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All numbers are equal

  1. Nov 11, 2011 #1
    All numbers are equal :D

    heres the latest proof and it doesnt involve 0 anywhere. no term like (a-b) is involved unlike other similar proofs

    -ab = -ab
    => a^2 - a^2 - ab = b^2 - b^2 - ab
    => a^2 - a(a+b) = b^2 - b(a+b)
    => a^2 - a(a+b) + (a+b)^2/4 = b^2 - b(a+b) + (a+b)^2/4

    as (x-y)^2 = x^2 - 2xy + y^2,

    [a - (a+b)/2]^2 = [b - (a+b)/2]^2

    taking square roots,

    a - (a+b)/2 = b - (a+b)/2
    a = b

    hence proved :D

    now with all numbers having been proved equal without involving "0" anywhere, whats the need to study anything :D

    just noticed, this is my 1st post here after 7 years :eek:
     
  2. jcsd
  3. Nov 11, 2011 #2
    Re: All numbers are equal :D

    You can't take square roots. x^2 = y^2 does not imply x = y
    If it did, this would have been your 7th post after 1 year.
     
  4. Nov 11, 2011 #3

    chroot

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    Re: All numbers are equal :D

    Your second line contains an error. You added a^2 - a^2 to one side of the equation, but b^2 - b^2 to the other side. The only way for this to leave the equality unchanged is for a to equal b... so it's no surprise that you find out later that a = b.

    - Warren
     
  5. Nov 11, 2011 #4
    Re: All numbers are equal :D

    This is the same as
    a^2 + b^2 - ab = b^2 + a^2 - ab , which don't think should create any problems.
     
  6. Nov 11, 2011 #5

    D H

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    Re: All numbers are equal :D

    This is your error. Taking square roots leads to

    a-(a+b)/2=b-(a+b)/2
    or
    a-(a+b)/2=-(b-(a+b)/2)

    The former yields a=b. The latter yields a+b=a+b.
     
  7. Nov 11, 2011 #6
    Re: All numbers are equal :D

    That step is correct. It's just adding zero to each side.
     
  8. Nov 11, 2011 #7
    Re: All numbers are equal :D

    it isnt an error. i added a^2 + b^2 to both sides

    -ab = -ab
    a^2 + b^2 - ab = a^2 + b^2 - ab

    a^2 - a^2 - ab = b^2 - b^2 - ab

    also a^2 - a^2 = 0 & b^2 - b^2 = 0, so adding that doesnt make a difference
     
  9. Nov 11, 2011 #8
    Re: All numbers are equal :D

    yes, you are right. a number always has 2 square roots, 1 positive & 1 negative
     
    Last edited by a moderator: Nov 14, 2011
  10. Jun 20, 2012 #9
    Re: All numbers are equal :D

    You got it wrong.
    2^2 = (-2)^2
    when you have sqr root you have to have either + or - sign. you only considered + sign. Consider - sign and you will get a+b = a+b
     
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