<s>[INST] 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?
 [/INST] Your argumentation is fine.
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 $x = y = \sqrt{2}$.
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