# Equivalence relation

1. Apr 7, 2015

### Suraj M

1. The problem statement, all variables and given/known data
If a relation R on N × N is
(a,b)R(c,d) iff
2. Relevant equations
--
3. The attempt at a solution
I got the reflexive and symmetric parts but not the transitive part....
here's what i have
$(a,b)(c,d)∈R and (c,d)(e,f)∈R$
To prove $(a,b)(e,f) ∈ R$ .i.e., $af(b+e)= be(a+f)$
i have
$$ad(b+c) = bc(a+d)$$and$$de(c+f) = cf(d+e)$$
my attempt was...
multiplying we get $$afcd(b+c)(d+e) = becd(c+f)(a+d)$$
$$af(b+c)(d+e) = be(c+f)(a+d)$$
by cancelling $afbe$ on both sides i get
$$af(bd+cd+ce) = be(ac+cd+fd)$$
stuck here :(
is this a wrong method, if not how do i proceed??,
Thank you

2. Apr 7, 2015

### MK5

what to proove

3. Apr 7, 2015

### haruspex

See if you can rearrange R into the form f(a,b) = f(c,d) for some function f.

4. Apr 8, 2015

5. Apr 8, 2015

### Suraj M

Im sorry, but could you please tell me how i could do that?

6. Apr 8, 2015

### Dick

Do some algebra. If $(a,b)R(x,y)$ then $ay(b+x)=bx(a+y)$. Try to rearrange that equation so $x$ and $y$ are on one side and $a$ and $b$ are on the other.

7. Apr 8, 2015

### Suraj M

Il get $ab(x-y)=xy(a-b)$
How will that help..?

8. Apr 8, 2015

### Suraj M

Oh brilliant!!
Got it thanks!!!!
All of you thank you..