\"Prove or Disprove: Let A and B be sets. Then the sets A x B and B x A are equi

Question

"Prove or Disprove:

Let A and B be sets. Then the sets A x B and B x A are equinumerous.

Note 1: For this problem, you may indicate that X is equinumerous to Y by writing X Y

without further comment.

Note 2. Be especially careful about how you handle the cases where either A or B is the

empty set. "

* I'm not sure if this is a proof or something I should consider disproving. State each case accordingly and thoroughly!

** If I can't read or understand your image/picture/solution, I won't rate. I'll ask questions and if you can guide me I'll rate you well! :) Thanks in advance!

Explanation / Answer

Let's deal with the empty set cases first.

Suppose A={ } or B={ } (empty set) . In this case A x B = B x A = { }.

Indeed if you want to take an element of A x B, then (a,b) cannot exist because either A or B is the empty set. A x B and B x A are then equinumerous (using the empty function from { } to { })

(http://en.wikipedia.org/wiki/Empty_function)

Suppose neither A and B are empty and define f : AxB -> BxA by f(a,b) = (b,a)

f is clearly one-to-one since f(a,b)=f(c,d) => (b,a)=(d,c) => (a,b)=(c,d)

f is also clearly onto since for (b,a) in BxA, f(a,b) = (b,a)

So we have made a bijection between AxB and BxA.

So AxB and BxA are equinumerous

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