# Thread: Proove x,y must be rational when m is rational

1. ## Proove x,y must be rational when m is rational

Let $m= \frac{y}{x+1}$. Prove that if m is rational, then x and y must also be rational.
I am a bit stuck on this problem. Here is how I've approached it:

Let $m= \frac{c}{d}$. Then $\frac{c}{d} = \frac{y}{x+1}$

Then from here I don't know what to do. We can't express x or y solely in terms of c and d. And expressing x in terms of c, d and y (or y in terms of c, d and x) is inconclusive because we cannot assume that y and x respectively are rational, as that is what we are trying to show.

2. This is false. For example, let
$x = \sqrt{2}$ and
$y = \sqrt{2} + 1$.

Then
$\frac{y}{x+1} = 1$
which is rational, but x and y are irrational.

3. Hmmmmmmmm. I shall post the original problem then. (The problem I posted was a smaller part of a larger problem).

Let (a,b) be a point in the first quadrant which is on the unit circle $x^2 + y^2 = 1$ and let m denote the slope of the line joining (-1,0) to (a,b). Prove that m is rational if and only if a and b are both rational.
Well m in terms of a and b is just $m=\frac{b}{a+1}$, right? Since the gradient of a line is given by change in vertical distance divided by change in horizontal distance?

And so the question evolves into:

Let $m=\frac{b}{a+1}$. Prove that m is rational if and only if a and b are both rational.
I did it one way; i.e. assuming a and b are rational, and showing that this leads to a rational m. But it's the reverse way I can't show.

And by way of counterexample, you've shown that m rational doesn't necessarily imply a,b rational.

So have I simplified the problem incorrectly? Or is the original problem just incorrect?

4. you have to prove the biconditional (both forward and backwards). I suggest using proof by contrapositive and contradiction

5. Finding all the Pythagorean triples, eh?

6. Originally Posted by r45
Hmmmmmmmm. I shall post the original problem then. (The problem I posted was a smaller part of a larger problem).

Well m in terms of a and b is just $m=\frac{b}{a+1}$, right? Since the gradient of a line is given by change in vertical distance divided by change in horizontal distance?

And so the question evolves into:

I did it one way; i.e. assuming a and b are rational, and showing that this leads to a rational m. But it's the reverse way I can't show.

And by way of counterexample, you've shown that m rational doesn't necessarily imply a,b rational.

So have I simplified the problem incorrectly? Or is the original problem just incorrect?
You haven't used the fact that $a^2+ b^2= 1$