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?
- SamwiseLv 77 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
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
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.