if a=b, then how is it possible?

Question

a=b
a2=a.b
a2 -b2=a.b- b2
(a+b)(a-b)=b(a-b)
a+b=b
2b=b
2=1

Is it correct? Where is the wrong?

Answer 1

Your fifth statement is wrong;
(a+b)(a-b)=b(a-b)
a+b=b

Because a=b, a-b=0. As you know 0 can not be on the denominator. When you attempted to cancel out a-b from both sides you are dividing by 0.

(a+b)(a-b)/(a-b)=b(a-b)/(a-b) is the same thing as (a+b)(a-b)/0=b(a-b)/0, which does not work.

Instead of a+b=b, your answer would have been 0=0

Answer 2

For A,B, and C the place they might equivalent any rational extensive variety which contain factions/decimals, the place A=(19-3B-5C)/2 B=(19-2A-5C)/3 C=(19-2A-3B)/5 for any fee of two of the three variables, the third will equivalent some absolute fee. If on the different hand the values of A,B, and C ought to be finished numbers, then a accessible mixture is A=a million, B=-a million, C=4, subsequently offering you with 2(a million) + 3(-a million) +5(4) =19 Or if the values of A,B, and C ought to be non-unfavourable A=8, B=a million, and C=0; 2(8) + 3(a million) + 5(0) = 19 Or if the values ought to be non-unfavourable and non-0 A=4, B=2, C=a million Your question is stable, besides the undeniable fact that, I solved this equation 4 techniques in below 5 minutes. finally it replaced into wager and consider, besides the undeniable fact that it wasnt assigning values at present day to A,B,and C, it replaced into extra of a job of subtracting the fee of two,3, and 5 from 19 until eventually you attain 0. case in point, 19-5-3-2-2-2-2-3=0

Answer 3

no, it's not right. 'cos
a - b = 0
then
( a + b ) ( a - b ) = b ( a - b )
( a + b ) ( 0 ) = b ( 0 )
simply
0 = 0 as a result
but we can think so
( a + b ) = b ( 0 ) / ( 0 )
this logic is wrong because the term " ( 0 ) / ( 0 ) " is undefined in mathematical rules.
ok.
the argue is given at question includes error as an anomaly. so the argue is complemantarily uncorrect.

Answer 4

You can't cancel (a-b) from both sides. This is because (a - b) = 0 when a = b and division by 0 is undefined.

Answer 5

a=b
a2=a.b
""""a2 -b2=a.b- b2""""

This is wrong because if you take b^2 on th eleft side then the RHS would be 0(Zero).

Answer 6

"(a+b)(a-b)=b(a-b)
a+b=b"

If a=b, then a-b=0. So basically, your argument amounts to 2*0=1*0, therefore 2=1. No, I'm afraid that doesn't work.

Answer 7

Actually, this is a proof to show that division by zero cannot take place. Mathematics can be used to prove observations but observations can't be used to prove mathematics

Answer 8

You divided by zero when you took away the (a-b) terms on line 4. This is not allowed.

Answer 9

This bit:

(a+b)(a-b)=b(a-b)
a+b=b

you divided through by a-b. But if a=b, then a-b=0. Dividing by zero is something you cannot do because the answer is undefined.

Answer 10

if a=b, then
a-b=0,
and you can't divide by zero

i actually saw this problem on a website, it was and is cool