State whether or not the relation on the set of reals is reflexive, symmetric, antisymmetric or transitive State whether or not the relation on the set of real numbers is reflexive, symmetric, anti-symmetric or transitive.
$$R= \{(x,y)\mid x=1\text{ or }y=1\}$$
This is what I have done up to now, not sure if I am right though.

i) it is reflexive.
for any arbitrary $x\in R$; $(x,x)\in R$
$x=1$ or $x=1$
ii) it is symmetric
since if $x=1$ or $y=1$ then $y=1$ or $x=1$
iii) it is not antisymmetric
counterexample.
$(1,5)\in R$ and $(5,1)\in R$
but $x\ne y$
IV) it is not transitive.
counterexample:
$(7,1)\in R$ and $(1,7)\in R$
But $(7,7)\notin R$

 A: The relation is clearly not reflexive: $\langle 2,2\rangle\notin R$. In fact, the only pair $\langle x,x\rangle$ that is in $R$ is $\langle 1,1\rangle$.
It is symmetric; your explanation isn’t very clear, but I suspect that you have the right idea. Here’s a more complete explanation:

Suppose that $\langle x,y\rangle\in R$; then either $\langle x,y\rangle=\langle 1,y\rangle$, or $\langle x,y\rangle=\langle x,1\rangle$. If $\langle x,y\rangle=\langle 1,y\rangle$, then $\langle y,x\rangle=\langle y,1\rangle\in R$, and if $\langle x,y\rangle=\langle x,1\rangle$, then $\langle y,x\rangle=\langle 1,x\rangle\in R$, so in all cases $\langle y,x\rangle\in R$, and $R$ is therefore symmetric.

Your argument that $R$ is not antisymmetric is basically right, though you really ought to conclude with $1\ne 5$ rather than with $x\ne y$. Your argument that $R$ is not transitive is fine. (Note that in the latter you observed that $\langle 7,7\rangle\notin R$: this should immediately tell you that $R$ is not reflexive.)
