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Cardinality of Complex vs. Real

  1. Nov 25, 2004 #1
    Prove that the set of complex numbers has the same cardinality as the reals.

    What I did was say that a + bi can be written as (a, b) where a, b belong to real. Which essentially means i have to create a bijection between (a, b) and z (where z belongs to real).

    a = 0.a1a2a3a4a5...
    b = 0.b1b2b3b4b5...


    z = 0.a1b1a2b2a3b3....

    Is there anything wrong with that?
  2. jcsd
  3. Nov 25, 2004 #2


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    You need to do a little more work. First, does that constitute a one to one correspondence between the reals and the complex numbers? Also, you need to deal with ambiguous representations of certain numbers such as 0.5 and 0.4999....
  4. Nov 25, 2004 #3
    Well essentially that shows that there is a one to one relationship between the reals and the cartesian plane (x, y). Also, there is a one to one relationship between complex and the cartesian plane. Thus, the cardinality for all 3 is the same.

    Pertaining to the ambiguosity of certain number, I'm not sure if i see how they pose a problem because if x is 0.5 or 0.499999 you get a different result.

    ps. pardon my ignorance. I just started set/number theory two days ago :tongue2: .
  5. Nov 26, 2004 #4


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    The problem is that [itex]0.5[/itex] and [itex]0.4\bar 9[/itex] are the same number and lead to different results!
  6. Dec 2, 2004 #5


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    he only needs an injection from the complexes to the reals, since there is an inclusion the other way, and if he assumes his reals are finite or infinite decimals not ending in all 9's, then his map never sends any pair of reals to a decimal ending in all 9's. so he does get an injection.
  7. Dec 28, 2004 #6
    There is no real number 4.9999....
    4.99... and 5 are just two metha-variables to denote the number 5 in R .Here denote means "interpretation." .
    Last edited: Dec 28, 2004
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