# If $x,y∉ℚ$ and $x^2-y^2∈ℚ_{≠0}$ then $x+y,x-y∉ℚ$?

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.

• what is "e" here? It seems to mean "and" – user253751 Jan 25 at 14:40
• 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". – 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.

• 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. – jMdA Jan 25 at 14:52
• @jMdA thanks, fixed. – Doug M Jan 25 at 18:15

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.

$$OE\Delta.$$