Prove that there is no smallest positive real number I have to prove the following: 
$$\text{Prove that there is no smallest positive real number}$$
Argument by contradiction
Suppose there is a smallest positive real number. Let $x$ be the smallest positive real number: 
$$x : x \gt 0, x \in \mathbb{R}$$
Let $y$ be $\frac{x}{10}$. Contradiction. This implies that $y < x$ which implies that you can always construct a number that is less than the "smallest positive real number". QED. 
Can someone please verify the write up of the proof and the proof itself? 
Thanks for your time!
P.S. I have seen this and this but I'm not looking for a way to approach the problem but rather verification and write up help.
P.P.S If there is another novel way of approaching this problem, I would like to know! 
 A: Another way to show this is using the Archimedian property of natural numbers. That is, the natural numbers don't have an upper bound in the reals. 
Consider any small real number, $\epsilon>0$. Since natural numbers are unbounded, there exists some $n \in \mathbb{N}$ such that $n>\frac{1}{\epsilon}$. Rearranging gives that $\epsilon>\frac{1}{n}$. Thus for any small positive real number $\epsilon$, there is a smaller positive real number $\frac{1}{n}$.
I like this proof because it shows the connection between really big numbers and really small numbers.
A: There is absolutely no need to use contradiction. Just prove the statement directly: there is no smallest positive real number. What this means is that if $r$ is a positive real number, then it isn't the smallest one. And indeed it's not because $\frac{r}{10}$ (or whatever) is smaller. 
A: I understand this is an old thread, but this is the first search result for this problem on google and I would like to add my own proof for the nonexistence of a smallest positive real number. 
Proof. 
Assume $a \in \mathbb{R}$ is the smallest positive real number. That is, for any positive real number $r \neq a$ , $0 \lt a \lt r$. Let $x \in \mathbb{R}$ be a real number greater than 1. Since $x$ is real, it has a real multiplicative inverse, $y \in \mathbb{R}$ such that $xy=1$. Therefore $y= \frac 1x$. Since we know $0 \lt 1 \lt x$, diving through by $x$, we see that $0 \lt \frac 1x \lt 1$. Multiplying by $a$ then yields $0 \lt a \cdot \frac 1x \lt a$. Thus ($a \cdot \frac 1x$) is a positive real number, as it is the product of two positive real numbers, and it is less than $a$. But there cannot be a positive real number less than $a$. This is a contradiction. Therefore there does not exist a smallest positive real number. $\blacksquare$
A: If $x$ is the smallest real,
then $0 < x < 1$.
But 
$1-x > 0$ so
$0 < x(1-x)
=x-x^2$
so $x^2 < x$.
(Just wanted to come up with a different answer.)
A: Here is a slightly different way to organize the proof.  What we will do is split it into two parts:


*

*For every positive real number there is another positive real number less than it.  Proof: Let $x>0$.  Then since $0<\frac{1}{2}<1$, we have $x>\frac{1}{2}x>0$, and so $\frac{1}{2}x$ is such a number.

*There is no smallest positive real number.  Proof: Assume for sake of contradiction that $x$ is the smallest such.  Then by 1 there is a smaller such number, contradicting minimality.
The idea with splitting the proof into two statements is that we have isolated the proof by contradiction into a very small part.  The risk with proof by contradiction is that, since you are in fact assuming something which is false from the beginning, any mistaken reasoning after that will look like a valid completion to the contradiction proof.
