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!

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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.

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.

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.

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.