Partial ordering of natural numbers

Question

How can I prove that the partial ordering of natural numbers has no least element? I genuinely have no idea how to do this. Q3c. This is not a homework/assignment task. It's a past exam paper which I am attempting to solve in preparation for my exam

Answer 1

By a "least element" I assume you mean an element $X$ such that every element $Y$ satisfies $Y≥X$? But given $x\in \mathbb N$ consider the element $x+1$. We can't have $x≤x+1$ in your system since $2x$ is never less than $x+1$ in $\mathbb N$.

I believe that some people use "least element" to refer to an element $x$ such that there is no element $y≠x$ with $y≤x$. That is a much weaker notion. In this case, $x=1$ works with your partial ordering.