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I got $(x,y)R(u,v) \Leftrightarrow x + v = y + u$

I have to prove that this is a transitive relation. We did not do any examples how to do this at school so as far as I came was:

$(x,y)R(a,b) \wedge (a,b)R(u,v) \Rightarrow (x,y)R(u,v)$

how do I go on from this point?

Matjaž Jerman
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2 Answers2

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You can't just write the implication. If that was the case you could just right that and you would be done.

You need to suppose that $(x,y) R (a,b)$ and $(a,b)R(u,v)$ then use the definitions of the relation to prove it.

$(x,y) R (a,b) \implies x+b=y+a$

$(a,b)R(u,v) \implies a+v=b+u$

you want to show $x+v=y+u$ now using the last two lines above can you do this. It shouldn't be too trick so I'll let you finish.

Also type in LaTeX much easier to read.

Sam D
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Suppose that $(x,y)\mathcal{R}(a,b)$ and also that $(a,b)\mathcal{R}(u,v)$. We wish to show that this implies that $(x,y)\mathcal{R}(u,v)$.

Since $(x,y)\mathcal{R}(a,b)$, by our definition of $\mathcal{R}$ that means that $x+b=y+a$.

Since $(a,b)\mathcal{R}(u,v)$, by our definition of $\mathcal{R}$ that means that $a+v=b+u$.

Now, try to combine those two equations in a convenient way to make it look at the end like $x+v=y+u$. Remember that this is what the final line should look like. Do not start with this as your first line.

Hint: What does $x+b+a+v$ look like?

JMoravitz
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  • I made it and came to x+v = y+u. So that is basically the end of proof? – Matjaž Jerman Nov 16 '15 at 00:08
  • Assuming that all steps in $x+v=\dots=\dots=y+u$ are correctly justified, yes. Since we did not say which $x,y,a,b,u,v$ these were (only that $(x,y)\mathcal{R}(a,b)$ and that $(a,b)\mathcal{R}(u,v)$), it must be true for all $x,y,a,b,u,v$ that satisfy the hypothesis, thereby proving that $\mathcal{R}$ is indeed transitive. – JMoravitz Nov 16 '15 at 00:10