Let E = $\left\{ x|\exists n \in N (x= {\frac{3n+4}{n}})\right\} $. Show that Inf E =3.
I need help understanding the last step involing the archimedean property.
We have 3 as a lower bound :
$3<\displaystyle\frac{3n+4}{n}$ for all n.
By the completeness axiom, There exist the greatest lower bound.
Let Inf E= $b_0$. We will show that $b_0$=3.
Suppose that $b_0>3 \Rightarrow b_0-3>0$
By the archimedean property
$n(b_0-3)>4 \Rightarrow \displaystyle\frac{3n+4}{n}<b_0$
The last inequality is a contradiction , so we conclude that $b_0=3$
I'm kinda confuse of which is the infinimum in the inequality and I would like to know how do we choose the number 4 in :
$n(b_0-3)>4$
is it because we want to arrive at this form of the equation ?
$\displaystyle\frac{3n+4}{n}<b_0$
By your assumption $b_0$ is the infimum. Yes basically that equality is found by looking at what you want to end up with. You want to get that an element in the set is less than the infimum to get your contradiction. It's often easier to start backwards. So ask yourself, how can I best use the Archimedean Postulate to show that: $$\frac{3n+4}{n}<b_0$$ You then rearrange this inequality in your head (or on paper) to the form where you can bring in the AP to say why there exists an $n$ such that the inequality holds.
It is very important not to state what you are aiming for as if it is true, as you haven't yet proven it. That must just be part of your working out on the side.