# Math Brain Teaser

#### chesart1

Joined Jan 23, 2006
269
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

#### thingmaker3

Joined May 16, 2005
5,083
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.

#### chesart1

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

#### FredM

Joined Dec 27, 2005
124
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..

#### recca02

Joined Apr 2, 2007
1,212
(A+B) (A-B) = B (A-B)
then this,
A + B = B
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?)

#### bloguetronica

Joined Apr 27, 2007
1,541
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
(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.

#### FredM

Joined Dec 27, 2005
124
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.. (?)

#### recca02

Joined Apr 2, 2007
1,212
But if A != B, then (A+B) (A-B) = B (A-B) could not be true.. (?)
yes,
true.

#### chesart1

Joined Jan 23, 2006
269
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?)

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.

#### chesart1

Joined Jan 23, 2006
269
The brain teaser has been solved.

#### Dave

Joined Nov 17, 2003
6,969
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
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