If $|a-1|<\epsilon$ and $|a-2|<\epsilon$ prove that $\epsilon>1/2$

Question

If $|a-1|<\epsilon$ and $|a-2|<\epsilon$ prove that $\epsilon>1/2$

I know that $\epsilon -2 \le \epsilon-1 \le a$ and $a \le \epsilon+1 \le \epsilon+2$ but can I use it somehow? How should I proceed?

Answer 1

We have that

$$|a-1|<\epsilon \iff -\epsilon<a-1<\epsilon\iff 1-\epsilon<a<1+\epsilon$$ $$|a-2|<\epsilon \iff -\epsilon<a-2<\epsilon\iff 2-\epsilon<a<2+\epsilon$$

therefore

$$2-\epsilon<1+\epsilon \implies 2\epsilon>1 \implies \epsilon >\frac12$$

Answer 2

Hint:

$$|a-1|+|a-2|=|a-1|+|2-a|\geq \left|(a-1)+(2-a)\right|...$$

Answer 3

If $\varepsilon\leqslant\frac12$, then$$\lvert a-1\rvert<\varepsilon\implies a<1+\varepsilon\leqslant\frac32\text{ and }\lvert a-2\rvert<\varepsilon\implies a>2-\varepsilon\geqslant\frac32.$$

Answer 4

The assumption tells you that $$ 2\epsilon>|a-1|+|a-2| $$ which is equivalent to $$ \epsilon>\frac{1}{2}(|a-1|+|a-2|). $$ Now it suffices to show that $$ |a-1|+|a-2|\geq 1. $$ which follows by triangle inequality or simply noticing that

• if $a<1$, then $|a-1|+|a-2|=(1-a)+(2-a)=3-a>3-1=2>1$;
• if $1\leq a\leq 2$, then $|a-1|+|a-2|=1$;
• if $a>2$, then $|a-1|+|a-2|=2a-3>2\times 2-3=1$.

Answer 5

$1-\epsilon <a<1+\epsilon$, and (!)

$2-\epsilon <a <2+\epsilon.$

Hence

$2-\epsilon <1+ \epsilon$ (why?), or

$1/2 < \epsilon.$

What happens if $\epsilon < 1/2$ ? (Draw it on number line).