[Math] Let a,b be rationals and x irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$.

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I'm trying to solve the following problems:

  1. Let $a$,$b$ be rationals and $x$ irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$

  2. Let $x$,$y$ be rationals such that $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ is also rational. Prove that either $x=y$, or $x+y=-1$.

Only thing I can think of is that if $\frac{x+a}{x+b}$ and $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ are rationals, then there's a number $\frac{m}{n}$ where integers $m$ and $n$ are co-prime.

I would appreciate any form of help.

Thank you.

Best Answer

For the first one, you can proceed directly: Notice that if the expression is equal to some rational $r$, then

$$x + b = r(x + a) \implies x(1 - r) = a - b$$

Now the right side is rational, but the left side is irrational unless.....

For the second, I'd suggest proceeding similarly. Write

$$x^2 + x + \sqrt 2 = r(y^2 + y + \sqrt 2)$$ and rearrange to get

$$x^2 + x - ry^2 - y = \sqrt2(r - 1)$$ From this, get $r$; then do some algebra to figure out when the first equality can hold.

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