# Prove for each r in R , Ar={(x,y): x^2 + y^2 = r^2} is a partition?

{Ar : r in R} is a family of subsets of RxR. Describe also geometrically and give the equivalence relation.

### 1 Answer

- kbLv 78 years agoFavorite Answer
Since r^2 = (-r)^2, we can assume without loss of generality that r ≥ 0.

For each fixed r ≥ 0, A(r) is a circle centered at (0, 0) with radius r.

(When r = 0, then A(0) is simply {(0, 0)}.)

Hence, {A(r) : r ≥ 0} is a family of concentric circles (same center at (0, 0)).

Equivalence relation:

(x, y) ~ (u, v) <==> x^2 + y^2 = u^2 + v^2.

(Under this relation, {A(r) : r ≥ 0} is the collection of distinct equivalence classes.)

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Proof that A(r) is a partition.

(i) Given (a, b) in R^2, note that (a, b) is in A(r), where r = √(a^2 + b^2).

Hence, (a, b) is in A(r) for some r ≥ 0.

(ii) Now, we show that A(r) ∩ A(s) = ∅ for distinct r, s ≥ 0.

If (a, b) is in A(r) ∩ A(s), then (a, b) is in both A(r) and A(s).

==> r = √(a^2 + b^2) = s, which contradicts r and s being distinct.

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I hope this helps1