Cardinal of C
in [Math]
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From: Pollux on 4 Jun 2010 20:11 Wait I'm confused! Non-uniqueness of decimal representations? I'm not sure I understand what that means. Can we sidestep that if we represent a and b in base 2? Pollux
From: Pollux on 4 Jun 2010 20:24 I've tried a few numbers, and I'm not yet seeing the problem with "non-unique decimal representations" causing to have only an injection. For the numbers below, it seems there is a unique real for each complex, and that I can uniquely decode that real back into a unique complex. What am I missing? Can you show numbers that would illustrate the problem? z1 = .1111 + .2222i -> .12121212 z2 = .1111 + .1111i -> .11111111 z3 = .1111 + .0000i -> .10101010 z4 = .0000 + .1111i -> .01010101 Thanks! Pollux
From: Pollux on 4 Jun 2010 20:59 Got it. Looked up "decimal representation" on wikipedia and realized that .999... and 1 were two representations for the same number. Hmmm... Pollux
From: David R Tribble on 5 Jun 2010 01:10 Daniel Giaimo wrote: >> Yes there is one. One simple way to get one is to map a complex number a >> + bi to the real number you get by interleaving the decimal >> representations of a and b. > Daniel Giaimo wrote: > Actually, I posted too soon. This doesn't quite work due to the > non-uniqueness of decimal representations. It does, however give you an > injection from C to R, so, at least assuming the axiom of choice, it > proves that |C| <= |R|. |R| <= |C| is clear, so |R| = |C|. Another observation is the theorem that any n-tuple of reals has the same cardinality as R. In other words, an n-dimensional real space R^n has the same cardinality as R. Thus C, which is essentially a set of pairs of reals <a,b> = a+bi, has cardinality |R^2| = |R|. -drt
From: Pollux on 4 Jun 2010 22:07
Is the proof difficult? Can I find it somewhere on the internet? Pollux |