2
$\begingroup$

Prove or refute that $\exists x, y, z \in$ with $\mathbb {Q}$ and $x\neq y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{z}$ and $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}\setminus\mathbb{Q}.$

If $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}$, we find that $\sqrt{1}+\sqrt{4}=\sqrt{9}$ or even $\sqrt{0}+\sqrt{1}=\sqrt{1}$ holds. Thus we are interested in the case where the roots are irrational numbers but the radicands are rational numbers.

After squaring both members of the equality we get $x+2\sqrt{xy}+y=z\implies\sqrt{xy}=\frac{z-x-y}{2}$. This means that the product of $\sqrt{x}$ by $\sqrt{y}$ is a rational number. We can only conclude that both $\sqrt{x}$ and $\sqrt{y}$ are rational or irrational.

I can't find any good aproach to this problem and I don't even know if this result is relevant. Thank you for advance.

$\endgroup$
  • $\begingroup$ Since $\sqrt x\sqrt y$ is rational, you conclude that $\sqrt x$ and $\sqrt y$ are both rational or irrational. However, you already know that both $\sqrt x$ and $\sqrt y$ are irrational in the question. $\endgroup$ – Toby Bartels Oct 28 '19 at 14:47
  • $\begingroup$ If you square again, then you get an equation involving only rational numbers, so the techniques of Diophantine equations could potentially be used to find all possible solutions, in addition to the ones given in the accepted answer and its comments. But it will be a 4th-degree equation, so probably not easy! $\endgroup$ – Toby Bartels Oct 28 '19 at 14:59
  • $\begingroup$ You need a stronger condition than simply saying $x\ne y$: For $x$ any positive nonsquare rational and $y=k^2x$ for any nonzero rational $k>0$, you have $z=(k+1)^2x$. $\endgroup$ – Andrés E. Caicedo Oct 28 '19 at 15:00
4
$\begingroup$

Letting $x = 2$, $y = 8$, and $z=18$ works: $$\sqrt 2 + \sqrt 8 = \sqrt 2 + 2 \sqrt 2 = 3 \sqrt 2 = \sqrt 18.$$ More generally, for any $a, b, c \in \mathbb Q$ we have $$ \sqrt{ab^2} + \sqrt{ac^2} = b\sqrt a + c \sqrt a = \sqrt{a(b+c)^2}.$$ This is an example whenever $b \ne \pm c$ and $\sqrt a$ is irrational.

$\endgroup$
  • 1
    $\begingroup$ @OscarLanzi I was just about to post the general solution: $$\sqrt{ab^2}+\sqrt{ac^2} = \sqrt{a(b+c)^2}$$ with rational $a$, $b$, $c$, and indeed, the ratio between $x$ and $y$ is a squared rational. $\endgroup$ – Jaap Scherphuis Oct 28 '19 at 15:00
0
$\begingroup$

If $a,b,c$ are positive rationals but not squares of rationals and $\sqrt c=\sqrt a +\sqrt b$ then $$c=(\sqrt a +\sqrt b)^2=a+b+2\sqrt {ab}\in \Bbb Q,$$ which requires $$\sqrt {ab}\in \Bbb Q,$$ which requires $ab=d^2$ where $0<d\in \Bbb Q,$ which requires $b=d^2/a,$ and hence $$c=a+b+2\sqrt {ab}=a+d^2/a+2d=(a+d)^2/a.$$ And so $\sqrt c=(a+d)/\sqrt a\not \in \Bbb Q.$

This is also sufficient. That is, if $a,d$ are positive rationals and $a$ is not the square of a rational then $\sqrt a$ and $\sqrt {d^2/a}=d/\sqrt a$ are irrational and so is $\sqrt a +\sqrt {d^2/a}=\sqrt {(a+d)^2/a}=(a+d)/\sqrt a.$

And to ensure $b\ne a$ it is necessary and sufficient that $b=d^2/a\ne a.$ That is, $a\ne d. $

For example $a=2, d=1.$ That is $(\sqrt 2+1/\sqrt 2)^2=9/2$ so $\sqrt 2 +\sqrt {1/2}=\sqrt {9/2}=3/\sqrt 2.$

$\endgroup$

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.