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 there is no smallest real number strictly greater than

1

2

.

Let S be the set of numbers strictly greater than

2

Notes Labelling the set makes it easier to structure the proof

.

Suppose S has a smallest element, x. Consider the number y = Since x >

1

2,

Also, since

1

1 2

2+x

< x,

1

2

Assume the proposition is false.

( 2 + x) . 1

y is the average of ½ and x

> 1, and so y = 1 2 (1 2 + x) > 1 2 is an element of S. 1

2 + x < 2 x, ,

and so y =

1

2

( 2 + x) < x . 1

Establish the contradiction, i.e. y is in S, and y is less than x.

Hence x cannot be the smallest element of S. This is a contradiction. Therefore there is no smallest number strictly greater than

1

2

.

2 Let a, b and c be odd integers. Show that ax 2 + bx + c = 0 has no solution in the set of rational numbers. Proof by contradiction Suppose there is a rational number, x, for which ax 2 + bx + c = 0 Assume x is in lowest terms, ie x = p/q where q ≠ 0, and p and q are coprime integers (ie they have no common factors greater than 1). Then 2

⎛ p ⎞ ⎛ p ⎞ a⎜⎜ ⎟⎟ + b⎜⎜ ⎟⎟ + c = 0 ⎝ q ⎠ ⎝ q ⎠ ⇒

ap 2 q

2

+

bp +c=0 q

⇒ ap 2 + bpq + cq 2 = 0 (*) We establish a contradiction by showing that the left side of equation (*) must be odd, and therefore cannot equal zero. We argue by cases, and use the following properties of integer multiplication and addition: odd × odd = odd, odd × even = even, even × even = even odd + odd = even, odd + even = odd, even + even = even. a, b and c are odd, and p and q cannot both be even, since they are coprime. Case 1:

p and q are both odd Then p2, pq and q2 are all odd and ap2, bpq and cq2 are all odd. The sum of 3 odd numbers is odd, so the left side of (*) is odd.

Case 2:

one is even, one is odd By symmetry we may assume p even, q odd. Then ap2 + bpq = p(ap + bq) is even and cq2 is odd. The sum of an even and an odd number is odd, so the left side of (*) is odd.

We have shown that the left side of the equation is always an odd number, and therefore cannot equal zero. Hence no such such rational number exists, and the result follows. Shirleen Stibbe

http://www.shirleenstibbe.co.uk

3 Prove that 2 is irrational. Proof by contradiction We first recall the multiplication table for modular arithmetic in ℤ3.

×

[0]

[1]

[2]

[0]

[0]

[0]

[0]

[1]

[0]

[1]

[2]

[2]

[0]

[2]

[1]

where [0] = { … , −6, −3, 0, 3, 6, …} [1] = { … , −5, −2, 1, 4, 7, …} and

[2] = { … , −4, −1, 2, 5, 8, …}.

The leading diagonal of this table tells us that if we square an integer divisible by 3 we obtain an integer divisible by 3 and that if we square an integer not divisible by 3, we obtain an integer whose remainder is 1 when divided by 3. Now suppose that than 1, such that

2 is rational, then we can find two positive integers p and q, having no common factor other p = 2 q

and so

p 2 = 2q2 .

Our table has shown that q2 must belong to either [0] or [1]. Since 2 ∈ [2], the third row of our table tells us that 2q2 must belong to either [0] or [2]. Of course p2 must also belong to [0] or [1] and, since p2 = 2q2 , we see that both p2 and 2q2 must belong to [0]. Our table shows that if p ∈ [1] or [2], then p2 ∈ [1] and so, since p2 ∈ [0], we see that p must be an element of [0]. Our table also shows that if q2 ∈ [1] or [2], then 2q2 ∈ [2] or [1] respectively, and so, since 2q2 ∈ [0], we see that q2 ∈ [0] and this means that q must be an element of [0] arguing as we did for p. So we have deduced that each of p and q is an element of [0]. This means that the positive integers p and q are both divisible by 3 which is a contradiction since p and q have no common factor other than 1. So we can conclude that Note:

2 is not rational and hence is irrational.

The same proof can be used to show that

2, 5,

Shirleen Stibbe

n is irrational for any positive integer n ∈ [2], i.e.

8, ... are all irrational. Good value for money!

http://www.shirleenstibbe.co.uk

4 Prove that the table below does not define a group.

*

e

a

b

c

d

e

e

a

b

c

d

a a

e

c

d b

b b d

e

a

c

c

c

b

d

e

a

d d

c

a b

e

Proof by contradiction Reminder: Lagrangeâ&#x20AC;&#x2122;s theorem says that the order of a subgroup of a group G divides the order of G. Assume that the table defines a group G. Then G has a subgroup {e, a} of order 2. This contradicts Lagrangeâ&#x20AC;&#x2122;s theorem, since the order of G is 5. So the table does not define a group. Note: the structure defined is closed, has an identity element, e, and every element is self-inverse. It is not, however, associative, since b(cd) = ba = d, whereas (bc)d = ad = b.