proof by contradiction

**AU11** · October 31st 2012, 12:08 PM

Hi,

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

I have started by saying let greatest value = x, where x is a member of S and x is greater than or equal to n

then let n=13k for some natural number k

x is greater than or equal to 13k

not sure where to go from here??

thanks

**Plato** · October 31st 2012, 12:48 PM

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

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

**AU11** · October 31st 2012, 12:54 PM

Hi,

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

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

**Plato** · October 31st 2012, 02:02 PM

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

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

**AU11** · November 2nd 2012, 07:12 AM

ohh right, I understand it now!

thanks for the help

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