# Equivalence relation help...

1. Feb 23, 2016

### Kingyou123

1. The problem statement, all variables and given/known data

2. Relevant equations
a/b=c/d

3. The attempt at a solution
I started out with transitivity and I figure it's transitive because a/b=e/f = af=be=(a,b)R(e,f).
It can't be symmetric because ad does not equal bc, 1*4 does not equal 2*3
I'm I correct or completely wrong?

2. Feb 23, 2016

### Staff: Mentor

Wrong. What does equivalence mean? Can you define all properties? (first in general)

3. Feb 23, 2016

### TeethWhitener

You have to show that the relation R is:

1) Reflexive: (a,b)R(a,b)
2) Symmetric: (a,b)R(c,d) iff (c,d)R(a,b)
3) Transitive: If (a,b)R(c,d) and (c,d)R(e,f), then (a,b)R(e,f)

Can you work the rest out from here?

4. Feb 23, 2016

### Staff: Mentor

Note that in the problem statement it says $ad = bc \Leftrightarrow \frac a b = \frac c d$. IOW, these two equations are equivalent. You should not have the 2nd and 4th "equals" there.
You're missing the point. R would be symmetric if (a, b) R (c, d) implies that (c, d) R (a, b).

Think about what the relation as defined means, relative to the set of numbers in the problem. Do you understand why (1, 3) R (2, 6)?

Last edited: Feb 24, 2016
5. Feb 23, 2016

### Kingyou123

Symmetric= if (a,b)R(C,d) then (c,d)R(a,b) since ad=bc
Reflexive =since a/b=a/b
My confusion was with that I thought it had to be for all x values.

6. Feb 23, 2016

### TeethWhitener

Nope, all pairs of values.

7. Feb 23, 2016

### Kingyou123

Could you explain part b?

8. Feb 23, 2016

### Staff: Mentor

For example 1/2 = 2/4 = 3/6, so (1,2)R(2,4) and (2,4)R(3,6). These three pairs are all equivalent and build together an equivalent class. One of them represents this class. But there are more classes.