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