# 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?

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?
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
11
(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.. (?)

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
147
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