This is false. For example, let
and
.
Then
which is rational, but x and y are irrational.
I am a bit stuck on this problem. Here is how I've approached it:Let . Prove that if m is rational, then x and y must also be rational.
Let . Then
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.
Any help, please? Many thanks!
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 , right? Since the gradient of a line is given by change in vertical distance divided by change in horizontal distance?Let (a,b) be a point in the first quadrant which is on the unit circle 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.
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.Let . Prove that m is rational if and only if a and b are both rational.
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?