2

Two real numbers $a$ and $b$ are equal iff $|a - b | \leq \epsilon$ for all $\epsilon >0$

I know that it's true if it says "iff $| a - b | < \epsilon$". But if I use $\leq$, isn't that statement false? Because intuitively, since $\epsilon$ can't be $0$, $| a - b |$ might not be $0$ for all $\epsilon$

Calvin Khor
  • 36,192
  • 6
  • 47
  • 102
jun
  • 661

2 Answers2

3

If you know that it is true for $<$, then you can either adapt the proof for $\le$ or simply argue that if $\mid a - b \mid \leq \epsilon $ $\forall \epsilon >0$, then also $\mid a - b \mid < \epsilon $ $\forall \epsilon >0$. In fact, given $\epsilon >0$, then we know that $\mid a - b \mid \leq \epsilon/2 < \epsilon $.

Paul Frost
  • 87,968
  • 1
    In other words, the inequality $|a-b| \le \epsilon$ holds for all positive numbers $\epsilon$, even ones that are not called "$\epsilon$". – GEdgar Feb 18 '21 at 11:56
0

Assume there is an $\epsilon$ such that $|a-b|=\epsilon$. Then with any smaller $\epsilon'$, you would have $|a-b|>\epsilon'$, which contradicts the hypothesis "for all".

In fact, equality never occurs because $\epsilon$ is nonzero.