# There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational.

Using the fact that $\sqrt{2}$ is irrational, prove the following:

There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational.

My attempt:

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

Suppose $y= \sqrt{2} + 1$ is rational. Than $y= \sqrt{2} + 1 = \frac mn$, where $m,n$ are integers. So, $y= \sqrt{2} = \frac mn - 1$ which is rational. This contradicts the original statement that $\sqrt{2}$ is irrational. Thus, $y= \sqrt{2} + 1$ is irrational.

Now, $x+y= \sqrt{2}+ \sqrt{2} + 1$ which equals $2\sqrt{2} + 1$.

Suppose $2\sqrt{2} + 1$ is rational. Then $2\sqrt{2} + 1 = \frac pq$ where $p,q$ are integers. So, $2\sqrt{2} = \frac pq - 1$ which equals $\sqrt{2}= \frac{p}{2q} + \frac 12$ which is rational. This contradicts the original statement that $\sqrt{2}$ is irrational. Thus, $2\sqrt{2} + 1$ is irrational.

So, $x+y$ is irrational if $x$ and $y$ are irrational.

Does this proof make sense? Is there a simpler way?

• Why not $\sqrt2$ and $2\sqrt2$, since nothing is asked about the product? – Parcly Taxel Oct 7 '16 at 11:26
• because choosing that would just make $x+y$ = $3\sqrt(2)$ instead of the $2\sqrt(2)$ that I have, which doesn't really make a difference, right? – SeesSound Oct 7 '16 at 11:33

However, one has to be careful that the statement "So, $x+y$ is irrational if $x$ and $y$ are irrational." does not hold for all $x,y \in \mathbb{R}$ (but, of course, for your choice).
An easier choice should be $x = y = \sqrt{2}$.
• The question does not ask for $x \ne y$. You can also take $y = 2 \, \sqrt{2}$ or, more generally, $y = q \, \sqrt{2}$ for any rational number $q$. – gerw Oct 7 '16 at 11:30
• And just in case it isn't clear to the OP, a trivial example where $x+y$ is rational and both $x$ and $y$ are irrational, contradicting the conclusion, is $x=\sqrt2$, $y=-\sqrt2$, $x+y=0$. – hvd Oct 7 '16 at 11:34