# Proving inequality For every integer n > 0 there is some real number x > 0 s.t. x < 1

• Feb 12th 2009, 10:14 PM
qtpipi
Proving inequality For every integer n > 0 there is some real number x > 0 s.t. x < 1
For every integer n > 0 there is some real number x > 0 s.t. x < 1/n..

I need a proof. Contrapositive? I don't know what to do.

Thanks.

Another question, too:

A fraternity has a rule for new members: each must always tell the truth or always lie. They know
which does which. If I meet three of them on the street and they make the statements below,
which ones (if any) should I believe?
A says: \All three of us are liars."
B says: \Exactly two of us are liars."
C says: \The other two are liars."
• Feb 13th 2009, 03:32 AM
clic-clac
Maybe you want to say an irrational number $x>0$ because $\frac{1}{n+1}$ is a real number such that $0<\frac{1}{n+1}<\frac{1}{n}$

Let $y>0$ be an irrational number. Then the sequence $\left(\frac{y}{k}\right)_{k>0}$ decreases and has $0$ for limit. So there is a $m\in\mathbb{N}$ such that $\frac{y}{m}<\frac{1}{n}.$
We just have to prove that $\frac{y}{m}$ is irrational: if that wasn't the case, since $m\mathbb{Q}=\mathbb{Q},$ then $m\times\frac{y}{m}$ would be a rational number, contradiction. Thus $\frac{y}{m}$ is an irrational number, and since there exist positive irrational numbers, that's ok.

An example: $\forall n\in\mathbb{N}-\{0\},\ 0<\frac{\sqrt{2}}{2n}<\frac{1}{n}$

At first sight, I would believe B :)

If A tells the truth, then he lies... Impossible
If C tells the truth (which implies that B lies), then B is right... Impossible
If B lies, then A tells the truth... Impossible