Determine whether the binary relation R on N given in each of the cases below is reflexive (r), symmetric (s), transitive (t) or antisymmetric (a), and state whether it is an equivalence relation, an order relation or neither of those.

  • (a) $a\operatorname Rb \iff a < b$;
  • (b) $a\operatorname Rb \iff b ≤ a$;
  • (c) $a\operatorname Rb \iff a ≤ b + 1$;
  • (d) $a\operatorname Rb \iff 3^m a = 3^n b$ for some $m,n ∈\Bbb N∪\{0\}$;

Hi, i'm a bit confused on question D on the worksheet above, The indices are throwing me off a bit, How would I go about in determining whether or not the relation is Reflexive, symmetric, transitive and antisymmetric? Thank you.

  • $\begingroup$ Linking to a worksheet is not an appropriate way to ask a question. Please at least write the question out yourself. $\endgroup$
    – Matt
    Oct 16, 2019 at 18:39
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    – Mefitico
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  • $\begingroup$ Consider $a=4$ and $b=5$. Are there any natural numbers $m$ and $n$ that you can pick that makes $3*4m=3*5n$? $\endgroup$
    – Kevin Long
    Oct 16, 2019 at 19:23

1 Answer 1


Hint: $3^m a= 3^nb$ means $\frac {3^m a}{3^nb}=1$ which means $\frac ab = \frac{3^n}{3^m} = 3^{n-m}$.

So this is just say that you can get from $a$ to $b$ multiplying or dividing by $3$ some number of times. Example $4R 36$ because $3^2*4 = 3^0*36$. But $24 \not R 36$ becuse $24 = 8*3$ while $36=4*3^2$ and there is no way you can make $8*3^k = 4*3^j$.

So reflexive.

Are there $m,n$ so that $3^n a = 3^m a$ for all $a$? Well, whenever $m=n$ that is true.


If $3^ma = 3^nb$ does that mean there are $m',n'$ so that $3^{m'}b = 3^{n'} b$. Well, of course! Just use the same values. $3^nb =3^ma$.

(Or you could use any values where $n'-m' = m-n$)


If $3^ma = 3^nb$ and $3^jb =3^kc$ does that mean there are $w,v$ so that $3^wa = 3^vc$?

Well, yes use my dividing/multiply by three to get from one to the other.

$3^ma = 3^nb$ so

$3^{m+j} = 3^{n}(3^jb)= 3^{n}(3^kc)= 3^{n+k}$. Yep.

And anti-symmetric.

Does $3^n a= 3^mb$ simultaneously $3^kb =3^ja$ imply $a=b$? Of course not. $a = 3^{m-n} b=3^{j-k} b$ and it's certainly possible to have $m-n=j-k$ be other values than $0$.


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