Prove that there is no smallest positive real number
Solution 1:
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.
Solution 2:
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.