# show the statements

I am writing this because they seem simple, but just to assure myself. Show that if a,b rational numbers show that:

a. b $\ge$ a if and only if for every $\epsilon$ > 0, we have a < b + $\epsilon$

b. For $\epsilon$ > 0, |a-b|< $\epsilon$ if and only if b- $\epsilon$ < a < b+ $\epsilon$

i was thinking that since a < b we add to both sides $\epsilon$ and we get a + $\epsilon$< b + $\epsilon$ which is smaller than a. I think i just need help expressing them, because they seem simple...

• This is false if $a=b$, so you need to assume $a\ne b$. Commented Apr 13, 2014 at 18:43
• @Alex I disagree. Commented Jun 12 at 13:48

For part a) we can prove this b contradiction.

Assume that $b\geq a$, and that $\exists~ \epsilon>0$ such that $a\geq b+\epsilon$. Since $\epsilon>0$, we must have that $b+\epsilon>b$.

So then we have $a\geq b+\epsilon>b$. This means that $a>b$, but we have assumed that $b\geq a$. Thus we have a contradiction.

For part b) just use the properties of absolute values:

$|x|<a\Longleftrightarrow-a<x<a$,

So

$|a-b|<\varepsilon\Longleftrightarrow-\varepsilon<a-b<\varepsilon$

And adding $b$ to all three expressions, we get

$b-\varepsilon<a<b-\varepsilon$.