Example proofs; Direct

Example Proofs: Direct

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 Prove that for all positive real numbers a, b,

1

2 ( a + b)

Direct proof

≥

ab

Note: We presented this the wrong way round – i.e. started by assuming the result

Note:

( a − b) 2 ≥ 0 ⇒ a 2 − 2ab + b 2 ≥ 0

This starts with a statement that is true and deduces the required result.

for all real a, b

⇒ a 2 + 2ab + b 2 ≥ 4ab adding 4ab to both sides ⇒ (a + b) 2 ≥ 4ab ⇒ (a + b) ≥ 2 ab ⇒ 1 2 (a + b) ≥ ab

since a and b are positive as required

It would be incorrect to write it the other way round, i.e. to assume the proposition is true and deduce a true statement - though that's how I found the true statement initially (on the back of an envelope).

2 a) Prove that if m − n is even, then mn is the difference of two squares, where m and n are integers. b) Is the converse true? a) Direct proof: If m − n is even, then m − n = 2k for some integer k. So, m = n + 2k.

Then mn = (n + 2k)n = n2 + 2kn = (n2 + 2kn + k2) − k2 = (n + k)2 − k2 .

Since n + k and k are both integers, this shows that if m − n is even, then mn is the difference of two squares. b) Is the converse true?

The converse is: If mn is the difference of two squares, then m − n is even.

Setting m = 4 and n = 3, we have mn = 12 = 16 − 4 = 42 − 22, so mn is the difference of two squares.

But m − n = 4 − 3 =1 is odd, so the converse is FALSE. Note: This shows that the proposition: m − n is even, if and only if mn is the difference of two squares is FALSE.

3 Prove that x +

1 ≥ 2 for all x > 0. x

Direct proof:

For all x > 0, we have

(x − 1)2 ≥ 0 ⇒ x 2 − 2x + 1 ≥ 0 2

⇒ x + 1 ≥ 2x 1 ⇒ x + ≥ 2 (since x > 0 ) x

Shirleen Stibbe

Presented 'the wrong way round'. To see how we tried to convince them that the approach is wrong, have a look under Common Errors on this page of my web site.

http://shirleenstibbe.co.uk/proofs

Note: This starts with a statement that is true and deduces the required result.

It would be incorrect to write it the other way round, ie to assume the proposition is true and deduce a true statement - though that's how I found the true statement initially (on the back of an envelope).

http://www.shirleenstibbe.co.uk

4 Let a and b be positive integers. If a is a multiple of b, then for any positive integer c, c and b are co-prime if c and a are co-prime.

Direct proof Since a is a multiple of b, we may write a = pb for some positive integer p.

Suppose c and a are co-prime.

Let q be a positive integer that is a factor of both c and b.

Then we may write b = qr for some positive integer r.

Then we have a = pqr, so q divides a.

Note: We presented this as a biconditional (if and only if), then showed them why it was wrong in the 'Were you convinced' session. See details under "Common Errors' here on my web site: http://shirleenstibbe.co.uk/proofs

Then q is a factor of both a and c.

But c and a are co-prime.

So q = 1.

Hence 1 is the only common factor of c and b and so c and b are co-prime.

Note: Two numbers are co-prime if their only common factor is 1.

5

Let a, b be integers. Then ab is even if a + b is odd.

Proof:

The proposition is of the form p ⇒ q, where p is the statement: a + b is odd, q is the statement: ab is even We prove the theorem in two different ways.

Direct proof:

Suppose a + b is odd. Then one of a or b must be even and the other odd, since the sum of two even or two odd integers is even.

Without loss of generality, we may assume that a is even and b is odd, so that a = 2j for some integer j, and b = 2k + 1, for some integer k.

Then ab = 2j(2k + 1), which is even, so we have proved that if a + b is odd, then ab is even (p ⇒ q).

Contrapositive proof:

We prove the logically equivalent statement: if q is false then p is false (¬q ⇒ ¬p), that is, if ab is odd, then a + b is even.

Suppose ab is odd. Then a and b must both be odd integers, since the product of an even integer and any other integer is even. We therefore have a = 2j + 1 and b = 2k + 1 for some integers j and k.

Then a + b = (2j + 1) + (2k + 1) = 2j + 2k + 2 = 2(j + k + 1), which is even.

We have proved that if ab is odd, then a + b must be even (¬q ⇒ ¬p).

Therefore if a + b is odd, then ab is even.

Note: These two approaches both prove the same thing, ie, ab is even if a + b is odd. The converse (q ⇒ p) of the

statement is a + b is odd if ab is even, and is clearly false.

(Conterexample: if a = b = 2, then ab and a + b are both even.)

So the statement ab is even if and only if a + b is odd is false.

Shirleen Stibbe

http://www.shirleenstibbe.co.uk