Math Brain Teaser

Discussion in 'Off-Topic' started by chesart1, Nov 10, 2007.

  1. chesart1

    Thread Starter Senior Member

    Jan 23, 2006
    269
    1
    What is wrong with the following algebraic manipulation?

    A = B

    AA = AB

    AA - BB = AB -BB

    (A+B) (A-B) = B (A-B)

    A + B = B
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Step 2 can be omitted. Simply go from step 1 to step 3 by multiplying both sides by A.

    Step 5 will draw the attention of the quadratic equation police.
     
  3. chesart1

    Thread Starter Senior Member

    Jan 23, 2006
    269
    1
    I understand. Step five usually has that effect. You are correct about step 2. I will edit it. Thanks.
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    a-b =0
    hence division by it is not allowed since the result becomes indeterminate.
    eg: how many times 0's are = 1?

    follow this link.:D
    http://forum.allaboutcircuits.com/showthread.php?t=6034
     
  5. FredM

    Senior Member

    Dec 27, 2005
    124
    1
    surely (A+B)(A-B) = (A+B)*0 = 0
    and B(A-B) = B*0= 0
    therefore (A+B)(A-B)=B(A-B) is true.. (?)
    .. Where does division come into this?
    As for A+B=B I do not understand this at all.. Surely this would be 2A or 2B ?
    Maths is my weakness, so I presume I am missing something basic..
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    then this,
    what happened here?
    the (A-B) got canceled .
    what math operation allows this cancellation?
    the answer is division.
    but for division there is a condition that divisor must not be zero (in maths at 'later' levels it becomes customary to mention that divisor is not zero b4 dividing)
    meaning cancellation is not allowed.

    edit:
    else: 5*0= 378975892749848929472*0
    dividing both sides by zero (note this ain't allowed)
    5= 378975892749848929472.

    or for that matter
    any number will become equal to any other number. eg;1= infinity,etc. (getting my point?)
     
  7. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    (A + B) (A - B) = 2B (A - B), but since in this case A = B, and therefore A - B = 0, we can say that (A + B) (A - B) = B (A - B). However, since A - B = 0, both sides will be 0, independently of the remaining multipliers. Therefore you have an equation that simply states that 0 = 0.

    You can't also assume that A + B = B by dividing both sides by A - B, which is 0. It is an absurd.
     
  8. FredM

    Senior Member

    Dec 27, 2005
    124
    1
    Hmmm.. ok, I think I see it now.. It looks 'silly' when one is given A = B in the first place.. but if one was only given (A+B) (A-B) = B (A-B), without knowing that A=B, and tried to reduce this by division, one would end up with an absurd.. (?)..

    But if A != B, then (A+B) (A-B) = B (A-B) could not be true.. (?)
     
  9. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    yes,
    true.:)
     
  10. chesart1

    Thread Starter Senior Member

    Jan 23, 2006
    269
    1

    Of course, you are correct. The algebra class, I was in, was given this little algebraic manipulation in my first year of high school [1962]. No one was able to figure it out. It was a creative way to remind the class of the rule: Never divide by zero.
     
  11. chesart1

    Thread Starter Senior Member

    Jan 23, 2006
    269
    1
    The brain teaser has been solved.
     
  12. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    Further to my thread which recca linked to, the problem lies in the incorrect final two lines:

    For:

    (A+B) (A-B) = B (A-B)

    Where:

    A = B

    Then:

    (A+B) (A-B) = B (A-B) equates to (A+B)*0 = B*0

    Which gives:

    0 = 0 and not the final line of A + B = B

    (See http://forum.allaboutcircuits.com/showpost.php?p=34509&postcount=3) It does confuse you at first.

    Dave
     
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