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
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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.] y
2 Prove or disprove: If x and y are both irrational numbers, then x is also irrational. Proof by contradiction – or is it a counterexample? Suppose the proposition is true. y
Let x = y = √2, so that x = √2
√2
is irrational (*)
√2
Now let x' = √2 , and let y' = √2, so that x' and y' are irrational, by (*). Then x'
y'
√2 √2
= (√2 )
√2 √2
But (√2 )
must be irrational.
2
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!
= √2 = 2, which is rational – a contradiction.
Therefore the proposition is false. Interesting thought: If we write the proposition in logic terms, i.e. y
∀ x, y ∈ ℝ: {(x, y ∉ ℚ) ⇒ (x ∉ ℚ)} then we have disproved it by showing that y
y
∃ x, y ∈ ℝ: ¬{(x, y ∉ ℚ) ⇒ (x ∉ ℚ)}, or equivalently, ∃ x, y ∈ ℝ: {(x, y ∉ ℚ) ⋏ (x ∈ ℚ)} So this is actually a proof by counterexample. What's interesting is that it's not clear which of the two expressions, y
√2
y
√2 √2
x = √2 or x = (√2 ) , provides the counterexample. So we've actually proved that a counterexample exists, but we haven't shown what it is.
Shirleen Stibbe
http://www.shirleenstibbe.co.uk
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)3 + 5(3k + 1) = (27k3 + 27k2 + 9k + 1) + (15k + 5)
3k + 1
= 27k3 + 27k2 + 24k + 6 = 3(9k3 + 9k2 + 8k + 2) 2
(3k + 2)3 + 5(3k + 2) = (27k3 + 54k2 + 36k + 8) + (15k + 10)
3k + 2
= 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 + a + a 2 + ... + a n −1 = Proof by inspiration Let
S = 1 + a + a 2 +...+ a n−1
Then
Sa = a + a 2 +...+ a n−1 + a n
and
S (a −1) = a n −1
Since a ≠ 1, a − 1 ≠ 0, so
S=
an − 1 , where a ≠ 1 is a real number. a −1
Note: We showed them this proof after we'd made them slog through a proof by induction.
a n −1 a −1
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: 2
Suppose
! p$ ! p$ a # & + b # & + c = 0 . We multiply by q2 to get "q% "q% ap2 + bpq + cq2 = 0
Then
2
(*)
2
ap = − bpq − cq = 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.
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, Since x ∈ A, we have y ∈ f(A), and
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.
since x ∈ B, y ∈ f(B).
Counterexample:
so x ∈ A and x ∈ B.
(ii)
Therefore y ∈ f(A) ∩ f(B).
f(x) = x2 and A = {2, 3}, B = {−3, 2, 4}
This shows that f(A ∩ B) ⊆ f(A) ∩ f(B).
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.
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).
Shirleen Stibbe
http://www.shirleenstibbe.co.uk