I understand. Step five usually has that effect. You are correct about step 2. I will edit it. Thanks.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.
a-b =0What is wrong with the following algebraic manipulation?
(A+B) (A-B) = B (A-B)
A + B = B
then this,(A+B) (A-B) = B (A-B)
what happened here?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.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
yes,But if A != B, then (A+B) (A-B) = B (A-B) could not be true.. (?)
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?)
Further to my thread which recca linked to, the problem lies in the incorrect final two lines: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
by Jake Hertz
by Jeff Child
by Jake Hertz