# Why is this proof on rational set valid?

Prove that if $y$ is rational and $x$ is irrational, then $y + x$ and $yx$ (assume y $\neq 0$) is irrational.

I kinda guessed the proof and it turned out to be right with the key, but it doesn't answer my question. Here is my quick and dirty proof

Assume $x + y$ is rational such that $x + y = r$ ($r$ is rational), then $x + y = r \iff x = r - y \iff x = r + (-y)$ But this is a contradiction since $\mathbb{Q}$ is closed under addition.

Q1. Okay so I showed that by contradiction that my claim is wrong, but showing that something isn't doesn't tell us what the original thing was now does it? Or does this work because rational and irrational are polar opposites and I don't need to fill in?

The proof for multiplication is similar:

Assume $xy$ is rational such that $xy = r$ ($r$ is rational), then $xy = r \iff x = r/y$ But this is a contradiction since $\mathbb{Q}$ is closed under division

-
add comment

## 1 Answer

These arguments are just fine. In each you’ve shown that the assumption that a certain number is rational leads to a contradiction. This means that the assumption cannot be true: the number in question cannot be rational. But a real number that is not rational is by definition irrational, so you’ve shown that the number in question is irrational.

-
Could you join the chat, professor? –  Pedro Tamaroff Sep 8 '12 at 18:48
add comment