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B Cardinality and Natural Numbers

  1. Apr 14, 2017 #1
    Hi,
    I hoping someone might be kind enough to possibly tell me where I have made an error :)
    I'm more of a recreational maths person, lol - and I'm trying to make a scheme that 'maps' any decimal number to a natural one.
    The method I have come up with is a bit odd, I'm hoping it works but still trying make sure :)
    It produces a long natural number based on the position and value of the decimal
    eg:
    If you take any decimal, say, 0.3189
    You can make it into a natural number 3108009000

    0.71 becomes 710

    0.2291865 would be 2209001000800006000005000000

    So any decimal, in theory, following that pattern - has a natural number equivalent?

    Tenths become 10s, hundreths become 100s etc, and I just put the numbers in sequence.

    If anyone finds this remotely interesting, I would appreciate any suggestions
    Many thanks :)
     
  2. jcsd
  3. Apr 14, 2017 #2

    fresh_42

    Staff: Mentor

    Only those decimals with a finite number of digits. And yes, these are equally many as the natural numbers.
     
  4. Apr 15, 2017 #3
    thanks for the response :)

    Ahh, so it only works for 'finite' numbers of digits...
    I was hoping it might work for any amount, ie endless numbers lol

    Thanks again for the help :)
    Regards
     
  5. Apr 15, 2017 #4

    PeroK

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    Gold Member

    It can't work for a non-terminating decimal, because every natural number has a finite number of digits. For example:

    ##3.1415926 \dots## is the decimal expansion for ##\pi##.

    But:

    ##31415926 \dots## is not a natural number.

    In your system, how would you handle numbers like ##0.01, 0.10203## etc?
     
  6. Apr 15, 2017 #5

    fresh_42

    Staff: Mentor

    The rational numbers and the natural numbers have the same cardinality, also finite many copies of them. To show why the real numbers are more, there is a famous "diagonal argument": Assume you have written down all real numbers in decimal form and numbered them. If you then change the diagonal digits, e.g. add one and change nine to zero, that is changing the first digit of the first real, the second digit of the second real, the third digit of third real and so on, then you get a new number built from these changed diagonal digits that you haven't numbered. Thus the reals are larger than the natural numbers.
     
  7. Apr 15, 2017 #6
    Thanks for your replies so far :)
    Sorry for responding sooner

    I'll be honest and say I have hard time with the first bit of the argument, "written down all real numbers in decimal form and numbered them".
    I just can't get my head around that, personally - So I guess that's my failing, that's where I'm going wrong :)

    +PeroK - As for 0.01, for example, I flip the digits after the decimal point, so it would be "10" in natural numbers
    0.001 would be 100. And for 0.10203, a bit of a cheat... but I'd have 1,200 and 30000 --> 120030000 - I guess :)

    Don't know about "PI", hehe - I'd have to leave the 3 'whole' numbers for now and just go:
    0.31415926 --> 3 10 400 1000 50000 900000 2000000 60000000 = 310400100050000900000200000060000000 to begin with...

    Thanks again for your answers :)
     
  8. Apr 15, 2017 #7

    WWGD

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    Fernando, like PeroK and Fresh have said, this can only work for finitely-many digits, otherwise, the sum will not converge and so will not be a natural number. The decimal expansion is "guaranteed" to be a Cauchy sequence, thus converging to a Real number, but not so the expansion in the opposite direction.
     
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