For the open interval (-1, 1), intuitively I know that there is no least element in the interval, since there is always a negative value that's very close to -1 but never reaches -1, and it's less than some "supposed" least element.
For example, I know how to show that the interval (0, 1) has no least element:
By contradiction, suppose that there is a least element $p \in (0,1)$, but $ \frac{p}{2} \in (0,1)$ and $\frac{p}{2} < p$, which shows that $p$ is not the least element, which contradicts what was supposed. Thus there is no least element in (0,1).
However, for the interval (-1,1), I can't use the above method, since we're looking at the negative values for the least element, and negative value divided by a positive value will be greater than that negative value. So in this case, does it make sense to use absolute value for some element in (-1,1) and show that there is no least element?
Thanks
Suppose $p\in (-1,1)$ is a least element.
Then $-1 \lt p \lt 1 \implies-2 \lt p-1 \lt 0\implies- 1\lt \frac{p-1}{2} \lt 0\implies \frac{p-1}{2} \in (-1,1)$.
But $\frac{p-1}{2} \lt p $ is a contradition