I suppose $2 - \sqrt{2} $ is rational. so $$2- \sqrt{2} = {a/b} $$ where a,b are integers and gcd(a,b) = 1.

$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$ $$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$ $$\text{Step 3. }2b^2 = (2k)^2$$ Step 3 since $a^2$ is even and so $a$ is even too. Also, k is an integer. $$\text{Step 4. }b^2 = (2k)^2 \text{ //We see $b^2$ is even }$$ Step 4 since $b^2$ is even and so $b$ is even too.

We see $ a$ and $b$ are even. We see $gcd(a,b) \neq 1$

Is this proof correct ?

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    $\begingroup$ Suppose otherwise that $2-\sqrt{2}$ is rational. Then as a rational number minus another rational number is rational, that would imply that $2-(2-\sqrt{2})=\sqrt{2}$ is rational. But then we're at our classical example and we know that $\sqrt{2}$ is not rational, so we are done. $\endgroup$ – JMoravitz May 22 '19 at 17:01
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    $\begingroup$ It's the standard proof that $\sqrt{2}$ is irrational, but it's difficult to see how it pertains to $2 - \sqrt{2}$. $\endgroup$ – Theo Bendit May 22 '19 at 17:03
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    $\begingroup$ As for if your "attempt" is correct or not... you essentially copied the proof for that $\sqrt{2}$ is irrational. But.... here you began with $2-\sqrt{2}=\frac{a}{b}$ and then moments later wrote that $2=(\frac{a}{b})^2$ which is incorrect. Instead, $(\frac{a}{b})^2 = (2-\sqrt{2})^2 = 6-4\sqrt{2}$, not $2$ $\endgroup$ – JMoravitz May 22 '19 at 17:04
  • $\begingroup$ Note that the sum of a rational number and an irrational number is irrational. $\endgroup$ – El Ectric May 22 '19 at 17:06
  • $\begingroup$ Sorry that was a stupid mistake on squaring both sides! Indeed I proved $\sqrt {2}$ is irrational, so this proof becomes very easy. Thanks everyone $\endgroup$ – Mandey May 22 '19 at 17:14

For the sake of Contradiction let us assume that $2-\sqrt{2} \in \Bbb Q$ , say $2-\sqrt{2}=r \in \Bbb Q \implies \sqrt{2}= 2-r$ but since both $2,r \in \Bbb Q$ and $(\Bbb Q,+)$ is a subgroup of $(\Bbb R,+)$ (i.e. basically addition of two rational numbers is rational) it follows that $\sqrt{2} \in \Bbb Q$ , a contradiction!


As an alternate approach, note that $2-\sqrt 2$ is a root of $x^2-4x+2$ (the conjugate root is $2+\sqrt 2$ so just compute $(x-(2-\sqrt 2))(x-(2+\sqrt 2))$.

By the Rational Root Theorem, this quadratic has no rational roots.


Step 1 of your proof is incorrect:


$$2- \sqrt{2} = {a/b} $$

it does not follow that:

$$2 = (a/b)^2$$

but rather that:

$$(a/b)^2=(2- \sqrt{2})^2=4-4\sqrt{2}+2=6-4\sqrt{2}$$

Now, I am sure you want to know: so how do you prove it then?! Well, likely you will get many Answers that will show you how ... but your question was whether your proof was correct ... it is not.


That proof is not correct.

You start with $2-\sqrt{2}=\frac{a}{b}$ but on your next line you jump to $2=\left(\frac{a}{b}\right)^2$.

What you instead want to do is show that $\sqrt{2}$ is irrational (which seems like what you've done) then prove the following property: a rational minus an irrational number yields an irrational number. This can be done by considering the contrapositive, which states that if $p,q$ are rationals, then $p+q$ is rational. Here's a hint on how to prove this:

Let $p=\frac{a}{b},q=\frac{c}{d}$. What is $p+q$ in terms of $a,b,c,d$?

  • $\begingroup$ There's no hint $\endgroup$ – Mandey May 22 '19 at 17:19
  • $\begingroup$ Hover over it/click on it; it's a spoiler in case you want to think more about it yourself without getting spoiled on the solution $\endgroup$ – auscrypt May 22 '19 at 17:20

All you have to do is prove that $\sqrt{2}$ is irrational.

This is because the sum of an irrational number and a rational number is irrational.

  • $\begingroup$ Not strictly true, since 0 is a rational number. The better approach is to subtract the 2 over and then you have irrational equal to rational. This of course would require a proof that $\sqrt{2}$ is irrational. $\endgroup$ – Jacob Cleveland May 22 '19 at 17:05
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    $\begingroup$ @JacobCleveland What are you suggesting isn't strictly true? It is most certainly true that irrational plus rational results in irrational, every time, even if the rational number were zero. $\endgroup$ – JMoravitz May 22 '19 at 17:07
  • $\begingroup$ You're absolutely right, I was probably thinking of irrationals not being closed under addition. $\endgroup$ – Jacob Cleveland May 22 '19 at 18:21

Yet another proof: $0<2-\sqrt{2}<1$, so $(2-\sqrt{2})^n\to 0$. Note that $(2-\sqrt{2})^n=a_n-b_n\sqrt{2}$ for some integers $a_n,b_n$ by binomial theorem, so $$0<(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$ If $2-\sqrt{2}$ were rational, say $=p/q$, $p,q\in\mathbb{N}$, then by equation (1) $(2-\sqrt{2})^n$ must be at least $1/q$, contradicting $(2-\sqrt{2})^n\to 0$.


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