Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a non-zero rational, then $x+y$ and $x-y$ are both irrational numbers.

I know that $\mathbb{Q}$ is closed under addition and product, however I have no clue how to solve this problem. I would really appreciate help. I tried assuming that the sums are rational to reach a contradiction, but I failed.

  • $\begingroup$ what is "e" here? It seems to mean "and" $\endgroup$ – user253751 Jan 25 at 14:40
  • $\begingroup$ ah yes, its "and" hahah i took this piece from my latex notes and translated it from my mother tongue, guess I missed this "e". $\endgroup$ – belwarDissengulp Jan 25 at 14:43

A rational number times a rational number gives a rational number and a rational number plus a rational number gives a rational number. $\mathbb Q$ is a field.

When we multiply two irrational numbers it is possible that we get a rational number. e.g. $(\sqrt 2)(\sqrt 8) = 4.$ But, a non-zero rational number times an irrational number will always be irrational. The similar rules apply for addition.

$x^2 - y^2 = (x+y)(x-y)$

If $x^2-y^2$ is rational then either $(x+y)$ and $(x-y)$ are both rational or both irrational.

If $(x+y)$ and $(x-y)$ are both rational, then $(x+y) + (x-y) = 2x$ must be rational. But the proposition states that $x,y$ are both irrational. So, it can't be the case that $(x+y)$ and $(x-y)$ are both rational, thus the must be both irrational.

  • 3
    $\begingroup$ There is a small error in this post, a rational number times an irrational number can be rational, if the rational number is zero. This explains why the question specifies that $x^2 - y^2$ is non-zero. $\endgroup$ – jMdA Jan 25 at 14:52
  • 1
    $\begingroup$ @jMdA thanks, fixed. $\endgroup$ – Doug M Jan 25 at 18:15


$x, y \notin \Bbb Q \tag 1$


$0 \ne x^2 - y^2 \in \Bbb Q, \tag 2$

first note that (2) implies that

$x - y, x + y \ne 0, \tag 3$


$x^2 - y^2 = (x - y)(x + y) = 0; \tag 4$

then in light of (3), we have either

$x - y, x + y \in \Bbb Q \tag 5$


$x - y, x + y \notin \Bbb Q, \tag 6$


$x + y = \dfrac{x^2 - y^2}{x - y} \tag 7$


$x - y = \dfrac{x^2 - y^2}{x + y}. \tag 8$

Now if (5) binds, then

$x = \dfrac{(x + y) + (x - y)}{2} \in \Bbb Q \tag 9$


$y = \dfrac{(x + y) - (x - y)}{2} \in \Bbb Q \tag{10}$

in contradiction to (1); therefore (6) must hold.



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.