MATHS: How is THIS Possible???

Question

see the following:

let a=b..
...multiplying 'a' on both sides....

a^2=ab. (^ means exponent or power...we all know)
...subtracting 'b^2' from both sides...
a^2 - b^2= ab - b^2

since a^2 - b^2=(a+b)(a-b)
then,
(a+b)(a-b)=ab -b^2
(a+b)(a-b)=b(a-b) (taking b common on right side)
....canceling (a-b) on both sides....
we get
a+b=b...

now how is this POSSIBLE??? earlier i showed that a=b, then WHERE is the mistake in between???

don't tell me that a=b=0 solves it, coz we can specify that earlier also that
a=b= any number like 1, then how come
a+b=b...ie.
1+1=1???

Answer 1

Unhandled Exception.
0x00FEEDBABE: Divide By Zero Error, press any key to quit.

Answer 2

Suppose y = x/x;
Now

y = 1 if x is not equal to zero
= undefined if x = 0;

So what you are doing is you are saying x/x = 1 when x = 0 which is not true. As I wrote x/x = undefined when x = 0. So after (a+b)(a-b)=b(a-b) your next step is wrong.

If you have studied Calculas you must know that left hand limit of x/x near zero is not equal to right hand limit of it near zero.

I hope it clears you doubt. There are many more examples of this kind. one is
a^2 = b^2
=> a = b
which is not true because if
a^2 = b^2
=> a^2 - b^2 = 0
=> (a-b)(a+b) = 0
=> (a-b) = 0 OR (a+b) = 0
=> a = b OR a = -b
=> a = +/-b

All the best.

Answer 3

Duh!!! You don't --EVER-- 'cancel' anything. What you do is --DIVIDE-- both sides by (a-b) so that you get a factor of 1 on both sides.

But, in this case, since you started off with the assumption that a = b then you divided 0 by 0 which is about as undefined as you can get.

BTW.... This little piece of legerdemain (along with several hundred common variants, all involving division by 0 in one form or another) is at least 2500 years old ☺

Doug

Answer 4

As soon as you subtracted 'b^2' from both sides, you made both sides of the equation equal to 0!

Answer 5

looking on your college, you will be able to correctly be waiting to locate an 'astronomy for non-scientists' direction, yet once you pick to examine actual astronomy heavily, then you somewhat choose all the math you will come across. the two astronomy and physics at degree point and above are in actuality branches of arithmetic, and the mathematics may well be very checking out, at that. you're unlikely to be waiting to slide actually into around trigonometry and the Schroedinger wave equation, to point merely 2 factors, on the inspiration of an O-point, or in spite of the present call is. in step with probability you would be extra useful sticking to astronomy as a interest.

Answer 6

It is possible only when a=b=0
Also, if a = b then a^2 - b^2 = 0

Answer 7

There is no mistake. This is true for a=b.
In fact base on your last statement:
a+ b = b
a = 0,

Now if a = 0, b = 0 which is pausible.

Answer 8

The problem is that when you are "....canceling (a-b) on both sides...." you are dividing by zero (since you specify that a=b so a-b=0). Obviously in maths, you can't divide by zero - if you do you get inconsistent results as you have seen.

cheers

Answer 9

it has to do with the fact that as you manipulate a and b, yu multiply the terms by (a-b), which, since a=b, is zero. Thus you can pretty much say anything you want since anything multiplied by 0 is 0

Answer 10

You are right, It's not possible.
So in mathematics this type of equation is said as "not difined."

Because in step :- (a+b)(a-b)=b(a-b) If u want to cancle then either u have to divide equation by (a-b) or shift (a-b) to L.H.S

Then it becomes (a+b)(a-b)/(a-b)=b(a-b)/(a-b)
Here (a-b)=0
there fore equation becomes 0/0=0/0. In maths 0/0 forms are said to be not defined forms, because 0/0(zero divided by zero) is neither equal to 1 nor equal to infinity.

Hence maths tell this equation as not defined form.