proof by contradiction

Oct 2012
3
0
Kent
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

MHF Helper
Aug 2006
22,507
8,664
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 \(\displaystyle n=13k~\&~n=13k<13(k+1)=n+13~?\)
 
Last edited:
Oct 2012
3
0
Kent
Hi,

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

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

Plato

MHF Helper
Aug 2006
22,507
8,664
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.
 
Oct 2012
3
0
Kent
ohh right, I understand it now!

thanks for the help