# Relations - Proving symmetry/anti-symmetry with a defined set

Struggling a bit on this question regarding relations. Would appreciate your help immensely.

Determine if a relation is reflexive, symmetric, anti-symmetric & transitive, for:

$$R = \{(x,y) : xy = 4\} \text{ defined by } A = {0,1,2,3}$$

I've determined that it is not reflexive, as $$x^2 = 4$$ is only true for $$(2,2)$$.

I think it is not symmetric as $$yx = 4$$ is only true for $$(2,2)$$

I don't know how to prove anti-symmetry or transitive in this case, either.

The third question is the same, but the set was $$\{0,1,2,3,4\}$$ so I proved it was anti-symmetric by using $$(1,4)\in \mathbb{R}$$ and $$(4,1)\in \mathbb{R}$$, where $$1\ne4$$.

Thank you.

• check the definition of symmetric – J. W. Tanner May 10 at 19:53
• Would it be 4 = x/y? @J.W.Tanner – Ian May 10 at 20:25
• It would be $(x,y)\in R \iff (y,x)\in R$ – J. W. Tanner May 10 at 20:29

Check the definitions.

Symmetric means $$(x,y)\in R\iff (y,x)\in R$$ ; i.e., $$xy=4 \iff yx=4,$$

which is true because multiplication is commutative, so $$R$$ is symmetric.

Anti-symmetric means if $$(x,y)\in R$$ and $$(y,x)\in R$$ then $$x=y.$$

When the set is {$$0,1,2,3$$} and $$(x,y)\in R$$ and $$(y,x)\in R$$, then $$x=y=2,$$ so $$R$$ is anti-symmetric,

but when the set is {$$0,1,2,3,4$$} then $$(1,4),(4,1)\in R$$ so $$R$$ is not anti-symmetric because $$1\ne4$$.

Transitive means if $$(x,y)\in R$$ and $$(y,z)\in R$$ then $$(x,z)\in R.$$

$$R$$ is not transitive for {$$0,1,2,3,4$$} because $$(1,4)\in R$$ and $$(4,1)\in R$$ but $$(1,1)\not\in R$$,

but $$R$$ is transitive for {$$0,1,2,3$$} because in that case only $$(2,2)\in R$$.