Is this a correct proof of $r+\sqrt{2}\in\Bbb{Q}$?

Question

My proof goes as follows:

Suppose that $r+\sqrt{2}$ is rational, then $r+\sqrt{2}=p/q\leftrightarrow2=(\dfrac{p-rq}{q})^2$. Hence $\dfrac{p-rq}{q} | 2$ thus $\dfrac{p-rq}{q}=2k$ where $k\in\Bbb{N}$. Therefore we have $2=(2k)^2=4k^2\leftrightarrow (k=\sqrt{1/2}\lor k=-\sqrt{1/2})$. Due to the contradiction $k\in\Bbb{N}\land k\notin\Bbb{N}$, we conclude that $r+\sqrt{2}$ is irrational.

If $r + \sqrt 2 \in \Bbb Q$, then $\sqrt 2 \in \Bbb Q$, and we have reduced the problem to a classic case. — Robert Lewis, Sep 09 '18 at 16:39
@Bright: which implication? That $\sqrt 2 \in \Bbb Q$ or that $\sqrt 2$ irrational? — Robert Lewis, Sep 09 '18 at 16:53
@RobertLewis The implication that if $r+\sqrt{2}\in\Bbb{Q}$ then $\sqrt{2}\in\Bbb{Q}$. — mechanicious, Sep 09 '18 at 16:54
@Bright, rational numbers are closed under addition/subtraction, i.e. difference of two rationals is rational. — Ennar, Sep 09 '18 at 17:08
@Bright well, if $r + \sqrt 2 \in \Bbb Q$ then $\exists s \in \Bbb Q$ with $r + \sqrt q = s$; then since $r, s \in \Bbb Q$, we have $\sqrt 2 = s - r \in\Bbb Q$ since $\Bbb Q$ is a field. — Robert Lewis, Sep 09 '18 at 17:11

score 3 · Answer 1 · answered Sep 09 '18 at 16:55

3

Suppose that $r+\sqrt{2}$ is rational, then $r+\sqrt{2}=p/q\leftrightarrow2=(\dfrac{p-rq}{q})^2$. Hence $\dfrac{p-rq}{q} | 2$...

No, because you don't know that $\dfrac{p-rq}{q}$ is even an integer. It's just a rational number.

answered Sep 09 '18 at 16:55

Jack M

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The square of $(p-rq)/q$ is divisible by 2 then $(p-rq)/q$ must be also divisible by 2. Am I missing something? – mechanicious Sep 09 '18 at 16:59
@Bright $\frac34=2\frac38$, therefore $\frac34$ is divisible by $2$? Divisibility only makes sense for integers. – Jack M Sep 09 '18 at 17:00
Right, good point. – mechanicious Sep 09 '18 at 17:08

score 2 · Accepted Answer · answered Sep 09 '18 at 17:10

You can actually use this property:

If $r+x\in\mathbb{Q}$ for $r\in\mathbb{Q}$ then also $x\in \mathbb{Q}$.

From this it is easy to see that for your theorem it is sufficient to prove that $\sqrt{2}$ is not rational, which is a well-known fact, that you can find here, for example Irrational number $\sqrt2$ proof.

Is this a correct proof of $r+\sqrt{2}\in\Bbb{Q}$?

2 Answers2