Proof that when $x + y$ are irrational then $x$ and $y$ are irrational I want to prove by contrapositive that:
Proof that if $x + y$ are irrational then $x$ and $y$ are irrational. $x,y \in \mathbb{R}$
I did the following:

Negation of the statement: $x + y$ are rational then $x$ and $y$ are also rational
$\exists m,n,i,j \in \mathbb{Z} $   $gcd(m,n)=1 $ $gcd(i,j)=1 $
Then $x = m/n$ and $y = i/j$
So when, $x + y = \frac{m}{n} + \frac{i}{j} = \frac{m*j + i*n}{n*j}$ 
Therefore if the gcd of $gcd(m,n,i,j)=1$ then we can conclude that the number is rational. 
q.e.d

Is this proof formally correct?
I appreciate your answer!
 A: The negation would not be the statement that if $x+y$ are rational, then $x$ and $y$ are also rational. $A$ implies $B$ is equivalent to "not $B$" implies "not $A$":
This would be: if either $x$ or $y$ is rational then $x+y$ is rational. Unfortunately this is wrong. Take $x=1$ and $y=\sqrt{2}$.
Edit: The second claim that $x+y\in \mathbb{Q}$ implies $x,y\in \mathbb{Q}$ is not true either. For $x=\sqrt{2}$ and $y=1-\sqrt{2}$ we have $x+y=1\in \mathbb{Q}$, but not $x$ and $y$ rational.  
A: The statement

If $x+y$ is irrational, then $x$ and $y$ are irrational.

is false (as witnessed by Dietrich Burde's counterexample).  Therefore, it shouldn't be possible to prove this statement.

However

If $x+y$ is irrational, then $x$ is irrational or $y$ are irrational (or both).

is true.  This is formally written: $$x+y \not\in \mathbb{Q} \implies (x \not\in \mathbb{Q}) \vee (y \not\in \mathbb{Q}).$$
In this case, the contrapositive would be $$\neg\left((x \not\in \mathbb{Q}) \vee (y \not\in \mathbb{Q})\right) \implies \neg(x+y \not\in \mathbb{Q}).$$ Or equivalently $$\neg(x \not\in \mathbb{Q}) \wedge \neg(y \not\in \mathbb{Q}) \implies \neg(x+y \not\in \mathbb{Q})$$ by de Morgan's Law.  This is equivalent to $$(x \in \mathbb{Q}) \wedge (y \in \mathbb{Q}) \implies x+y \in \mathbb{Q}$$ since a non-irrational number is rational (assuming that $x$ and $y$ were taken as real numbers all along).
Proof: If $x \in \mathbb{Q}$ and $y \in \mathbb{Q}$, then, by definition of a rational number, we can write $$x=\frac{a}{b} \qquad\text{and}\qquad y=\frac{b}{c}$$ for some integers $a,b,c,d$ with $b \neq 0$ and $c \neq 0$.
Hence
\begin{align*}
x+y &= \frac{a}{b}+\frac{c}{d} \\
    &= \frac{ad}{bd}+\frac{bc}{bd} \\
    &= \frac{ad+bc}{bd}.
\end{align*}
Since $ad+bc$ is an integer, and $bd$ is a non-zero integer, we have that $x+y \in \mathbb{Q}$.
