Suppose that $x$ and $y$ are irrational, but $x + y$ is rational. Prove that $x -y$ is irrational.

i was wondering if someone could check my proof $Q= \{a/b , c,d : a,c ∈ \mathbb Z , b,d ∈ N>0\}$ $a/b =x+y$

$a/b -y=x$

proof by contradiction.

Let $x-y$ is rational

$c/d = x-y$

sub $a/b -y = x$ in for $x$

$c/d = (a/b -y) - y$

$a/b - c/d = 2y$

This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.

there it can be said $x-y$ is irrational

3 Answers

Yes, your proof seems right.

I like the following writing.

If $x-y\in\mathbb Q$ so $x-y+x+y=2x\in\mathbb Q$, which says $x\in\mathbb Q$, which is a contradiction.

Id est, $x-y\not\in\mathbb Q$.

Hint: $a,b \in\mathbb Q\implies \dfrac{a\pm b}2\in\mathbb Q$.

Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.

Sum of rational and irrational numbers is irrational. See here.

If $x,y$ are irrational and $x+y$ is rational, then: $$x-y=(x+y)-2y \ \text{is irrational.}$$