If $|a-b| \leq \frac{\epsilon}{2}$ and $|a| \gt \epsilon$, constructively prove that $|b|\geq \frac{\epsilon}{2}$.

Question

I have a proof for the claim, but it's ugly and involves cases and a contradiction. Is there a constructive proof for the statement:

If $|a-b| \leq \frac{\epsilon}{2}$ and $|a| \gt \epsilon$, then $|b|\geq \frac{\epsilon}{2}$

I'm guessing it involves the triangle inequality theorem, but I am struggling to pin it down.

Thank you!

Answer 1

Hint

$$|a-b|\le\frac{\epsilon}{2}\Leftrightarrow -\frac{\epsilon}{2}\le a-b\le \frac{\epsilon}{2}\Leftrightarrow a-\frac{\epsilon}{2}\le b \le a+\frac{\epsilon}{2}\qquad (1)$$

$$|a|>\epsilon \Leftrightarrow a <-\epsilon \text{ or } a>\epsilon \Leftrightarrow \left(a+\frac{\epsilon}{2} <-\frac{\epsilon}{2}\right) \text{ or } \left(a-\frac{\epsilon}{2}>\frac{\epsilon}{2}\right)\qquad(2)$$

Now, put both inequalities, $(1)$ and $(2)$, together. Can you finish?

Answer 2

Using $$ |x-y|\ge|x|-|y| $$ one has $$ |b|=|a-(a-b)|\ge|a|-|a-b|>\epsilon-\frac{\epsilon}{2}=\frac{\epsilon}{2}. $$

Answer 3

You have to use the inverse triangle inequality, that is for any real numbers $a$ and $b$ we have $$|a|-|b| \leq |a-b|.$$

On the one hand, $$|a|-|b| \leq |a-b|\leq \frac{\varepsilon}{2}.$$ Hence, $$|a|\leq |b|+\frac{\varepsilon}{2}.$$

Also $|a|>\varepsilon$, so $$\varepsilon<|a|\leq |b|+\frac{\varepsilon}{2}.$$ In particular, $$\varepsilon<|b|+\frac{\varepsilon}{2},$$ where the inequality is strict because of the fact that $\varepsilon<|a|$, and the result is proven: $$|b|>\frac{\varepsilon}{2}.$$