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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.

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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
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    $\begingroup$ @jMdA thanks, fixed. $\endgroup$ – Doug M Jan 25 at 18:15
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With

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

but

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

first note that (2) implies that

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

lest

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

then in light of (3), we have either

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

or

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

for

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

and

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

Now if (5) binds, then

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

and

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

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

$OE\Delta.$

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