0

I am trying to show the following:

Assume $\alpha=\sqrt{2}$. Prove that there exists a number $c>0$ such that for all numbers $p,q$ with $q\neq 0$ we have $\mid q\alpha-p\mid>\frac{c}{q}$.

Here is what I have done so far: Wlog $q>0$. Since $\mid (q\alpha-p) (q\alpha+p) \mid=\mid 2q^2-p^2\mid\neq 0 $. So $\mid (q\alpha-p) (q\alpha+p) \mid=\mid 2q^2-p^2\mid \geq 1$

So $\mid q\alpha-p\mid\geq \frac{1}{\mid q\alpha +p\mid}$. There are two cases to consider $\mid q\alpha\mid >\mid p\mid$ and $\mid q\alpha\mid <\mid p\mid $. For the first we have,

$$ \frac{1}{\mid q\alpha +p\mid}\geq \frac{1}{\mid q\alpha \mid +\mid p\mid }>\frac{1}{2\mid q\alpha \mid} $$ So set $c=\frac{1}{2 \mid\alpha\mid}$. I couldn't find something useful for the second case i.e., $ \mid q\alpha \mid < \mid p\mid $. Thanks for any help/hint.

user64066
  • 2,480

1 Answers1

1

Both $q\alpha$ and $p$ are positive, so $|q\alpha + p|=q\alpha+p$ and you don't have a problem. If you had considered $q \lt 0$ you could just say that $|a\alpha -p| \gt 0 \gt \frac cq$ so this is not a problem either.

Ross Millikan
  • 374,822