Re: proof by contradiction

Quote:

Originally Posted by

**AU11** I have to prove the following by contradiction

the set S = [all natural numbers n such that n is a multiple of 13] has no greatest element

We know that $\displaystyle (\forall k\in\mathbb{N})[k<k+1]~.$

If $\displaystyle n=\max(S)$ is it true $\displaystyle n=13k~\&~n=13k<13(k+1)=n+13~?$

Re: proof by contradiction

Hi,

I don't get why you have written k=13k?

plus do I not talk about x (the greatest element) anymore?

Re: proof by contradiction

Quote:

Originally Posted by

**AU11** Hi,

I don't get why you have written k=13k?

plus do I not talk about x (the greatest element) anymore?

See my edit.

If $\displaystyle x$ is a real number $\displaystyle x+1$ is a real number and $\displaystyle x<x+1$ so there cannot a greatest real number.

If $\displaystyle 0<x$ then $\displaystyle 0<\frac{x}{2}<x$ so there can be no smallest positive number.

Re: proof by contradiction

ohh right, I understand it now!

thanks for the help