Shirleen Stibbe http://www.shirleenstibbe.co.uk

Example Proofs: Miscellaneous methods Proofs Workshop Note: the proofs in this handout are not necessarily in the same form as they were presented at the workshop. In particular, any errors you spot here are entirely accidental, not deliberate.

1 Is it possible to choose 55 different integers between 1 and 100 inclusive, so that no two numbers differ by 12?

The pigeonhole principle. If m letters are placed in n pigeonholes and m > n, then at least one pigeonhole must contain more than one letter. This is known as the pigeonhole principle. This may seem obvious, but there are occasions when this sort of argument helps us with a proof.

Solution:

No. To prove this, we arrange the numbers 1 to 100 in a rectangular grid with 12 columns, as follows:

1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32 33 34 35 36

37 38 39 40 41 42 43 44 45 46 47 48

49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70 71 72

73 74 75 76 77 78 79 80 81 82 83 84

85 86 87 88 89 90 91 92 93 94 95 96

97 98 99 100

Any two numbers in the same column differ by a multiple of 12, and no two numbers from different columns differ by a multiple of 12.

If no two numbers can differ by 12, I must avoid choosing two numbers next to each other in the same column. The shading shows that I can choose a maximum of 5 numbers from each of the first 4 columns, and 4 numbers from each of the last 8 columns, giving a total of 20 + 32 = 52 numbers that satisfy the criterion.

So if I choose 55 numbers, then at least two will differ by 12. [I cannot fit 55 letters into 52 pigeonholes so that no pigeonhole contains more than one letter.]

2 Prove or disprove: If x and y are both irrational numbers, then xy is also irrational. Proof by contradiction – or is it a counterexample? Suppose the proposition is true.

Let x = y = √2, so that xy = √2√2 is irrational (*)

Now let x' = √2√2, and let y' = √2, so that x' and y' are irrational, by (*).

Then x' y' = (√2√2)√2 must be irrational.

But (√2√2)√2 = √22 = 2, which is rational – a contradiction. Therefore the proposition is false.

Interesting thought: If we write the proposition in logic terms, i.e.

∀  x,  y  ∈  ℝ:  {(x,  y  ∉  ℚ)  ⇒  (xy ∉  ℚ)}  then  we  have  disproved  it  by  showing  that      

∃  x,  y  ∈  ℝ:  ¬{(x,  y  ∉  ℚ)  ⇒  (xy ∉  ℚ)},  or  equivalently,    ∃  x,  y  ∈  ℝ:  {(x,  y  ∉  ℚ)  ⋏  (xy ∈  ℚ)}  So this is actually a proof by counterexample. What's interesting is that it's not clear which of the two expressions,

xy = √2√2 or xy = (√2√2)√2, provides the counterexample. So we've actually proved that a counterexample exists, but we haven't shown what it is.

Note: the students really liked this proof. In the 'Were you convinced' session that followed the Example proofs, they rated this one almost as highly as the (incorrect) direct proof!

Shirleen Stibbe http://www.shirleenstibbe.co.uk

11

−

−=aaSn

3 Prove that n3 + 5n is divisible by 3, where n is an integer. Proof by exhaustion:

Any integer leaves a remainder of 0, 1 or 2 when divided by 3. So we may write n = 3k + m, where k is an integer and m = 0, 1 or 2.

The table below shows all possible cases:

m n n3 + 5n

0 3k (3k)3 + 5(3k) = 27k3 + 15k

= 3(9k3 + 5k)

1 3k + 1 (3k + 1)3 + 5(3k + 1) = (27k3 + 27k2 + 9k + 1) + (15k + 5)

= 27k3 + 27k2 + 24k + 6

= 3(9k3 + 9k2 + 8k + 2)

2 3k + 2 (3k + 2)3 + 5(3k + 2) = (27k3 + 54k2 + 36k + 8) + (15k + 10)

= 27k3 + 54k2 + 51k + 18

= 3(9k3 + 18k2 + 17k + 6)

Since the terms in the final bracket of each row of the table is an integer, n3 + 5n is divisible by 3.

Note: it is easy to show that n3 + 5n is always even, so we may conclude that n3 + 5n is divisible by 6.

4 Prove that for all positive integers n, 1 11

2 1+ + + + =−

−−a a a a

an

n... , where a ≠ 1 is a real number.

Proof by inspiration

Let 12 ...1 −++++= naaaS

Then and Since a ≠ 1, a − 1 ≠ 0, so 5 Let f(x) = ax2 + bx + c, where a, b, c are integers, a ≠ 0, c ≠ 0, and let r = p/q, where p and q are integers, with

no common factors. Prove that if r is a solution to the equation f(x) = 0, then q must divide a, and p must divide c.

Direct proof:

Suppose a pq!

"#

$

%&

2

+b pq!

"#

$

%&+ c = 0 . We multiply by q2 to get

ap2 + bpq + cq2 = 0 (*)

Then ap2 = − bpq − cq2 = q(− bp − cq).

q divides the right hand side, so q must divide ap2, and p and q have no common factors. Therefore q must divide a.

Similarly, rewriting (*) as ap2 + bpq = −cq2, it is easily shown that p must divide c.

Extension: This is called the Rational Zeros Theorem. Prove it for a polynomial of degree n.

1)1( −=− naaS

nn aaaaSa ++++= −12 ...

Note: We showed them this proof after we'd made them slog through a proof by induction.

Shirleen Stibbe http://www.shirleenstibbe.co.uk

6 Let f be a function and let A and B be sets in the domain of f. Prove that: (i) f(A ∩ B) ⊆ f(A) ∩ f(B)

(ii) f(A ∪ B) = f(A) ∪ f(B) Proof (i) Suppose y ∈ f(A ∩ B).

This means that y = f(x) for some x ∈ A ∩ B, so x ∈ A and x ∈ B. Since x ∈ A, we have y ∈ f(A), and since x ∈ B, y ∈ f(B).

Therefore y ∈ f(A) ∩ f(B). This shows that f(A ∩ B) ⊆ f(A) ∩ f(B). (ii) Suppose y ∈ f(A ∪ B)

This means that y = f(x) for some x ∈ A ∪ B, so either x ∈ A or x ∈ B. If x ∈ A then y ∈ f(A) and if x ∈ B then y ∈ f(B).

Therefore y ∈ f(A) ∪ f(B). This shows that f(A ∪ B) ⊆ f(A) ∪ f(B). (1) Now suppose that y ∈ f(A) ∪ f(B).

Then y = f(x) for some x ∈ A or y = f(x) for some x ∈ B. In either case, x ∈ A ∪ B, so y ∈ f(A ∪ B)

and this shows that f(A) ∪ f(B) ⊆ f(A ∪ B). (2) It follows from (1) and (2) that f(A ∪ B) = f(A) ∪ f(B).

In the initial presentation, we claimed that the result of part (i) was that f(A ∩ B) = f(A) ∩ f(B) (which is false) to point out a common error, i.e. forgetting to show that each set is a subset of the other to prove equality. Counterexample: f(x) = x2 and A = {2, 3}, B = {−3, 2, 4} f(A ∩ B} = {4} and f(A) ∩ f(B) = {4, 9} We invited them to do part (ii) during the following session, to bring home the point.