# My proof that the square root of 2 multiplied by r is irrational

1. Jul 18, 2014

### airbusman

Here it is, for you to critique. This is a proof by contradiction. This is a good example of how I usually go about doing proofs, so if you give me tips on how to improve this particular proof, I'll be able to improve all my other proofs.

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Jul 18, 2014

### PeroK

There is a much simpler proof. Hint: the product (or quotient) of two rational numbers is rational.

(I didn't look carefully at your proof as it's way more complicated that it needs to be.)

3. Jul 18, 2014

### thelema418

I would recommend writing your proof as a narrative, giving explicit information about what you are doing.

I think you need to state that $n \neq 0$, $b \neq 0$, etc. Or, $m, n, a, b \in \mathbb{Z}^*$. I understand the restrictions on the variables contextually, but I like to be certain when dividing by a variable that I am not dividing by 0.

Step 2 is not clear. Also, I think you need to say something about a and b being relatively prime. The same for m and n.

There is an issue just before step 4. You don't give the information, you just said you solved for $x$. But then, how do you know:

$x^2 = 2 \frac{y^2 m^2}{n^2}$

is even? An integer times 2 is even... but here you have a rational number times 2.

I do think you can work from this information, but you have to do something different algebraically.