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

Which partial ordering? If you use the "normal" way to order natural numbers 0 (or 1 depending on what your set of natural numbers contain) will be the least element. Or do you mean integers? — Ove Ahlman, Nov 03 '15 at 10:06

score 0 · Accepted Answer · answered Nov 03 '15 at 10:50

0

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.

answered Nov 03 '15 at 10:50

lulu

70,402

that makes sense! the first paragraph is the one i'll use. – Azza Nov 03 '15 at 11:38
how can i solve 3b? – Azza Nov 03 '15 at 11:39
Same example works. In this order, $x$ and $x+1$ are not comparable. – lulu Nov 03 '15 at 12:07

Partial ordering of natural numbers

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