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.

  • 1
    $\begingroup$ Should the numerator in the top question be $x+a$? $\endgroup$ – Brian Tung Jan 20 '16 at 22:04
  • 1
    $\begingroup$ Don't you mean $\frac{x+a}{x+b}$ in the first question? $\endgroup$ – Wojowu Jan 20 '16 at 22:04

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.


1. There are $p,q\in\mathbb{Z}$, with $gcd(p,q)=1$ such that $\frac{x+a}{x+b}=\frac{p}{q}$. Then $$(x+a)q=(x+b)p\Leftrightarrow x(q-p)=bp-aq\Leftrightarrow x=\frac{bp-aq}{q-p}\in\mathbb{Q}\Leftrightarrow x=0.$$ This is absurd. Therefore $p=q$ and $x+a=x+b$, so $a=b$.

2. Let $r\in\mathbb{Q}$ such that $\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r$ and suppose that $\frac{x}{y}$ is a irreducible fraction. Then $$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r\Leftrightarrow x²+x+\sqrt{2}=(y²+y+\sqrt{2})r\Leftrightarrow x²+x-y²r-yr=\sqrt{2}(r-1).$$ If $r\neq 1$ we have $\sqrt{2}=\frac{x²+x-y²r-yr}{r-1}\in\mathbb{Q}$, but this is absurd. It follows that $r=1$ and $$x²+x+\sqrt{2}=y²+y+\sqrt{2}\Rightarrow y(y+1)=x(x+1).$$ If $x=-1$, we have $y(y+1)=0$ and so $y=0=x+1$ or $y=x=-1$. If $x\neq -1$ we have $$x=y(\frac{y+1}{x+1}).$$ If $y=-1$, it is analogous. We can suppose that $y\neq -1$. If $y=0$, $x=0$. Suppose $-1\neq y\neq 0$. We have $$\frac{x}{y}=\frac{y+1}{x+1}.$$ Since $\frac{x}{y}$ is irreducible, follows the last equality that $x=y$.


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.