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#1 2006-04-16 11:10:38

Registered: 2006-01-24
Posts: 37

Numerically Equivalent Proof

I need to the prove the following:

Prove that if A and B are finite sets, then A ≈ B if and only if |A| = |B|.

This is confusing me because doesn't the definition of numerically equivalent actually say that the cardinality of A = the cardinality of B? If so, does that mean I can just say "By the definition of numerically equivalent"?


#2 2006-04-16 15:21:03

Registered: 2005-12-04
Posts: 3,791

Re: Numerically Equivalent Proof

Nope, A ≈ B iff there exists a function θ that is 1-1 and onto.

It's actually a lot easier than it seems.  What you need to do is define the sets A and B without a loss of generality.  So just make up names for them.  a0, a1, a2, ..., a_n are all in A and b0, b1, b2, ..., b_n are all in B.  You know they each go to n since they are finite and have the same number.  Then let θ:A->B such that θ(a_k) = b_k.  It should be really easy to show 1-1 and onto with this map.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."


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