#### AU11

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
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 $$\displaystyle n=13k~\&~n=13k<13(k+1)=n+13~?$$

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#### AU11

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
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.

#### AU11

ohh right, I understand it now!

thanks for the help