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# Let R be the relation on Z defined by mRn if and only if mn>0 or m=n=0 1 Prove R is an equivalence relation 2 How many distinct equivalence?

### 1 Answer

- SamwiseLv 711 months agoFavorite Answer
1. We're going to need to prove each of the various characteristics of an equivalence relation for two different cases:

mn > 0

and

m = n = 0

But that doesn't complicate things too much. I am assuming the conventional use of Z, denoting the set of integers.

R is reflexive, because for any integer a, either

a^2 > 0

or

a = 0

and therefore aRa.

Now consider any integers a and b such that aRb.

Then either ab > 0, in which case ba > 0; or

a = b = 0, in which case b = a = 0.

Thus if aRb, then bRa, and R is symmetric.

Now consider any integers a, b, c such that aRb and bRc.

It's possible that aRb because a = b = 0,

and in that case since b = 0,

bRc tells us that c = 0 as well,

so a = c = 0 and aRc.

If a and b are nonzero, then aRb tells us that ab > 0

and so a and b have the same sign.

In that case, bRc tells us that b and c have the same sign,

so a and c have the same sign and ac > 0,

and so aRc.

These two cases establish that R is transitive.

We've shown R is reflexive, symmetric, and transitive,

and so R is an equivalence relation.

2. I assume this should have asked, "How many distinct equivalence CLASSES?"

There are three classes: positive integers, negative integers, and zero.

For any two positive integers m and n, mn > 0 and so mRn.

For any two negative integers m and n, mn > 0 and so mRn.

If m = n = 0, then mRn.

So each of these classes has all its elements related by R.

But if only one of integers f and g equals zero,

then fg = 0 and f ≠ g and so fRg is not true.

If one of f and g is positive and one negative,

then fg < 0 and f ≠ g, and so fRg is not true..

So no two members of two different classes, as described above,

is related by R.