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Even vs. Odd

  1. Jul 21, 2003 #1
    Here is a question I suddnely thought of.

    The natural number series: 1,2,3...

    If we double the series, we get: 2,4,6...

    In other words get all the even numbers.

    My question is there any ways to determine all odd number series?
     
  2. jcsd
  3. Jul 21, 2003 #2
    If you add one more step and substract 1 after multpliying by 2 to the series you obtain the odd number series.
     
  4. Jul 21, 2003 #3

    selfAdjoint

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    Or add 1 and treat 1 as a special case. So every even number is of the form 2n and every odd number is of the form 2n+1. Where n can be read as "something".

    Now look what you can do. (2n)^2 = 4n^2 = 2(2n^2) so the square of an even number is of the form 2 times something, and so it is even too.

    And (2n+1)^2 = 4n^2 + 4n +1
    = 2(2n^2+2n) + 1.
    So the square of an odd number is one more than an even number, so it is odd.

    You have just proved a theorem: The square of an even number is even and the square of an odd number is odd.
     
  5. Jul 21, 2003 #4
    Take this another step.

    Pretend that you are IBM Corporation{1} in the mid-1950s. Start your sequence with 0 instead of 1.

    0 1 2 3 4 5 .....

    Now, square the members of this sequence.

    0 1 4 9 16 25 .....

    This is the derived sequence of perfect squares.

    Now, subtract adjacent members of this sequence.

    Voilá!

    1 3 5 7 9 .....

    ---

    {1}IBM computer systems numbered peripherals devices and their plug ports beginning with 0 instead of with 1 (which is what other computer manufacturers started with at that time).
     
    Last edited: Jul 21, 2003
  6. Jul 22, 2003 #5
    One more bit.

    Start with

    0 1 2 3 4 5 .....

    This time, instead of squaring each member, multiply adjacent neighbors.

    0 2 6 12 20 30 .....

    Now, subtract adjacent terms of this.

    2 4 6 8 10 .....

    whoopie!
     
  7. Jul 22, 2003 #6

    Integral

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    n(n+1) - n(n-1)=n2+n - n2+n =2n
     
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