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Recall that 1 isn't prime or composite.

I cannot think of any combinations. I am trying to find a counterexample to show that this relation is not transitive.

Arctix
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  • Do you mean "that divides $xy$, $yz$ but not $xz$"? Otherwise I have a hard time understanding the question. – frog Feb 19 '16 at 08:13
  • If the prime number is different then $x=2\cdot3$, $y=3\cdot5$, $z=5\cdot 7$. If the prime number must be the same, then it is impossible because for the hypothesis it follows that $x=p\cdot a$, $y=p\cdot b$, $z=p\cdot c$, thus they are all divisible by $p$ – Giovanni Resta Feb 19 '16 at 08:15
  • What is "and" in this case? $+$? – Arthur Feb 19 '16 at 08:15

1 Answers1

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Assuming you don't expect it to be the same prime in every case, then $x=2, y=6, z=3$ is a solution.

πr8
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  • This is the proper solution. There is a prime number p that divides both 2 and 6; namely, p = 2. There is a prime number p that divides both 6 and 3; namely, p = 3. There is not a prime number p that divides both 2 and 3. Thanks! I apologize for the confusion in the question; it's a tough one to word! – Arctix Feb 19 '16 at 08:18
  • @Arctix No worries. Given what I think you're trying to prove, the ideal statement might be: "Show that the relation R given by $xRy \iff \text{hcf}(x,y)>1$ is not transitive". – πr8 Feb 19 '16 at 08:20