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Let $R$ be the relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(w, x)R(y,z)$ if and only if $w + x \leq y + z$. Let $S$ be the relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(w, x)S(y,z)$ if and only if $w \leq y$ and $x \leq z$.

(a) Is $R$ antisymmetric?

(b) Is $S$ antisymmetric?

I said no for $R$ because it is symmetric therefore cannot be antisymmetric and yes for $S$ because $(w,x)$ must always be less then or equal to $(y,z)$. Just wanted to make sure that I was correct, or am on the right track.

Gibbs
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1 Answers1

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If $(w,x)R(y,z)$ then $w+x \leq y+z$. For example, $(1,2)R(3,4)$. Suppose you want to check the symmetry in this case: you should get that this implies $(3,4)R(1,2)$, that is $3+4\leq 1+2$. This is clearly false. So $R$ is not symmetric.

Your discussion on $S$ looks weak: assume $(w,x)S(y,z)$ and $(y,z)S(w,x)$. Is it true that $(w,x) = (y,z)$? Well, the first means $w \leq y$ and $x \leq z$. The second $y \leq w$ and $z \leq x$. So $w \leq y \leq w$ and $x \leq z \leq x$, hence...

Gibbs
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  • So is R also antisymmetric? I was just taking into consideration of the equals part and not the less then.. –  Apr 29 '18 at 23:34
  • Try to apply a similar reasoning as the one I used for $S$ to check if $R$ is antisymmetric. – Gibbs Apr 29 '18 at 23:45
  • If (w,x)S(y,z) and (y,z)S(w,x) for w+x ≤ y+z, then y+z ≤ w+x is for the second one, then w+x ≤ y+z ≤ w+x, but because the numbers can switch such that it can be either w+x or x+w, it is not antisymmetric? I feel like I'm going in a completely different direction.. –  Apr 29 '18 at 23:51
  • Are you using $R$ or $S$? – Gibbs Apr 29 '18 at 23:52
  • I'm using S as a guide to answer R –  Apr 29 '18 at 23:54
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    Then your conclusion is unclear to me, but looks like you are almost there. To fix this, take an example: $(1,2)R(2,1)$ because $1+2 \leq 2+1$. Analogously, $(2,1)R(1,2)$. However $(1,2) \neq (2,1)$. So $R$ is not antisymmetric. – Gibbs Apr 30 '18 at 00:00