Prove that both number are rational numbers

Are positive rational numbers x and y, for which number

$\displaystyle \frac {x+\sqrt{y}}{y+\sqrt{x}}=Q$

is rational. Prove that both x and y are squares of rational numbers.

what I did:

$\displaystyle a= \sqrt{x}$

$\displaystyle b= \sqrt{y}$

$\displaystyle \frac{a ^{2}+b}{b^{2}+a}=\frac{a ^{2}+b}{b^{2}+a}* \frac{b^{2}-a}{b^{2}-a}= \frac{a^{2}b^{2}-a^{3}+b^{3}-ab}{b^{4}-a^{2}}$

I don't know what to do next ;<

Re: Prove that both number are rational numbers

Hey bigos.

You might want to take an alternative approach and prove that sqrt(y) is irrational if y is not the the square of a rational number and similarly for sqrt(x).

Then you can use the fact that sum of rationals is rational and division of one rational with respect to another is also rational if the denominator is non-zero.

Re: Prove that both number are rational numbers

Perhaps there's a way to continue there, but I don't see it.

Suppose $\displaystyle u = \sqrt{y} - Q\sqrt{x}$, and $\displaystyle v = \sqrt{y} + Q\sqrt{x}$.

What can you say about $\displaystyle u$ by itself? And then what can you say about $\displaystyle u$ and $\displaystyle v$?

(It's very clean, except must account for the special case $\displaystyle u = 0$.)

Re: Prove that both number are rational numbers

chiro, I don't know how I prove that that sqrt(y) is irrational. but I will try:

I assume that:

$\displaystyle a= \sqrt{x}$

$\displaystyle a \in irrational$

$\displaystyle x \in rational$ because $\displaystyle x=b^{2}$ and what next?... I think I'm too stupid for this :P

Re: Prove that both number are rational numbers

Dude - it's not about stupid (stupid people wouldn't be asking the question!), but about dead ends. Like wandering through a maze, you don't know for sure that it's a dead end until you try it and find out it's a dead end. To stretch the analogy: sometimes, after you've understood some general features of the maze, you can make a credible educated guess that "that isn't likely to work", which certainlly can save time. Of course, in a maze, the other options available to you are obvious, but with math, that's very much not the case.

(FYI: My previous suggestion isn't a dead end, and is an option open to you. It's actually a very short path to solve the maze - if you can figure out how to use it.)

Also: the conclusion about x and y follows if you can show that their square roots are rational. That fact is a theorem unto itself, and so I'd assume it's permissible to simply assert it here as a known theorem. In other words, if you can show that the square root of x (or y) is rational, then you can say "Thus x must be the square of a rational number."

Re: Prove that both number are rational numbers

$\displaystyle \frac {x+\sqrt{y}}{y+\sqrt{x}}=Q$ $\displaystyle \in$ rational

and I transform:

$\displaystyle x+ \sqrt{y}=Qy+Q \sqrt{x}$

$\displaystyle \sqrt{y}-Q \sqrt{x}=Qy-x$

I have your $\displaystyle u$. The equation shows that the $\displaystyle u$ must be rational.

I think this shows that $\displaystyle \sqrt{y}, \sqrt{x}$ are rational number? I'm right?

Re: Prove that both number are rational numbers

Your reasoning about u is correct: u is rational. But it doesn't follow that those square roots are rational. (It's not that easy.) For instance, suppose y=12, Q = 2, and x = 3. Then u = 0, which is rational, but both those square roots are IRrational.

I was thinking, now that you've proven that u is rational, look at the product uv. If u not 0, you'll then be able to learn something about v. And then you'll be able to make another short argument, combining what you know about u and v, to show that the square roots are rational (which finishes the problem for the u not 0 case).

The special case u=0 remains to be checked. Just look at what u is, and plug the consequences of u=0 back into the original statement, and you should be able to show, again, that those square roots are rational.

Re: Prove that both number are rational numbers

When $\displaystyle u \neq 0$ $\displaystyle \Rightarrow \sqrt{x} , \sqrt{y} \in$ Rational

So $\displaystyle v$ is rational too.

When $\displaystyle u=0$

$\displaystyle \sqrt{y}-Q \sqrt{x}=0$

$\displaystyle \sqrt{y} =Q \sqrt{x}$

(Worried) Argh.. it looks like a simple... :(

Re: Prove that both number are rational numbers

Are you familiar with the proof of irrational square roots (whose number is not square)?

Quadratic irrational - Wikipedia, the free encyclopedia

In terms of the rational case we know that SQRT(a/b) = SQRT(a)/SQRT(b) where a >= 0 and b > 0.

So once you prove a and b are square numbers then it means SQRT(a/b) is also a rational number since SQRT(a) is integer and SQRT(b) is integer if a and b are integers.

Re: Prove that both number are rational numbers

A few comments: First, the statement as given is false. x = y = 3 is a counterexample. There's an infinite number of these counter examples, and they all look like x = y and Q = 1. So we'll have to assume that Q isn't 1. Also, because of the square roots, we'll need both x and y non-negative. Also, x = y = 0 is excluded to keep the denominator from being 0. Also, Q wll have to be non-negative since everything else is. Since the numerator is 0 only when x = y = 0, an excluded case, will have Q is positive.

Claim: Let $\displaystyle x, y,$ and $\displaystyle Q$ be rational numbers, with $\displaystyle x, y \ge 0, Q>0$, and not both 0, and $\displaystyle Q \ne 1$.

Then $\displaystyle \frac{x+\sqrt{y}}{y+\sqrt{x}} = Q$ implies that $\displaystyle \sqrt{x}$ and $\displaystyle \sqrt{y}$ must both be rational.

Proof: Let $\displaystyle u = \sqrt{y} - Q \sqrt{x}, v = \sqrt{y} + Q \sqrt{x}$.

If $\displaystyle u = 0$, then $\displaystyle \sqrt{y} = Q \sqrt{x}$, so $\displaystyle Q = \frac{x+\sqrt{y}}{y+\sqrt{x}} = \frac{x + (Q\sqrt{x}) }{(Q^2x)+\sqrt{x}}$, so

$\displaystyle Q(Q^2x+\sqrt{x}) = x + Q\sqrt{x}$, so $\displaystyle Q^3x = x$, so $\displaystyle (Q^3-1)x = 0$.

But since $\displaystyle Q \ne 1$, have that $\displaystyle u = 0$ implies that $\displaystyle x = 0$.

So $\displaystyle u = 0$ implies $\displaystyle \sqrt{y} = Q \sqrt{x}$ and $\displaystyle x = 0$, and so $\displaystyle y = 0$ also.

Thus $\displaystyle u = 0$ is impossible, because it would imply that $\displaystyle x = y = 0$, contrary to $\displaystyle x$ and $\displaystyle y$ cannot both be 0.

Therefore $\displaystyle u \ne 0$ and we have:

$\displaystyle u = \sqrt{y} - Q \sqrt{x} = Qy -x$ is rational.

Also, $\displaystyle uv = (\sqrt{y} - Q \sqrt{x})(\sqrt{y} + Q \sqrt{x}) = y - Q^2x$ is rational.

Thus $\displaystyle v = (uv)/(u)$ is a quotient of rationals, and hence is rational. (Here's where $\displaystyle u \ne 0$ was required.)

Since $\displaystyle u$ and $\displaystyle v$ are both rational, so is $\displaystyle \frac{u+v}{2} = \sqrt{y}$.

Since $\displaystyle u$ and $\displaystyle v$ are both rational, and $\displaystyle Q$ is rational, so is $\displaystyle \frac{v-u}{2Q} = \sqrt{x}$. (Note $\displaystyle Q \ne 0$.)

Therefore both $\displaystyle \sqrt{x}$ and $\displaystyle \sqrt{y}$ are rational.