# Thread: Relations, reflexive and transitive

1. ## Relations, reflexive and transitive

The relation R on real numbers given by

xRy <=> x-y E(belongs) to Q

can you help me with Symmetric and Transitive?

Someone told me this is how to show its symmectric

x-y = a/b

-(x-y) = y-x = -a/b

cant figure out transitive?

2. Here is something that you could do almost without thinking.

The definition of a transitive relation applied to R: for any three numbers x, y, z, if $\displaystyle x-y\in\mathbb{Q}$ and $\displaystyle y-z\in\mathbb{Q}$, then $\displaystyle x-z\in\mathbb{Q}$. One has to prove this. Let's fix some numbers x, y and z and assume that $\displaystyle x-y\in\mathbb{Q}$ and $\displaystyle y-z\in\mathbb{Q}$. This means there are rational numbers a / b and c / d such that x - y = a / b and y - z = c / d. One has to prove that x - z is a rational number.

Here you actually turn on the brain: x - z = (x - y) + (y - z). Is this number rational?

3. I suppose, since the sum of to rational numbers will be a rational number?

4. Yes.

5. I was wondering if you could tell me how to see if the relation R on real numbers given by

xRy <=> x*y > 0

is transitive?

I'd so no, since I can pick out numbers that will not satisfy
x*y > 0 and y*z > 0 => x*z > 0.

6. $\displaystyle x\cdot y>0$ if and only if they both positive or both negative. i.e have same sign not zero.

7. So the relation is transitive? if they share same sign and one of them or both isnt zero?

8. Yes it is transitive.
But neither can be zero. Both must be positive or both negative.