I read these proofs and many times they use the triangle inquality to prove something. For example on this proof can you explain why it is used and how it proves the case. I have a test coming up this week and more understanding of all this would help. Thanks in advance.

[The Archimedean Principle] To each $M \epsilon\Re$ there corresponds and integer $n \epsilon\aleph$ such that $n \geq M$.
PROOF Suppose the result is false. Then there is an $M_o \epsilon\Re$ such that $n\leq M_o$ for all $n \epsilon\aleph$. Thus {n} is bounded by $M_o$ and increasing. It follows from the Monotone Sequence Property that $n \rightarrow a$ for some real number a. In particular given $\varepsilon = 0.25$, there is an N such that $n \geq N \rightarrow |n - a| \leq \varepsilon$. Apply this inquality to n=N and n=n+1. By the triangle inquality we have
$
1=|(N+1) - a - (N-a)|\leq |(N+1) - a | + |N - a|\leq 0.25 + 0.25 =0.5
$

So we have $1\leq 0.5$ Since this last statement is false, it follow that {n} is not bounded. Therefore, some n is greater than M.

On this proof I would like to know why they chose $N>\frac{1}{\varepsilon}$.

Prove $\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$
Let $\varepsilon >0$. Use the Archimedean Property to choose $N\epsilon\aleph$ such that $N>\frac{1}{\epsilon}$. Multiplying this inequality by $\frac{\varepsilon}{N}$, we see that $\frac{1}{N}<\varepsilon$. In particular, $n\geq N$ implies $\frac{1}{n} \leq \frac{1}{N} < \varepsilon$. Since $\frac{1}{n}$ are all positive, it follows that $| \frac{1}{n} |<\varepsilon$for all $n\epsilon\aleph$

2. Bump in case someone who didnt see this can answer this...