Prove by contradiction that a real number that is less than every positive real number cannot be positive

Question

"Prove by contradiction that a real number that is less than every positive real number cannot be positive"

[This is what I did, but it is definitely missing something...

Proof: 1)Assume a real number, n, is less than every positive real and cannot be negative

But if n is less than every positive real number then n is less than the smallest positive real number and thus n can only be less than or equal to zero i.e $n\le0$

$n\le0$ is a contradiction as n cannot be negative

therefore proven by contradiction]

I'm sure that I'm missing something...

Answer 1

By contradiction, suppose that the number $x$ less than every positive real number is positive therefore let $y=\frac x 2>0$ and we have $y<x$.

Answer 2

Assuming that the nonnegative $x\in\mathbb{R}$ is less than every positive real number, then there are two possibilities:

1.) $x=0$,

or

2.) $x>0$.

If $x>0$, then $\exists y\in\mathbb{R}$ such that $y=\frac{x}{a}$ for some $a>1$. Conventionally $a=2$ is used, and therefore $y<x$ where $y>0$. However, we have assumed that $x$ is smaller than every positive real number, which is in contradiction with the fact that $y<x$ and $y\in\mathbb{R}$. Therefore it must be the case that $x=0$.