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"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...

stochasticmrfox
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    "then $n$ is less than the smallest positive real number" You seem to be ignoring the fact that there is no smallest positive real number. Zero is not positive. "and thus $n$ can only be less than or equal to zero" This needs proof. – JMoravitz Jul 21 '18 at 19:40
  • To prove this you should start with whatever properties of the order structure of the real numbers you have proved or assumed as axiomatic. Please edit the question to include an attempt of that kind and we may be able to help. – Ethan Bolker Jul 21 '18 at 19:43

2 Answers2

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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 $y<x$.

user
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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 $x=0$.

BoisterousLemma
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