I have the set $V = \{ -n, -(n-1), -(n-2), ... , -2, -1, 0, 1, 2, ... , n-2, n-1, n\}$ and the relation over it $xRy \iff x + y$ is a power of $3$ and I need to prove that it is symmetrical and anti-reflexive.
So $x + y$ is the same as $y + x$, and if $x + y$ is a power of three then so is $y + x$, so then $xRy$ is the same as $yRx$ which would mean that the relation must be symmetrical.
For the relation to be reflexive then $x + x$, would need to be a power of $3$, but $x + x = 2x$, which is an even number and so it can't be a power of $3$, and so $xRx$ doesn't hold, which means the relation is anti-reflexive.
All of this is rather simple to work out in my head, but how would I prove it mathematically? Some help would really be appreciated.
$\endgroup$ 42 Answers
$\begingroup$It looks perfectly mathematical as you've written it; where do you think there are gaps?
One minor fix -- as written you haven't shown that it's anti-reflexive, just that it's not reflexive (since you started with "For the relation to be reflexive" and then derived a contradiction). Instead, start with "Suppose $xRx$ for some $x \in V$" and derive a contradiction from there.
$\endgroup$ $\begingroup$You have done a very good job with your proof that $R$ is symmetric on $V$. For anti-reflexivity, you need to show that no element $x$ of of $V$ satisfies$ xRx$. You may prove that by contradiction. Suppose there is an element $x$ in $V$ for which $xRx$ is true. By definition of $ R$ that means $2x$ is a power of $3$ which is impossible because no power of $3$ is even.Therefore no element of $V$ satisfies the reflexivity condition.
$\endgroup$
