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.


Is this proof formally correct?

I appreciate your answer!

  • 1
    $\begingroup$ let $x=a+\sqrt b,y=c-\sqrt b$ $\endgroup$ – lab bhattacharjee Sep 12 '13 at 18:32
  • $\begingroup$ it is obviously false. Let $x\in \mathbb{R}\setminus \mathbb{Q}$ and $y=0 \in \mathbb{Q}$ $\endgroup$ – Dominic Michaelis Sep 12 '13 at 18:32
  • $\begingroup$ Only one invariant : irrational+ rational =irrational $\endgroup$ – lab bhattacharjee Sep 12 '13 at 18:33
  • $\begingroup$ The correct negation is that when $x+y \in \mathbb{R}\setminus \mathbb{Q}$, then at least one of $x,y$ are irrational $\endgroup$ – Dominic Michaelis Sep 12 '13 at 18:33
  • $\begingroup$ Thx a lot for your answers!!! $\endgroup$ – Kare Sep 12 '13 at 18:37

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.

  • $\begingroup$ Sorry that I cannot upvote you, I have to less votes yet. However, how to prove by contrapositive then? $\endgroup$ – Kare Sep 12 '13 at 18:40
  • $\begingroup$ Yeah it was definitely helpful. However, please would you be so kind to add how to prove by contrapositive in the formal way? $\endgroup$ – Kare Sep 12 '13 at 18:41
  • $\begingroup$ What do you want to be proved ? $\endgroup$ – Dietrich Burde Sep 12 '13 at 18:43
  • $\begingroup$ My problem is, that I do not know how to write this prove by contrapositive. I tried, $x+y = \frac {m}{n} + y$ which gives me a rational number. Is this the correct formal prove by contrapositive? $\endgroup$ – Kare Sep 12 '13 at 18:47

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.


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 $x+y \in \mathbb{Q}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.