# Partial ordering on Natural numbers

Preparing for exams, and came across this past year question. Any ideas?

I know that for partial order, it must be reflexive, transitive and anti-symmetric, but how exactly do i show this?

• Next step is to recall what those terms actually mean. Can you show at least some of them? (one of them should be completely trivial). Oct 26, 2015 at 11:37
• reflexive means a<=a so in this case, since x=y, we can say that the relation is reflexive, coz x=y is effectively saying a<=a
– Azza
Oct 26, 2015 at 11:38
• Great (though you would probably want to phrase it a bit more precisely on a test). What about the other two? Oct 26, 2015 at 11:40
• is it antisymmetric for the same reason. So anti-symm means, if a<=b and b<=a, then a=b. So if x<=y, and y<=x, x=y (which is what we were told)
– Azza
Oct 26, 2015 at 11:42
• Not quite. $x\leq y$ might be because $x=y$ but it could also be because $3x\leq y$, so you need to account for this. Oct 26, 2015 at 11:43

• anti-symmetric: let's see what happens if $x \preceq y$, $y \preceq x$, but $x\neq y$ : we have $3x \le y$ and $3y \le x$, so $9x \le x$, so $x=y=0$, impossible
• transitive: if $x \preceq y$ and $y \preceq z$, and x!=y and y!=z (otherwise the relation is trivial), then $3x \le y$ and $3y \le z$ so $(3x \le) 9x \le z$ , $3x \le z$ so $x \preceq z$
• @Azza you have $9x \le x$ so $x=0$ and $3y \le x$ Oct 26, 2015 at 12:31