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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.

Sarvesh Ravichandran Iyer
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N3X
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    Why is this being voted down? The user stated a clear problem and even provided a complete solution. (+1) for the question. – lulu Jan 09 '18 at 00:20
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    Spruce up what you've said(which is absolutely correct) with a few dashes of symbolic language.(That's what they say in the cook book). To me, this is understandable, and does not need further change.(+1). – Sarvesh Ravichandran Iyer Jan 09 '18 at 00:20
  • To your question itself: I find the arguments you gave to be complete and accurate. I don't think any more should be required. – lulu Jan 09 '18 at 00:22
  • To make it more "mathematical", perhaps you could just say that $\forall x,y\in V$ before you proceed. For the proving that even numbers cannot be powers of $3$, you could simply write out a general form of the prime factorisations (maybe that seems a bit more convincing than reasoning through it with language) – Harry Alli Jan 09 '18 at 00:23

2 Answers2

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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.

BallBoy
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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.

Mohammad Riazi-Kermani
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