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

[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 learned 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.

 

[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.)

 

[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.