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Can anyone spot the error in this fallacy

  1. Dec 2, 2009 #1
    <prepared>a + b = t
    <prepared>(a + b)(a - b) = t(a - b)
    <prepared>a^2 - b^2 = ta - tb
    <prepared>a^2 - ta = b^2 - tb
    <prepared>a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
    <prepared>(a - t/2)^2 = (b - t/2)^2
    <prepared>a - t/2 = b - t/2
    <prepared>a = b

    Therefore all numbers are the same!
  2. jcsd
  3. Dec 2, 2009 #2
    When you take the square root from the 6th to 7th line.
  4. Dec 2, 2009 #3
    Could you explain why that is erroneous?
  5. Dec 2, 2009 #4
    the formulas seem legit, but if you put numbers in each, as i have done a =1 b =2 t =3

    the 6th line shows inconsistency. BOTH 5th lines would appear to be -1.4375=-1.4375 and the next line you would expect the same, but one appears to be 0.5625 = 1.5625

    i still dont know why is that. but i would like to know
  6. Dec 2, 2009 #5
    I'm perfectly happy with this. All numbers are equal.
  7. Dec 2, 2009 #6
    You need to take +/- when you take a square root unless it's an absolute value.
  8. Dec 2, 2009 #7
    lets just argue for sakes here that it is absolute value because it really doesn't matter, if you put the -/+ in front of the numbers no way will it still equal each other. the magnitude is still different.

    ok might be a bit confusing, assume absolute for our case. the magnitudes should equal in theory but by placing numbers its not. but there is nothing wrong with the square root.
  9. Dec 2, 2009 #8


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    Why in the world would think that?

    This does not say anything about a and b beyond what you have done. To satisfy the process you have defined you must have a = b . It does not say anything in general about anything.
  10. Dec 2, 2009 #9
    EDIT: I think you misread what I wrote. The solutions to a square root must be +/- or you just give ABSOLUTE VALUE.

    I think my wording may have confused you. What I mean is that if you give an absolute value of 3 it can be +/-3. So I was saying that the solutions have to be +/- unless you give your answer as an absolute (which we are not doing in this with numbers.)

    All that this proves as of now with letters is that a=b, if you assume that all letters must be different numbers then YES the square root is the error.
    Last edited: Dec 3, 2009
  11. Dec 3, 2009 #10
    I don't think you are doing your math correctly anyways (from what I read up above) You do remember BEDMAS correct?


    We end up with this correct? Now what we have is (-0.5)^2=(0.5)^2 which is true.

    But if we take the square root in the previous step we end up with:

    [tex]1-\frac{3}{2}[/tex] which = -0.5 or we have [tex]-1+\frac{3}{2}[/tex] which gives us 0.5.

    Same for the other side. If you end up with -0.5=0.5 Then you are wrong in your taking of the square root and this is not a valid solution.
  12. Dec 3, 2009 #11
    oh okay, i see your point, but this still goes back to my first post on this, the calculation i done from 5th to 6th line goes haywire, where squareroot hasnt been applied yet.
  13. Dec 3, 2009 #12
    You're not doing your math correctly.
  14. Dec 3, 2009 #13
    a = 1 b = 2 t = 3

    5th line
    1^2 -3(1) + 3^2/4 = 2^2 - 3(2) + 3^2/4
    0.25 = 0.25

    6th line
    (1-3/4)^2 = (2-3/4)^2
    0.5625 = 1.5625

    how is that wrong math?
  15. Dec 3, 2009 #14
    5th line:
    [tex]a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4[/tex]



    Both sides equal -0.5

    6th line:
    [tex](a - t/2)^2 = (b - t/2)^2[/tex]



    Both sides equal 0.25... This is way off what you got. If you post your steps maybe I can help you?

    EDIT: I noticed you have it as 3/4... why? As well I'm quite certain (1-3/4)^2 does not equal 0.5625
    Last edited: Dec 3, 2009
  16. Dec 3, 2009 #15
    OH, i left my t at t/4 istead of t/2. my bad.
  17. Dec 3, 2009 #16
    Well even so your answer for line 5 is still wrong and so is the one half of line 6... I'll help you out with that if you wanted.
  18. Dec 3, 2009 #17
    haha cheers, but how i showed you is exactly how i calculated. should have brought a calculator. but i think i seem to know why it's wrong. thanks anyway!
  19. Dec 3, 2009 #18
    Heh sure, I'm not trying to make you look 'stupid' or anything just if you need help with how to calculate it I'd be glad to show you. (I'm not sure how old you are but I'm assuming still in highschool? No offense.)

    Welcome to the forums by the way :smile: Did you get hit with a fish yet?
  20. Dec 7, 2009 #19
    Square root issues aside I do not see where you are saying anything other than a and b are the same number and t is any number which is the sum of those numbers.
  21. Dec 7, 2009 #20
    Without giving too much away,

    a+b = t.

    For two terms

    a - t/2 and b - t/2,

    either both terms are equal and a=b=t/2,

    or one of 'a or 'b is less than t/2.
  22. Dec 8, 2009 #21
    Basically, if what is being stated were true, then for any numbers that satisfy a+b=t, a=b. So, we can define a and b arbitrarily, and set the value of t to a+b, and a should equal b, which is clearly wrong. For instance, I could pick a = 24 and b = 68, and then set t = 92. And by this "proof", 24 = 68. Of course, the proof isn't valid thanks to the square root issues, so obviously it isn't true. But that's the claim, which is akin to other logical fallacies which prove 1 = 2 or 1 = 0 or other impossibilities.

  23. Dec 9, 2009 #22
    Here is the corrected version:
    <prepared>a + b = t
    <prepared>(a + b)(a - b) = t(a - b)
    <prepared>a^2 - b^2 = ta - tb
    <prepared>a^2 - ta = b^2 - tb
    <prepared>a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
    <prepared>(a - t/2)^2 = (b - t/2)^2
    <prepared>a - t/2 = [tex]\pm[/tex](b - t/2)
    <prepared>a = b or a = t - b

    Therefore all numbers are the same or they are different!
  24. Dec 17, 2009 #23
    You cannot take the square root of real negative numbers.
  25. Dec 17, 2009 #24
    Well first this isn't true and second it's not why the square root is the error in this.

    The reason the square root is the error has already been explained:

    When you take the square root your answer must be +/-. The OP does not do this and so the wrong square root value ended up being chosen to continue to the final answer.

    When we take the square root of:

    (a - t/2)^2
    Our answer will be
    a-t/2[/tex] OR [tex] -a+t/2[/tex]
  26. Dec 17, 2009 #25
    I shoudl have completed the thought: "You cannot take the square root of real negative numbers and wind up with a single real number."

    As you explained, you wind up with an either/or, and there the equality fails.
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