- #1

- 11

- 0

I just learnt how to do proof by contradiction yesterday, so forgive any ugliness in my proof.

## Homework Statement

1) Prove: r√2 is irrational, where r is rational and r ≠ 0

variables involved:

r, rational

m, integer

n, integer

a, integer

b, integer

a', integer

b' integer

x, integer

y, integer

## Homework Equations

Knowledge of the proof that √2 is irrational was used. That is all.

## The Attempt at a Solution

1) r√2 = (m/n)√2 = a/b (this is assuming that Q in the implication P→Q is negated, which is the basis of the proof by contradiction)

2) (m√2)/n = 2a'/2b' = a'/b'

a'/b' = x/y, where at least one of x or y is odd (this is where the contradiction will lie later on)

3) (m√2)/n = x/y

m√2 = xn/y

√2 = xn/ym

2 = ((x^2)(n^2))/((y^2)(m^2))

2y^2m^2 = x^2n^2

..............after solving for x..............

x^2 is even therefore x is even

4) x = 2k

2y^2m^2/n^2 = 4k^2

..............after solving for y.............

y^2 is even so y is even

So there is a contradiction; both x and y are even when at least one of x or y should be odd. Therefore, the statement that was required to be proved is true.