Prove there's always a larger element

Question

We have a number $ 0 < x < 1 $. We also have the function $1-\dfrac{1}{n}$ with $n \in \mathbb{N}$. How can I prove that for any $x \in \mathbb{R}$, there exists an $n \in \mathbb{N}$ such that $ 1-\dfrac{1}{n} > x$?

Of course, my intuitive problem is that you should in theory also be able to prove the reverse, because we have $x \in \mathbb{R}$.

Does there exist a simple proof for this?

Simply take $ n > \frac{1}{1-x}$. – Crostul Sep 15 '14 at 22:56 — Crostul, Sep 15 '14 at 22:56
@Crostul How does that prove anything? – skaf Sep 15 '14 at 23:01 — skaf, Sep 15 '14 at 23:01

score 2 · Accepted Answer · answered Sep 15 '14 at 23:01

2

Let $n \in \mathbb{N}$ any natural number such that $n > \frac{1}{1-x}$. Then $$\frac{1}{n}<1-x$$ $$1- \frac{1}{n}>x$$

answered Sep 15 '14 at 23:01

Crostul

36,738
4
36
72

But is this really a proof? You just rewrote it. I know the statement is really trivial, but is the proof too? – skaf Sep 15 '14 at 23:09
Yes, it is trivial. – Crostul Sep 15 '14 at 23:56

Prove there's always a larger element

1 Answers1