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In the book they solve it by taking examples. Let x be this,y that,z and y those and they just put in condition. But all values do not give same result.

I took $x$, $y$, $z$ as $\sqrt 3$, $\sqrt 2$ and $4$. Let $(x,y),\ (y,z) \in R$, $xy\in R$ and $yz\in R$ that gives $xy= \sqrt 6$ and $yz=4\sqrt 2$. On solving for $x$ and $z$ from these equations and they multiplying $x$ and $z$, $xz= \sqrt 6$ which is irrational and $(x,z)\in R$ but in answer they solved it like Solution

How different results from different values?

Nothing special
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Dingus45191
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  • What's the definition of a transitive relation? – dvdgrgrtt Nov 15 '23 at 06:16
  • Either it is true or it is not. If it is not the easiest way to prove is by example. Note that $1\ R \sqrt{2}$ and $\sqrt{2} R 1$ but it is not true that $1 R 1$. – copper.hat Nov 15 '23 at 06:17
  • To disprove something, you need to find a counterexample right? It may happen that there are tons of examples but that does not prove your statement. – Soumik Mukherjee Nov 15 '23 at 06:20
  • You are asked to check if $R$ is transitive for $all$ numbers. Your examples shows that the transitivity property is verified for a certain triplet of numbers but your task is really to either show that it is verified for all triplets or to find a triplet for which it is not verified. I like the counterexample $ (\Pi, 1, {1\over \Pi})$ – Daniele Nov 15 '23 at 06:21
  • Ok how to decide which examples to check – Dingus45191 Nov 15 '23 at 07:21
  • In questions they never say if it is transitive or not. They say to check. Sometimes it is Sometimes it isn't – Dingus45191 Nov 15 '23 at 07:22

1 Answers1

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A relation R on a set X is transitive if, for all elements a, b, c in X, whenever R relates a to b and b to c, then R also relates a to c.

Note that,

if transitivity holds,it is true for $\textbf{all}$ elements a,b,c in X.

Here,in your example (x,z)$\in$ R but,in your book's example,and also in the example in 2nd comment, (x,z)$\notin$ R.

So, (x,z)$\in$ R does not hold for all x,y,z $\in$ R.

So R is not transitive.

Math solver
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  • Ok how do I know which example and which values to take – Dingus45191 Nov 15 '23 at 07:21
  • You have to check some examples definitely.But if R is transitive,then giving an example supporting to it ,is not sufficient. Because, according to definition,you have to check for all x,y,z.In this case,you have to prove it in general.But to disprove it,one counter example is sufficient.,i,e,you can say that,the condition is not true for all x,y,z, i,e, you have at least one pair not satisfiing the condition of transitivity. – Math solver Nov 15 '23 at 07:37