# Equivalence relation

Gold Member

## Homework Statement

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

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## 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

## Answers and Replies

MK5
what to proove

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

Gold Member
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)

Gold Member
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?

Science Advisor
Homework Helper
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.

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

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