Given a relation $R$, is it reflexive? Symmetric? Transitive?

Question

Define the relation $R$ on the set $\mathbf Z^+$ of all positive integers by: for all $a,b \in \mathbf Z^+,aRb$ if and only if $gcd(a,b)\gt 1$.

(a) is $R$ reflexive? Symmetric? Transitive?

so here is my work:

$R$ is not reflexive. A counterexample: let $a=1\in\mathbf Z^+$ then $gcd(1,1)=1 \not\gt1$

$R$ is symmetric. Proof: let $a,b \in \mathbf Z^+$ so that $aRb$ . Then $gcd(a,b)\gt1$. Since $gcd(a,b)=gcd(b,a)$ then $gcd(b,a)\gt1$ which means $bRa.$

$R$ transitivity? so I am stuck on this. I can't come up with a counterexample but I can't prove it either.

(b) Find the number of integers $a\in \{1,2,3,...,100\}$ so that $aR4$.

ok, so this would mean that $gcd(a,4)\gt 1$,so this would the number of even numbers in the above set? so $\frac {100}{2}=50$?

(c) Find the number of integers $a \in \{1,2,3,...,100\}$ so that $aR10$.

This would be the same as (b) wouldn't it?

Answer 1

For transitivity in (a) try $2,6$, and $3$. Your answer to (b) is fine. For (c) remember that anything divisible by $5$ is related to $10$, and not all of those are even.

Answer 2

Hint for transitive: to show a counterexample you need $aRb$ and $bRc$ but $a \not R c$. Suppose $\gcd(a,b) \neq \gcd(b,c)$ but both are greater than $1$

Part b is correct, but c is not. There are some odd numbers $a$ such that $aR10$