descriptive statistics that you will sometimes see reported in the psychological literature, known as skew and kurtosis. In practice, neither one is used anywhere near as frequently as the measures of central tendency and variability that we’ve been talking about. Skew is pretty important, so you do see it mentioned a fair bit; but I’ve actually never seen kurtosis reported in a scientific article to date.
## [1] -0.9174977
## [1] 0.009023979
## [1] 0.9250898
Since it’s the more interesting of the two, let’s start by talking about the skewness . Skewness is basically a measure of asymmetry, and the easiest way to explain it is by drawing some pictures. As Figure 5.4 illustrates, if the data tend to have a lot of extreme small values (i.e., the lower tail is “longer” than the upper tail) and not so many extremely large values (left panel), then we say that the data are negatively skewed . On the other hand, if there are more extremely large values than extremely small ones (right panel) we say that the data are positively skewed . That’s the qualitative idea behind skewness. The actual formula for the skewness of a data set is as follows
\[\text { skewness }(X)=\dfrac{1}{N \hat{\sigma}\ ^{3}} \sum_{i=1}^{N}\left(X_{i}-\bar{X}\right)^{3} \label{skew}\]
where N is the number of observations, \(\bar{X}\) is the sample mean, and \(\hat{\sigma}\) is the standard deviation (the “divide by N−1” version, that is). Perhaps more helpfully, it might be useful to point out that the
psych
package contains a
skew()
function that you can use to calculate skewness. So if we wanted to use this function to calculate the skewness of the
afl.margins
data, we’d first need to load the package
library( psych )
which now makes it possible to use the following command:
skew( x = afl.margins )
## [1] 0.7671555
Not surprisingly, it turns out that the AFL winning margins data is fairly skewed.
The final measure that is sometimes referred to is the kurtosis of a data set. Put simply, kurtosis is a measure of the “tailedness”, or outlier character, of the data. Historically, it was thought that this statistic measures “pointiness” or “flatness” of a distribution, but this has been shown to be an error of interpretation. See Figure 5.5.
## [1] -0.9631805
## [1] 0.02226287
## [1] 1.994329
By mathematical calculations, the “normal curve” (black lines) has zero kurtosis, so the outlier character of a data set is assessed relative to this curve. In this Figure, the data on the left are less outlier-prone, so the kurtosis is negative and we call the data platykurtic . The data on the right are more outlier-prone, so the kurtosis is positive and we say that the data is leptokurtic . But the data in the middle are similar in their outlier character, so we say that it is mesokurtic and has kurtosis zero. This is summarised in the table below:
| informal term | technical name | kurtosis value |
|---|---|---|
| just pointy enough | mesokurtic | zero |
| too pointy | leptokurtic | positive |
| too flat | platykurtic | negative |
The equation for kurtosis is pretty similar in spirit to the formulas we’ve seen already for the variance and the skewness (Equation \ref{skew}); except that where the variance involved squared deviations and the skewness involved cubed deviations, the kurtosis involves raising the deviations to the fourth power: 75
\[\text { kurtosis }(X)=\dfrac{1}{N \hat{\sigma}\ ^{4}} \sum_{i=1}^{N}\left(X_{i}-\bar{X}\right)^{4}-3\]
The
psych
package has a function called
kurtosi()
that you can use to calculate the kurtosis of your data. For instance, if we were to do this for the AFL margins,
kurtosi( x = afl.margins )
## [1] 0.02962633
we discover that the AFL winning margins data are just pointy enough.
Contributors
- Danielle Navarro (Associate Professor (Psychology) at University of New South Wales)
- Peter H. Westfall (Paul Whitfield Horn Professor and James and Marguerite Niver Professor, Texas Tech University)


8.12B: Deinococcus and Thermus
- Compare Deinococcus and Thermus bacteria
DEINOCOCCUS
This is the one genus of three of the Deinococcales group from the Deinococcus-Thermus phylum, and is highly-resistant to environmental hazards . It has several species that are resistant to radiation (they have become famous for their ability to eat nuclear waste and other toxic materials), survive in the vacuum of space, and in extremes of heat and cold. There are 47 species of Deinococcus described according to NCBI on 25 August 2011.
These bacteria have thick cell walls that give them Gram-positive stains, but they include a second membrane and so are closer in structure to those of Gram-negative bacteria. Cavalier-Smith calls this clade Hadobacteria (from Hades, the Greek underworld).
They are also characterized by the presence of the carotenoid pigment Deinoxanthin that give them their pink color, and a high resistance to gamma and UV radiation. They are usually isolated according to these two criteria.
D. RADIODURANS
Deinococcus radiodurans is an extremophilic bacterium, one of the most radioresistant organisms known. It can survive cold, dehydration, vacuum, and acid, and is therefore known as a polyextremophile and has been listed as the world’s toughest bacterium in The Guinness Book of World Records .
D. radiodurans is a rather large, spherical bacterium, with a diameter of 1.5 to 3.5 µm. Four cells normally stick together, forming a tetrad. The bacteria are easily cultured and do not appear to cause disease. Colonies are smooth, convex, and pink to red in color. D. radiodurans does not form endospores and is nonmotile. It is an obligate aerobic chemoorganoheterotroph, i.e., it uses oxygen to derive energy from organic compounds in its environment.
It is often found in habitats rich in organic materials, such as soil, feces, meat, or sewage, but has also been isolated from dried foods, room dust, medical instruments and textiles. It is extremely resistant to ionizing radiation, ultraviolet light, desiccation, and oxidizing and electrophilic agents. Its genome consists of two circular chromosomes, one 2.65 million base pairs long and the other 412 thousand base pairs long, as well as a megaplasmid of 177 thousand base pairs and a plasmid of 46 thousand base pairs. It has about 3,195 genes. In its stationary phase, each bacterial cell contains four copies of this genome; when rapidly multiplying, this increases to eight to 10 copies.
D. radiodurans is capable of withstanding an acute dose of five thousand Gy (five hundred thousand rad) of ionizing radiation with almost no loss of viability, and an acute dose of 15 thousand Gy with 37% viability. A dose of five thousand Gy is estimated to introduce several hundred double-strand breaks (DSBs) into the organism’s DNA (~0.005 DSB/Gy/Mbp (haploid genome)).
Several bacteria of comparable radioresistance are now known, including some species of the genus Chroococcidiopsis (phylum cyanobacteria) and some species of Rubrobacter (phylum actinobacteria); among the archaea, the species Thermococcus gammatolerans shows comparable radioresistance.
D. radiodurans also has a unique ability to repair damaged DNA. It isolates the damaged segments in a controlled area and repairs it. This bacteria can also repair many small fragments from an entire chromosome.
THERMUS
A genus of thermophilic bacteria that can tolerate high temperatures, it is one of several bacteria belonging to the Deinococcus-Thermus group and includes the following three species: T. aquaticus, T. antranikianii, and T. igniterrae . Thermus aquaticus is the source of the heat-resistant enzyme Taq DNA polymerase , one of the most important enzymes in molecular biology because of its use in the polymerase chain reaction (PCR) DNA-amplification technique.
It thrives at 70°C (160°F), but can survive at temperatures of 50°C to 80°C (120°F to 175°F). This bacterium is a chemotroph — it performs chemosynthesis to obtain food. However, since its range of temperature overlaps somewhat with that of the photosynthetic cyanobacteria that share its ideal environment, it is sometimes found living jointly with its neighbors, obtaining energy for growth from their photosynthesis.
The Deinococcus-Thermus are a small group of bacteria composed of cocci that are highly-resistant to environmental hazards. There are two main groups: the Deinococcales include two families, with three genera, Deinococcus and Truepera.
The Thermales include several genera resistant to heat (Marinithermus, Meiothermus, Oceanithermus, Thermus, Vulcanithermus).
Though these two groups evolved from a common ancestor, the two mechanisms of resistance appear to be largely independent.
Key Points
- The Deinococcales include two families with three genera, Deinococcus and Truepera , the former with several species that are resistant to radiation; they are famous for their ability to eat nuclear waste and other toxic materials, survive in the vacuum of space and in extremes of heat and cold.
- The Thermales include several genera resistant to heat, including Thermus .
- These bacteria have thick cell walls that give them Gram-positive stains, but they include a second membrane and so are closer in structure to those of Gram-negative bacteria.
Key Terms
- radioresistance : Any form of resistance that an organism has to protect itself against the harmful effects of ionizing radiation
- clade : A group of animals or other organisms derived from a common ancestor species.
- polyextremophile : An organism which can tolerate two or more extreme environmental factors.


2.3: DNA is Packaged into Chromatin
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- Ying Liu
- City College of San Francisco
DNA can be highly compacted
If stretched to its full length, the DNA molecule of the largest human chromosome would be 85mm. Yet during mitosis and meiosis, this DNA molecule is compacted into a chromosome approximately 5µm long. Although this compaction makes it easier to transport DNA within a dividing cell, it also makes DNA less accessible for other cellular functions such as DNA synthesis and transcription. Thus, chromosomes vary in how tightly DNA is packaged, depending on the stage of the cell cycle and also depending on the level of gene activity required in any particular region of the chromosome.
Levels of compaction
There are several different levels of structural organization in eukaryotic chromosomes, with each successive level contributing to the further compaction of DNA (Figure \(\PageIndex{2}\)). For more loosely compacted DNA, only the first few levels of organization may apply. Each level involves a specific set of proteins that associate with the DNA to compact it. First, proteins called the core histones act as spool around which DNA is coiled twice to form a structure called the nucleosome . Nucleosomes are formed at regular intervals along the DNA strand, giving the molecule the appearance of “beads on a string”. At the next level of organization, histone H1 helps to compact the DNA strand and its nucleosomes into a 30nm fibre . Subsequent levels of organization involve the addition of scaffold proteins that wind the 30nm fibre into coils, which are in turn wound around other scaffold proteins.
Chromatin Packaging Varies inside the Nucleus: Euchromatin & Heterochromatin
Chromosomes stain with some types of dyes, which is how they got their name (chromosome means “colored body”). Certain dyes stain some regions along a chromosome more intensely than others, giving some chromosomes a banded appearance. The material that makes up chromosomes, which we now know to be proteins and DNA, is called chromatin . Classically, there are two major types of chromatin, but these are more the ends of a continous and varied spectrum. Euchromatin is more loosely packed, and tends to contain genes that are being transcribed, when compared to the more densely compacted heterochromatin, which is rich in repetitive sequences and tends not to be transcribed. Heterochromatin sequences include short, highly-repetitive sequences called satellite DNA , which acquired their name because their bouyant density is distictly different from the main band of DNA following ultracentrifugation.
Morphological features of Chromosomes
Chromosomes also contain other distinctive features such as centromeres and telomeres. Both of these are usually heterochromatin. In most cases, each chromosome contains one centromere . These sequences are bound by centromeric proteins that link the centromere to microtubules that transport chromosomes during cell division. Under the microscope, centromeres of metaphase chromosomes can sometimes appear as constrictions in the body of the chromosome (Figure \(\PageIndex{3}\)) and are called primary (1°) constrictions. If a centromere is located near the middle of a chromosome, it is said to be metacentric , while an acrocentric centromere is closer to one end of a chromosome, and a telocentric chromsome is at, or near, the very end. In contrast, some species have a holocentric centromere, where no single centromere can be defined and the entire chromsome acts as the centromere. Telomeres are repetitive sequences near the ends of linear chromosomes, and are important in maintaining the length of the chromosomes during replication, and protecting the ends of the chromosomes from alterations.
It is essential to describe the similarity between chromosomes using appropriate terminology (Fig 2.4). Homologous chromosomes are typically pairs of similar, but non-identical, chromosomes in which one member of the pair comes from the male parent, and the other comes from the female parent. Homologs contain the same gene loci but not necessarily the same alleles. Non-homologous chromosomes contain different gene loci, and are usually distinguishable based on cytological features such as length, centromere position, and banding patterns.
An unreplicated chromosomes can undergo replication, to produce a replicated chromosome that has two sister chromatids, which are physically connected to each other at the centromere and remain joined until cell division. Because a pair of sister chromatids is produced by the replication of a single DNA molecule, their sequences are essentially identical (same alleles), differing only because of DNA replication errors. On the other hand, non-sister chromatids come from two separate, but homologous chromosomes, and therefore usually contain the same gene loci in the same order, but do not necessarily have identical DNA sequences (allelic differences).
The decondensed chromosomes are not randomly arranged in within the interphase nucleus. They often have specific locations within the nucleus and relative to one another (Figure \(\PageIndex{5}\))
Chromosome and DNA Replication
When the cell enters S-phase in the cell cycle (G1-S-G2-M) all the chromosomal DNA is replicated. This is done by enzymes called DNA polymerases . All DNA polymerases synthesize new strands by adding nucleotides to the 3'OH group present on the previous nucleotide. For this reason they are said to work in a 5' to 3' direction. DNA polymerases use a single strand of DNA as a template upon which it will synthesize the complementary sequence. This works fine for the middle of chromosomes - DNA-directed DNA polymerases travel along the original DNA strands making complementary strands (Figure \(\PageIndex{6}\)a).
DNA replication in both prokaryotes and eukaryotes begins at an Origin of Replication (Ori). Origins are specific sequences on specific positions on the chromosome. In E. coli , the OriC origin is ~245 bp in size. Chromosome replication begins with the binding of the DnaA initiator protein to an AT-rich 9-mer in OriC and melts the two strands. Then DnaC loader protein helps DnaB helicase protein extend the single stranded regions such that the DnaG primase can initiate the synthesis of an RNA primer, from which the DNA polymerases can begin DNA synthesis at the two replication forks. The forks continue in opposite directions until they meet another fork or the end of the chromosome (Figure \(\PageIndex{7}\)).
The ends of linear chromosomes present a problem – at each end one strand cannot be completely replicated because there is no primer to extend. Although the loss of such a small sequence would not be a problem, the continued rounds of replication would result in the continued loss of sequence from the chromosome end to a point were it would begin to loose essential gene sequences. Thus, this DNA must be replicated. Most eukaryotes solve the problem of synthesizing this unreplicated DNA with a specialized DNA polymerase called telomerase, in combination with a regular polymerase. Telomerases are RNA-directed DNA polymerases. They are a riboprotein, as they are composed of both protein and RNA. As Figure \(\PageIndex{6}\)b shows, these enzymes contain a small piece of RNA that serves as a portable and reusable template from which the complementary DNA is synthesized. The RNA in human telomerases uses the sequence 3-AAUCCC-5' as the template, and thus our telomere DNA has the complementary sequence 5'-TTAGGG-3' repeated over and over 1000’s of times. After the telomerase has made the first strand a primase synthesizes an RNA primer and a regular DNA polymerase can then make a complementary strand so that the telomere DNA will ultimately be double stranded to the original length (Figure \(\PageIndex{8}\)). Note: the number of repeats, and thus the size of the telomere, is not set. It fluctuates after each round of the cell cycle. Because there are many repeats at the end, this fluctuation maintains a length buffer – sometimes it’s longer, sometimes it’s shorter – but the average length will be maintained over the generations of cell replication.
In the absence of telomerase, as is the case in human somatic cells, repeated cell division leads to the “ Hayflick limit ”, where the telomeres shorten to a critical limit and then the cells enter a senescence phase of non-growth. The activation of telomerase expression permits a cell and its descendants to become immortal and bypass the Hayflick limit. This happens in cancer cells, which can form tumours as well as in cells in culture, such as HeLa cells , which can be propagated essentially indefinitely. HeLa cells have been kept in culture since 1951.
Eukaryote Chromosomes have Multiple Origins of Replication
In prokaryotes , with a small, simple, circular chromosome, only one origin of replication is needed to replicate the whole genome. For example, E. coli has a ~4.5 Mb genome (chromosome) that can be duplicated in ~40 minutes assuming a single origin, bi-directional replication, and a speed of ~1000 bases/second/fork for the polymerase.
However, in larger, more complicated eukaryotes , with multiple linear chromosomes, more than one origin of replication is required per chromosome to duplicate the whole chromosome set in the 8-hours of S-phase of the cell cycle. For example, the human diploid genome has 46 chromosomes (6 x 10 9 basepairs). The shortest chromosomes are ~50 Mbp long and so could not possibly be replicated from one origin. Additionally, the rate of replication fork movement is slower, only ~100 base/second. Thus, eukaryotes contain multiple origins of replication distributed over the length of each chromosome to enable the duplication of each chromosome within the observed time of S-phase (Fig 2.9).


3.3: Devices to Aid Manual Inspection
The previous section outlined procedures for conducting manual inspections of infrastructure components. There are a variety of devices that can be used to aid such manual inspections.
Rulers and gauges are often useful for measuring lengths and depths of components or cracks. Mobile (battery) powered drills, wrenches and wire brushes also can be useful.
Devices to aid visual inspection in difficult or impossible to reach vantage points can be particularly useful. Hand-held mirrors and lamps are a simple example of such aid, but devices that are more elaborate exist. Figure 3.4 shows a video camera that suited for use in a small pipe that can provide pictures of the interiors so that corrosion or root intrusions may be identified. More elaborate pipeline ‘pigs’ can be stabilized in the desired location within the pipe and can have self-locomotion.
Unmanned flying vehicles (commonly called ‘drones’) provide another means of aiding manual inspections (Figure 3.5). These devices can simplify inspection of a variety of infrastructure components such as bridges or power lines. Drones are regulated for safety reasons and inspectors using such devices must determine the acceptable use of drones in any particular location. In addition to unmanned flying devices, underwater aids for visual inspection are also available, but visibility for such devices may be an issue.


19.8: Checklist for Blood Glucose Monitoring
Use the checklist below to review the steps for completion of “Blood Glucose Monitoring.”
Steps
Disclaimer: Always review and follow agency policy regarding this specific skill.
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Prepare before completing the procedure:
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Review the patient’s medical history and current medications.
- Note if the patient is receiving anticoagulant therapy. Anticoagulant therapy may result in prolonged bleeding at the puncture site and require pressure to the site.
- Assess the patient for signs and symptoms of hyperglycemia or hypoglycemia to correlate data to pursue acute action due to an onset of symptoms.
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Determine if the test requires special timing, for example, before or after meals.
- Blood glucose monitoring is typically performed prior to meals and the administration of antidiabetic medications.
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Review the patient’s medical history and current medications.
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Gather supplies: nonsterile gloves, alcohol swab, lancet, 2″ x 2″ gauze or cotton ball, reagent strips, and blood glucose meter.
- Determine if the blood glucose meter needs to be calibrated according to agency policy to ensure accuracy of readings.
- Read and understood the manufacturer’s instructions and agency policy for the blood glucose meters.
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Perform safety steps:
- Perform hand hygiene.
- Check the room for transmission-based precautions.
- Introduce yourself, your role, the purpose of your visit, and an estimate of the time it will take.
- Confirm patient ID using two patient identifiers (e.g., name and date of birth).
- Explain the process to the patient and ask if they have any questions.
- Be organized and systematic.
- Use appropriate listening and questioning skills.
- Listen and attend to patient cues.
- Ensure the patient’s privacy and dignity.
- Assess ABCs.
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Have the patient wash their hands with soap and warm water, and position the patient comfortably in a semi-upright position in a bed or upright in a chair. Encourage the patient to keep their hands warm. Washing reduces transmission of microorganisms and increases blood flow to the puncture site.
- Agency policy may require use of an alcohol swab to clean the puncture site.
- Ensure that the puncture site is completely dry prior to skin puncture.
- Remove a reagent strip from the container and reseal the container cap to keep the strips free from damage from environmental factors. Do not touch the test pad portion of the reagent strip.
- Follow the manufacturer’s instructions to prepare the meter for measurement.
- Place the unused reagent strip in the glucometer or on a clean, dry surface (e.g., paper towel) with the test pad facing up, based on manufacturer recommendations.
- Apply nonsterile gloves.
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Keep the area to be punctured in a dependent position. Do not milk or massage the finger site:
- Dependent position will increase blood flow to the area.
- Do not milk or massage the finger because it may introduce excess tissue fluid and hemolyze the specimen.
- Warm water, dangling the hand for 15 seconds, and a warm towel stimulate the blood flow to the fingers.
- Avoid having the patient stand during the procedure to reduce the risk of fainting.
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Select the appropriate puncture site. Cleanse the site with an alcohol swab for 30 seconds and allow it to dry. Perform the skin puncture with the lancet, using a quick, deliberate motion against the patient’s skin:
- The patient may have a preference for the site used. For example, the patient may prefer not to use a specific finger for the skin puncture. However, keep in mind their preferred site may be contraindicated. For example, do not use the hand on the same side as a mastectomy.
- Avoid fingertip pads; use the sides of fingers.
- Avoid fingers that are calloused, have broken skin, or are bruised.
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Gently squeeze above the site to produce a large droplet of blood.
- Do not contaminate the site by touching it.
- The droplet of blood needs to be large enough to cover the test pad on the reagent strip.
- Wipe away the first drop of blood with gauze.
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Transfer the second drop of blood to the reagent strip per manufacturer’s instructions:
- The test pad must absorb the droplet of blood for accurate results. Smearing the blood will alter results.
- The timing and specific instructions for measurement will vary between blood glucose meters. Be sure to read the instructions carefully to ensure accurate readings.
- Apply pressure, or ask the patient to apply pressure, to the puncture site using a 2″ x 2″ gauze pad or clean tissue to stop the bleeding at the site.
- Read the results on the unit display.
- Turn off the meter and dispose of the test strip, 2″ x 2″ gauze, and lancet according to agency policy. Use caution with the lancet to prevent an unintentional sharps injury.
- Remove gloves.
- Perform hand hygiene.
- Assist the patient to a comfortable position, review test results with the patient, ask if they have any questions, and thank them for their time.
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Ensure safety measures when leaving the room:
- CALL LIGHT: Within reach
- BED: Low and locked (in lowest position and brakes on)
- SIDE RAILS: Secured
- TABLE: Within reach
- ROOM: Risk-free for falls (scan room and clear any obstacles)
- Document the results and related assessment findings. Report critical values according to agency policy, such as values below 70 or greater than 300, and any associated symptoms. Read more about hypoglycemia and hyperglycemia in the “ Blood Glucose Monitoring ” section of this chapter.


2.4.14: Soundness with Identity predicate
\(\Log{LK}\) with initial sequents and rules for identity is sound.
Proof. Initial sequents of the form \({} \Sequent \eq[t][t]\) are valid, since for every structure \(\Struct M\) , \(\Sat{M}{\eq[t][t]}\) . (Note that we assume the term \(t\) to be closed, i.e., it contains no variables, so variable assignments are irrelevant).
Suppose the last inference in a derivation is \(=\) . Then the premise is \(\eq[t_1][t_2], \Gamma \Sequent \Delta, A(t_1)\) and the conclusion is \(\eq[t_1][t_2], \Gamma \Sequent \Delta, A(t_2)\) . Consider a structure \(\Struct M\) . We need to show that the conclusion is valid, i.e., if \(\Sat{M}{\eq[t_1][t_2]}\) and \(\Sat{M}{\Gamma}\) , then either \(\Sat{M}{C}\) for some \(C \in \Delta\) or \(\Sat{M}{A(t_2)}\) .
By induction hypothesis, the premise is valid. This means that if \(\Sat{M}{\eq[t_1][t_2]}\) and \(\Sat{M}{\Gamma}\) either (a) for some \(C \in \Delta\) , \(\Sat{M}{C}\) or (b) \(\Sat{M}{A(t_1)}\) . In case (a) we are done. Consider case (b). Let \(s\) be a variable assignment with \(s(x) = \Value{t_1}{M}\) . By Proposition 5.12.3 , \(\Sat[,s]{M}{A(t_1)}\) . Since \(\varAssign{s}{s}{x}\) , by Proposition 5.13.3 , \(\Sat[,s]{M}{A(x)}\) . Since \(\Sat{M}{\eq[t_1][t_2]}\) , we have \(\Value{t_1}{M} = \Value{t_2}{M}\) , and hence \(s(x) = \Value{t_2}{M}\) . By applying Proposition 5.13.3 again, we also have \(\Sat[,s]{M}{A(t_2)}\) . By Proposition 5.12.3 , \(\Sat{M}{A(t_2)}\) . ◻


2.5: Contributions to Science
There are also a number of
Jewish
and
Christian
traditions. The Bible’s Old and New Testaments contain numerous astronomical references. The Jewish Bible, or Old Testament, refers to stars, constellations, and eclipses. The Christian New Testament has astronomical references, such as the Star of Bethlehem, denoting the birth of Jesus. The
Greeks
were credited with development of scientific principles, starting around 500 BC. Alexandria, Egypt was a great library and research center that opened around 300 BC, and survived until it was destroyed by fire. A couple of Greeks, to note, included Eratosthenes, who determined Earth’s circumference by measuring the Sun’s shadow at two points on Earth around 240 BC, and Hipparchus, who developed a stellar brightness magnitude scale that is still in use today. There were also a number of significant
Islamic
contributions from around the 8th and 9th Centuries AD. Muslims kept and translated historical records, developed Algebra, and developed many constellation and star names. Some examples of star names still used today are Aldebaran and Algol (Al: “the”). The researchers were not only Muslims, but also included Jews and Christians. Hindus also worked with the team of researchers and indirectly with the Chinese.


1.5: Types of Statistical Studies (3 of 4)
Learning Objectives
- Based on the study design, determine what types of conclusions are appropriate.
We now focus more closely on studies that investigate a relationship between two variables. In these types of studies, one variable is the explanatory variable , and the other is the response variable . To establish a cause-and-effect relationship, we want to make sure the explanatory variable is the only thing that impacts the response variable. We therefore get rid of all other factors that might affect the response. Then we manipulate the explanatory variable. Our goal is to see if it really does affect the response.
In an observational study, researchers may take steps to reduce the influence of these other factors on the response. But it is difficult in an observational study to get rid of all the factors that may have an influence. In addition, the researchers do not manipulate the explanatory variable to see if it affects the response. They just collect data and look for an association between the two variables. For these reasons, observational studies do not give convincing evidence of a cause-and-effect relationship.
In an experiment, researchers use a variety of techniques to eliminate the influence of these other factors. Then they manipulate the explanatory variable to see if it affects the response. For this reason, experiments give the strongest evidence for a cause-and-effect relationship.
Example
Hormone Replacement Therapy
When women go through menopause, the production of hormones in their bodies changes. The hormonal changes can cause a variety of symptoms that may be reduced by hormone replacement therapy. In the 1980s, hormone replacement therapy was common in the United States.
In the early 1990s, observational studies suggested that hormone replacement therapy had additional benefits, including a reduction in the risk of heart disease. In these observational studies, researchers compared women who took hormones to those who did not take hormones. Health records showed that women taking hormones after menopause had a lower incidence of heart disease. But women who take hormones are different from other women. They tend to be richer and more educated, to have better nutrition, and to visit the doctor more frequently. These women have many habits and advantages that contribute to good health, so it is not surprising that they have fewer heart attacks. But can we conclude from these studies that the hormones caused the reduction in heart attacks? No. The results are confounded by the influence of these other factors.
In 2002, the Women’s Health Initiative sponsored a large-scale, well-designed experiment to study the health implications of hormone replacement therapy. In this experiment, researchers randomly assigned over 16,000 women to one of two treatments. One group took hormones. The other group took a placebo . A placebo is a pill with no active ingredients that looks like the hormone pill. The experiment was double-blind . Blind means that women did not know if they were receiving hormones or the placebo. Double-blind means that the information was coded, so researchers administering the pills did not know which treatment the women received. After 5 years, the group taking hormones had a higher incidence of heart disease and breast cancer. This is exactly the opposite result from the result found in the observational studies! In fact, the differences were so significant that the researchers ended the experiment early. The National Institutes of Health declared that the observational studies were wrong. Hormone replacement therapy to treat menopausal symptoms is now rarely used.
What’s the Main Point?
An observational study can provide evidence of a link or an association between two variables. But other factors may also influence the results. These other factors are called confounding variables . The influence of confounding variables on the response variable is one of the reasons that an observational study gives weak, and potentially misleading, evidence of a cause-and-effect relationship. A well-designed experiment takes steps to eliminate the effects of confounding variables, including random assignment of people to treatment groups, use of a placebo, or blind conditions. Using these precautions, a well-designed experiment provides convincing evidence of cause-and-effect.
Contributors and Attributions
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


2.1: Introduction to Streak Plate Techniques
- Discuss the importance of creating streak plates in microbiology.
-
Define "colony" and understand how isolated colonies are formed.
- Discuss binary fission.
- Learn how to calculate generation times.
- Characterize colony morphology based on colony shape, elevation, margin, size, and color.
- Complete a Quadrant Streak, "Quad Streak", and describe the results.
How Microbes Grow
Microbes, like bacteria, are ubiquitous and found everywhere throughout life. We can support the growth of microbes through growth media, as described in "1.4: Aseptic Technique", and control their growth cycles. Manipulation of bacterial growth allows for clinical diagnosis and microbiology research.
Bacteria increase their cell population via a replication cycle called binary fission. Binary fission is characterized by 1 parent cell increasing their cell size, duplicating structural components, and copying their bacterial genome (DNA). Once a parent cell has copied all cell components and DNA, the center of the large parent cell will begin to constrict until a physical wall of separation, called a septum, is formed. The septum will cleave the cytoplasm, cell membrane, and cell wall forcing the enlarged parent cell into two new, separate daughter cells. Binary fission will increase a bacteria's population on growth media, creating colonies. .
Figure 1. A) Bacteria replicate through a process called binary fission. B) One parent cell copies its genome and internal structures, splits its cell body with a septum, and produces two identical daughter cells.
A colony is a macroscopic mass of bacterial cells undergoing binary fission that is easily visible with the human eye.  Bacterial colonies from the same species should all have the same appearance as they contain exact copies of the same bacterial genome. Observation of these colony characteristics, or colony morphology, assists with identification of bacterial species in clinical samples. To properly observe colony morphologies, pure, isolated colonies are necessary. A pure, isolated colony is a single colony that does not touch other cell growth and is not contaminated with other microbial species. Pure colonies can be produced by adhering to the rules of aseptic technique and using proper streak plate techniques.
Figure 2. This plate shows bacteria streaked across 4 different quadrants. The bacteria at the top of the plate are in an area of lower cell density and are capable of forming colonies. Pure, isolated colonies should not touch other bacterial growth or be contaminated.
The rate of bacterial growth, binary fission, in a bacterial colony can be determined by a formula called Generation Time. Generation time in bacteria is the amount of time is takes for a bacterial population to go through one round of binary fission and double the total population number (which is why it may also be called the doubling time). All bacteria species have their own, unique generation time. For example, E. coli's generation time is 20 minutes whereas M. tuberculosis' generation time is 15 to 20 hours. The speed of population doubling has impacts on the pathogenicity of the bacterial species and potential host colonization. The rate of population growth and generation time are determined with the following formula:
Step 1:
Total Growth Time/Generation Time= Number of Generations
Step 2:
2^Number of Generations X (Number of Original Cells)= Total Number of Cells
For Example:
Escherichia coli , under optimum conditions, has a generation time of 20 minutes. If an experiment started with only 10 E. coli cells and allowed them to grow for 2 hours, what would be the total cell population at the end of the experiment?
**Make sure units of time match, hours to hours, minutes to minutes, etc. Convert units of time prior to using the growth formula.
2hours x 60mins = 120mins
120mins / 20mins = 6 generations
2^6 x 10 cells = 640 cells, The total cell population after 2 hours in 640 E. coli cells.
Colony Morphology
As bacteria replicate and produce colonies, these colonies will have unique growth patterns that can be used to identify the bacteria's species. The characteristics of these growth patterns are called the colony morphology and include:
- Colony Form: overall general shape of the colony from an overhead view.
- Colony Elevation: shape of the colony from a side-view.
- Colony Margin: magnified observation of the colony's edge.
- Additional Characteristics: size, color, and smell.
Figure 3. Colony morphology describes the appearance of one macroscopic bacterial colony. The above illustrations show common colony morphologies for overall colony shape (colony form), height of the colony (elevation), and specific pattern of the colony's edge (margin).
Figure 4. The above images show examples of real life colony morphologies. We can also compare colony color and size in addition to form and elevation to best identify unknown species.
Quadrant Streak Plate Technique
In order to produce pure, isolated colonies so colony morphologies can be observed, streak plate techniques are used. The purpose of streak plating is to spread bacteria across a media surface, decrease cell density as the bacteria are spread, and produce individual colonies. There are multiple methods of streak plating including zig-zag streaks, T-streaks, and quad streaks. In this course, we will complete the quadrant or quad streak plate technique.
The quad streak technique decreases bacterial cell density by streaking 4 separate zones on an agar plate.
| Figure 5. To decrease bacterial cell density and produce isolated colonies, the quad streak can be completed following these 4 steps: Step 1 : Bacteria from a stock plate should be spread in 1 quadrant of the plate in a back-and-forth (zig- zag) pattern. Step 2 : A sterile loop is used to drag a small amount of bacteria from Quad 1 into Quad 2 and spread in a zig-zag pattern. Step 3 : A sterile loop is used to drag a small amount of bacteria from Quad 2 to Quad 3 and spread in a zig-zag pattern. Step 4 : A sterile loop is used to drag a small amount of bacteria from Quad 3 to Quad 4 and spread in a zig-zag pattern. Quad 4 should have the lowest bacterial density and produce isolated colonies. |
To complete a Quad Streak , use a sterile inoculation loop or swab to collect a small sample of bacteria from a stock plate and spread it in 1 area (quadrant 1) of an agar plate in a tight zig-zag pattern. No more bacteria should be collected from the stock plate. Sterilize the inoculation tool and drag a small area of bacteria from quadrant 1 into a new, un-inoculated area of the plate (quadrant 2). Spread the bacteria across quadrant 2 in a tight zig-zag pattern. Sterilize the inoculation tool again, drag a small amount of bacteria from quadrant 2 into the new quadrant 3 area, and spread the bacteria in a less-tight zig-sag pattern to start and create space for individual colonies to form. Finally, sterilize the inoculation tool one last time, drag a small amount of bacteria from quadrant 3 into the new quadrant 4, and spread the bacteria in a loose zig-zag pattern to ensure proper space for isolated colony growth. After incubation, there should be a lot of bacterial growth in quadrant 1 as this region has the highest cell density (coming from the original stock plate sample). Quadrant 4 should have the least amount of bacterial growth as cell density has been significantly decreased by streaking throughout the other quadrants and individual, isolated colonies should form.
Diagnosis of Strep Throat
Have you ever gone to the doctor’s office with a sore throat and was diagnosed with strep throat? You are not alone as strep throat is the most common cause of bacterial pharyngitis in humans. This disease is caused by Streptococcus pyogenes , a Group A strep bacterium. This bacterium thrives in the warm, moist environment of the throat, and often induces sore throat, difficulty swallowing, and fever in patients. Despite being more common in children, it can affect people of all ages.
But how does a doctor determine whether your sore throat is caused by strep? There are two main types of tests that are used to check for strep A infections: the Rapid Strep Test and the Throat Culture Test. The Rapid strep test looks for strep A antigens, or the substances in the bacteria that initiate an immune response. This test can provide results in 10-20 minutes, but it is less reliable than the throat culture test. The throat culture test is considered the “gold standard” diagnostic test for strep throat. This test involves swabbing the back of a patient’s throat and streak plating the swab on an agar plate. Doctor’s can then determine if the patient is infected with Streptococcus pyogenes, based on the characteristics of the bacterial culture that grows in the plate. This often results in more accurate diagnosis; however, the culture must be incubated for 24 to 48 hours resulting in a slower return time for the diagnosis. This highlights the importance of microbiology techniques when diagnosing patients with a common bacterial infection, such as strep.
Figure 6. The above image shows Streptococcus pyogenes colonies grown on a blood agar plate. Observing the characteristics of the bacterial colonies can help identify an organism.
Attributions
"Microbiology Laboratory Manual: Labs, 1.8 Plating on Petri Plates for Isolation" by Dr. Rosanna Hartline , West Hills College Lemoore is licensed under, LibreTexts: Biology is licensed under CC BY-NC-SA 4.0
"Microbiology Labs II: Microbial Growth, Generation Time" by Dr. Gary Kaiser , Community College of Baltimore County, Catonsville Campus, LibreTexts: Biology is licensed under CC BY 4.0
"Microbiology Lab Manual: Experiment 4, Colony and Cell Morphology; Simple Stains" by Sheri Miraglia , City College of San Francisco , LibreTexts: Biology is licensed under CC BY-SA 4.0
"Microbiology Labs 1: 8, Bacterial Colony Morphology " by Jackie Reynolds , Dallas College, Richland Campus , LibreTexts: Biology is licensed under CC BY-NC-SA 4.0
"Medical Tests: Strep A Test" by MedlinePlus , Use of MedlinePlus Materials Statement is in the Public Domain
"Group A Strep Infection: Testing for Strep Throat or Scarlet Fever" by Center for Disease Control , CDC Material Statement is in the Public Domain
"Group A Strep Infection: Clinical Guidance for Group A Streptococcal Pharyngitis" by Center for Disease Control , CDC Material Statement is in the Public Domain
Image Citations
Figure 1, "Microbiology Textbook: Chapter 9, Microbial Growth" by OpenStax , Digital ISBN 13: 978-1-947172-23-4 is licensed under CC BY 4.0
Figure 2, "Microbiology Laboratory Manual: Labs, 1.8 Plating on Petri Plates for Isolation" by Dr. Rosanna Hartline , West Hills College Lemoore is licensed under, LibreTexts: Biology is licensed under CC BY-NC-SA 4.0
Figure 3, Modified From:
- "Microbiology Lab Manual: Experiment 4, Colony and Cell Morphology; Simple Stains" by Sheri Miraglia , City College of San Francisco , LibreTexts: Biology is licensed under CC BY-SA 4.0
Figure 4, Modified From:
- "Microbiology Labs 1: 8, Bacterial Colony Morphology " by Jackie Reynolds , Dallas College, Richland Campus , LibreTexts: Biology is licensed under CC BY-NC-SA 4.0
Figure 5, Modified From:
- "Microbiology Laboratory Manual: Labs, 1.8 Plating on Petri Plates for Isolation" by Dr. Rosanna Hartline , West Hills College Lemoore is licensed under, LibreTexts: Biology is licensed under CC BY-NC-SA 4.0
- Streak plates 1.svg by Reytan, Wikimedia Commons is in the Public Domain
- Streak plates 2.svg by Reytan, Wikimedia Commons is in the Public Domain
- Streak plates 3.svg by Reytan, Wikimedia Commons is in the Public Domain
- Streak plates 4.svg by Reytan, Wikimedia Commons is in the Public Domain
Figure 6, "Streptococcus_pyogenes_(Lancefield_Group_A)_on_Columbia_Horse_Blood_Agar_-_Detail_-_(1).jpg" by Nathan Reading , Wikimedia Commons is in the Public Domain