0
$\begingroup$

Definition: Let $\epsilon > 0$, and $x,y$ be rational numbers. We say that $y$ is $\epsilon$ close to $x$ iff we have $d(y,x) \leq \epsilon.$

Question: Let $\epsilon > 0.$ If $x$ and $y$ are $\epsilon$ close, and $z$ is nonzero, then show that $xz$ and $yz$ are $\epsilon |z|$ close.

I am studying real analysis from terry tao's lecture notes and I am looking for an elegant proof to this question. My way would be case by case analysis of $x,y$ and $z$, such as if $x >y$, or else, if $z$ is postive or negative.

Is there a shorter and better method to show this ?

1 Answers 1

2

HINT: $|xz-yz|=|x-y||z|$. $\qquad\qquad\qquad$