Supremum/Infimum proof: $\sup \{1/x; x\in A\} = 1/\inf(A)$ Assume that $\inf(A)>0$ and let $A'=\left\{\frac{1}{x} : x\in A\right\}$. I need to show that $\sup(A') = \dfrac{1}{\inf(A)}$.
I think this is quite simple, $\sup(A')$ must be $\dfrac{1}{\inf(A)}$ since $\inf(A)$ is the smallest number to divide by, that is $x > \inf(A)$ for all $x\in A$.
Maybe I made a mistake, but if not is this sufficient? Or, how could I make this more formal? 
EDIT I could probably make this formal by proving this by contradiction...

What you’ve said is a reasonable intuitive explanation of the result, but it’s not a proof. In order to show that $\sup A'=\dfrac1{\inf A}$, you must show two things:


*

*$x\le\dfrac1{\inf A}$ for each $x\in A'$, and  

*if $u<\dfrac1{\inf A}$, then there is an $x\in A'$ such that $x>u$.


In other words, you must show that $\dfrac1{\inf A}$ is an upper bound for $A'$, and that no smaller number is an upper bound for $A'$. Neither of these is difficult, but you need to do them in order to have a proof.
For (1), for instance, if $x\in A'$, then $x=\dfrac1a$ for some $a\in A$, and $a\ge\inf A>0$, so $$x=\frac1a\le\frac1{\inf A}\;.$$
I’ll let you take a stab at (2) on your own.
