Question

Prove the following statement:

Suppose that (a, b),(c, d),(m, n),(pq,) ∈ S. If (a, b) ∼ (c, d) and (m, n) ∼ (p, q) then (an + bm, bn) ∼ (cq + dp, pq).

Answer 1

●Here (a,b) means (a/b)

●i.e.,S is the subset of rational numbers.

● Given that (a,b)~(c,d) , this implies that

a/b~c/d implies that ad=bc

similarly (m,n)~(p,q) implies that mq=pn

●Thus we have ad=bc. .....( i )

and mq=pn .......( ii )

● Now multiplying by "nq" to equation ( i ) and by "bd" to equation ( ii ).

● We get nqad=nqbc ...( iii)

and. bdmq= bdpn. .....( iv)

● Adding (iii ) and (iv).

●we have   

nqad+ bdmq= nqbc+ bdpn

●Implies that (dq)(an+ bm)= (bn)(qc+dp)

●Hence (an+bm,bn)~(qc+dp, dq)