Let N the set of all natural numbers and R be the relation on N * N defined by (a,b)R(c.d)=ad(b+c)=bc(a+d). Check whether R is an equivalence relation on N*N.

Asked by ELLIYAS | 25th Apr, 2013, 06:11: PM

Expert Answer:

Answer : Given : Let N the set of all natural numbers and R be the relation on N * N defined by (a,b)R(c.d)=ad(b+c)=bc(a+d). ToCheck : whether R is an equivalence relation on N*N.
 
 
We observe that ad(b+c) = bc(a+d)
=> (b+c)/ bc = (a+d)/ad
=> (1/c) +(1/b) = (1/a) +(1/d)
=> (1/a) - (1/b) = (1/c) - (1/d)
therefore  (a,b)R(c,d) if and only if  
(1/a) - (1/b) = (1/c) - (1/d)
 
let (a,b) belongs to N *N
=>(1/a) - (1/b) = (1/a) - (1/b)
therefore (a,b)R(a,b) 
=> R is reflexive
 
let (a,b)R(c,d)
=>(1/a) - (1/b) = (1/c) - (1/d)
=>(1/c) - (1/d) = (1/a) - (1/d)
therefore (c,d)R(a,b) 
=> R is symmetric
let (a,b)R(c,d) and (c,d)R(e,f)
=>(1/a) - (1/b) = (1/c) - (1/d)  and (1/c) - (1/d) = (1/e) - (1/f) 
=>(1/a) - (1/b) = (1/e) - (1/f)
therefore (a,b)R(e,f) 
=> R is transitive
 
 
=> R is an equivalence relation 

Answered by  | 25th Apr, 2013, 07:05: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.