Relation defined on the set of real numbers by xRy when $x^2 + y^2 = 1$. Show whether or not R is reflexive, symmetric, antisymmetric or transitive.

Question

Let R be the relation defined on the set of real numbers by $xRy$ whenever $x^2 + y^2 = 1.$ Show whether or not $R$ is reflexive, symmetric, antisymmetric or transitive.

All right so I think I've got this right and just want to make sure.

• Symmetric - Yes for all $x$ and $y$, if $x^2 + y^2 = 1$ then $1 = x^2 + y^2 $

• Reflexive - Yes for all $x$ and $y$, if $x^2 + y^2 = x^2 +y^2$

• Anti-Symmetric - Yes for all $x$ and $y$, if $x^2 + y^2 = 1$ and $1 = x^2 + y^2$ then $x^2 + y^2 = 1$ is true.

• Transitive - Yes, if $x^2 + y^2 = x_2^2 + y_2^2 $ and $x^2 + y^2 = 1$ then $x_2^2 + y_2^2 =1$

I think I've done all these right, unless I've misunderstood the definitions for the relations.

Answer 1

You misunderstood. Instead of using the $R$ notation, say that $x \sim y$ if and only if $x^2+y^2=1$. So

• Not reflexive, as we have elements $x$ such that $x \not\sim x$. For example, $1 \not\sim 1$, as $1^2+1^2 = 2 \neq 1$.
• It is symmetric, because if $x \sim y$, then $x^2+y^2 = 1$. So $y^2+x^2=1$ says that $y \sim x$.
• Not transitive for reasons similar to reflexivity. For example, $1 \sim 0$ and $0 \sim 1$, but $1 \not\sim 1$.

And so on...

Answer 2

Unfortunately your answer is mostly wrong. For example; that $R$ is reflexive means that $xRx$ holds for all real numbers $x$. But clearly $x^2+x^2=1$ does not hold for all real numbers $x$, so the relation is not reflexive.

Transitivity and antisymmetry also fail, and symmetry does hold though you may want to verify this again with a proper understanding of what the relation is.