Originally Posted by

**Matter_Math** Yep, I seen your edit.

Now instead of starting a new thread. I have 1 more question. Its part 2, from the question above. Here is the question :

**Part (II). Define the relation ~ on Z x Z by the rule (a,b) ~ (a`,b`) if and only if there is a rational number r, so that a = a` * r and b = b` * r . Is an equivalence relation? If so, prove it; if not, give a counter example.**

It is an equivalence relation. Here is the proof that I am asking you to check please:

~ is reflexive because :

(a,b) ~ (a,b) since a = a * 1, b = b * 1, thus r = 1

~ is symmetric because

if (a,b) ~ (a`,b`) then a = a`*r , b = b`*r, for some rational r.

then a` = 1/r * a , b` = 1/r * b.

thus if (a,b) ~ (a`,b`) for some r

then (a`,b`) ~ (a,b) for some 1/r

~ is transitive because :

[a] if (a,b) ~ (a1,b1) iff a = a1 * r1, b = b1 * r1, for some rational r1

[b] if (b,c) ~ (b2,c1) iff b = b2 * r2, c = c1 * r2, for some rational r2

[c] the (a,c) ~ (a1,c1) iff a = a1 * r3, c = c1 * r3, for some rational r3

since :

a = a1 * r1 //From [a]

a1 = 1/r1 * a

a = a1 * r3 //From [b]

a = (1/r1)*a * r3

r1*a = r3*a

r1 = r3

c = c1 * r2 //From [b]

c1 = 1/r2 * c

c = c1 * r3 //From [c]

c = (1/r2) * c * r3

r2*c = r3*c

r2 = r3

Since r1 = r3, and r2 = r3, thus r1 = r2.

Thus (a,b) ~ (a1,c1) for some rational r2.

Is the above correct? I wasn't sure how to close the proof. Thank you again.