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.
(e) Let
“if z has the property that
is an integer, then z + 1 z z m + 1 is also an integer for all m , z m ”. 0 ≤ m ≤ n
•
• Suppose that
“if z has the property that
is an integer, then z + 1 z z m + 1 is also an integer for all m , z m ”. 0 ≤ m ≤ k
We wish to prove that
If
“
z m + 1 is also an integer for all m , z m ”. 0 ≤ m ≤ k
So to prove that
“
z k + 1 + 1 z k is an integer”. + 1
By the Binomial Theorem:
The LHS is an integer (since
Hence
If we combine these two bullet points, we have proved that “
Note:
If
242 .
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”.
Then link the two cycles into a single cycle of length
, by eliminating the final step 2 k + 1 in the first cycle, and v 1 v 2 ⋯ v k 0 → 00 ⋯ 00 v 1 v 2 ⋯ v k 1 → 00 ⋯ 01 in the second cycle, reversing the first cycle, and inserting the joins
00 ⋯ 00 → 00 ⋯ 01 and v 1 v 2 ⋯ v k 1 → v 1 v 2 ⋯ v k 0 to produce a single cycle of the required kind joining all
sequences of length 2 k + 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-)