# Prove or find a counterexample: For all real numbers x and y it holds that x + y is irrational if, and only if, both x and y are irrational.

The following is a Homework Question that I've been working on and I would like some feedback on my answer: Prove or find a counterexample: For all real numbers $x$ and $y$ it holds that $x + y$ is irrational if, and only if, both $x$ and $y$ are irrational.

So I've got (or at least like to think I've got) a counter example.

Let $x = 1$ and $y = \sqrt{2}$.

Thus $x$ is rational and $y$ is irrational. Adding $x$ and $y$ gives us $1 + \sqrt{2}$ which cannot be simplified any further (right?) and is an irrational number.

Thereby proving that $x + y$ can be irrational without both x and y being irrational.

Is this okay? Any feedback is greatly appreciated thank you!

• Hint: $\sqrt{2}-\sqrt{2}=0$ is rational. – user140943 Apr 7 '14 at 23:55
• oh yeah, you already have a counterexample... good job. – user140943 Apr 7 '14 at 23:55
• @user140943 Yes, the OPs counter example is in one direction, you got one for the other, which is useful. If you post it as an answer, I'll up vote. – Git Gud Apr 7 '14 at 23:56
• Yes, it's a fine counterexample provided you can prove that $1 + \sqrt 2$ is irrational. But it's easier to just change $x$ to be, say, $0$. – user61527 Apr 7 '14 at 23:57
• I would have but Im a little confused as to whether or not 0 is a natural number :/ – user2710184 Apr 8 '14 at 0:05

Hint: $\sqrt{2}-\sqrt{2}=0$ is rational.

Your counterexample works fine. I think a better version would be $$(\sqrt{2}-1)+(1)=\sqrt{2}$$ is clearly irrational, whereas $1$ is rational. Otherwise you would have to justify why the sum of an irrational and a rational number is irrational.

• Justifying that $1+\sqrt 2$ is irrational is the same as proving that $\sqrt 2-1$ is, so your second counter example doesn't help. – Git Gud Apr 8 '14 at 0:02
• But if I used (√2−1) wouldnt I have to show why THAT is irrational? – user2710184 Apr 8 '14 at 0:02
• @GitGud It doesn't matter what $\sqrt{2}-1$ is, all that matters is that the sum is irrational and at least one of the numbers in the sum is rational. – user140943 Apr 8 '14 at 0:04
• @user140943 You're right. – Git Gud Apr 8 '14 at 0:04

Note that an "if and only if" statement is really two statements:

• if $x$ and $y$ are both irrational then $x+y$ is irrational, and
• if $x+y$ is irrational then $x$ and $y$ are both irrational.

To prove the "iff" statement false you need an example where one of these is false: you could use the example given by @mathse or @LAcarguy.

It's not necessary to prove that both parts are false, but if you want to, you could simply use both those examples: @mathse's example shows that the second part of the statement is false, while @LAcarguy's example does it for the first part.

• False, since x=√2 and y=√2 then both x and y are irrational and x + y is also irrational ; √2 + √2 = 2.828427 – KRISSH Apr 11 '14 at 1:35
• @KRISSH, I don't understand. What statement are you claiming to be false? – David Apr 11 '14 at 2:04
• sorry,please ignore false. – KRISSH Apr 11 '14 at 2:06

It is not true. Example is: $x = 1-\sqrt{2}$ and $y = \sqrt{2}$, then both $x$ and $y$ are irrationals but $x + y = 1$ a rational number.

• That's some hint! – Adam Bjorndahl Apr 11 '14 at 0:52
• In fact neither implication is true. Taking $x=0$ and $y=\pi$ shown that $x+y$ is irrational, but it is not the case that both $x$ and $y$ are irrational. – MPW Apr 11 '14 at 1:30

Here's a hint, although the question has already been answered. Try turning the statement around, and seeing if you can find a rational number (or even a whole number...) and an irrational number whose difference is irrational.

Another counterexample is $x=0$ and $y=$ something irrational.