Equivalence relation

[Suraj M]

Homework Statement

If a relation R on N × N is
(a,b)R(c,d) iff
ad(b+c) = bc(a+d)

Homework Equations

--

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

 

[MK5]

what to proove

 

[haruspex]

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

 

[Suraj M]

what to proove

To prove (a,b)(e,f)∈R(a,b)(e,f) ∈ R .i.e., af(b+e)=be(a+f)af(b+e)= be(a+f)

 

[Suraj M]

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

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

 

[Dick]

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

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.

 

[Suraj M]

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

 

[Suraj M]

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