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By a contradiction. Then prove that there is also no nonzero rational solution.

I'm lost in my first proof course. I appreciate any help.

Xizel
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    This is essentially an assertion that the square root of $2$ is irrational. There is a standard proof of this by contradiction. Are there any restrictions on the kind of proof you need to produce? And how much help did you want? – Brian Tung Jan 27 '17 at 00:37
  • My professor had a solution without square root that I didn't catch. It went something along the line of $x^2 = 2y^2$, $x^2$ is even, so $x^2$ is divisible by 2 and $x$ is divisible by 4. I'm not sure where he was going with that. The full proof would be appreciated. – Xizel Jan 27 '17 at 00:42
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    Try google. This is a standard – Aganju Jan 27 '17 at 00:48
  • Your question is a duplicate of a gazillion questions about the proof that $\sqrt{2}$ is irrational. – Rob Arthan Jan 27 '17 at 01:11
  • @Xizel: The proof doesn't involve square roots, perhaps, but the proposition is still equivalent to the assertion that the square root of $2$ is irrational. That was meant as a hint as to what to google. – Brian Tung Jan 27 '17 at 02:59

3 Answers3

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Rearrange the equation so that you have $$\frac{x^2}{y^2} = 2.$$ Now since $$\frac{x^2}{y^2} = \left( \frac{x}{y} \right)^2$$ the question becomes is there a rational number, $x/y$, whose square is equal to 2?

Oiler
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If there is a solution, then there is one where x has the smallest possible absolute value. Clearly y has a smaller absolute value than x.

$2y^2$ is even, therefore $x^2$ is even, therefore x = 2z for some z. So $2y^2 = 4z^2$ and $y^2 = 2z^2$. Rename y to x an z to y, then we have $x^2 = 2y^2$. Voila - we have a solution where x has a smaller absolute value then the smallest possible absolute value.

gnasher729
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Rewrite it as

$$x^2=2y^2$$

thus, we can see that $x^2$ is even, and likewise, $x$ is even. This implies that $x=2x_1$, so

$$(2x_1)^2=4x_1^2=2y^2$$

$$\implies 2x_1^2=y^2$$

thus, $y^2$ is even, so $y$ is even. This implies that $y=2y_1$, so

$$2x_1^2=(2y_1)^2=4y_1^2$$

$$\implies x_1^2=2y_1^2$$

Hm, now doesn't this seem a bit familiar with what we started with? By induction...

Simply Beautiful Art
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