how is this possible:
1^2 - 1^2 =1^2 - 1^2
1x1 - 1x1 = (1+1)(1-1) [:-a^2- b^2= (a-b)(a+b)]
1(1-1) =(1+1)(1-1)
1=[(1+1)(1-1)]/(1-1)
1=(1+1)
1=2
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Answered by
2
the error is in the 4th step... in which u are dividing the RHS by (1-1) which is 0,
You cannot divide it by zero as the value becomes non-determined value..
You cannot divide it by zero as the value becomes non-determined value..
devraj:
but both the (1-1) gets cancelled
you cannot carry on with calculations with a value which is not defined
but (1-1)/(1-1)= 1 if (1-1)is taken as any variable you would cancel it
(1-1)is not a variable... its like this.. u can prove 2 =7.. 1st consider the equation, multiply both sides by 0, u'l get 0=0.. LHS=RHS hence 2=7.. but its wrong.. u cannot multiply or divide with zero..
thanx for explaining
Answered by
1
1^2 - 1^2 =1^2 - 1^2
1x1 - 1x1 = (1+1)(1-1) [:-a^2- b^2= (a-b)(a+b)]
1(1-1) =(1+1)(1-1)
1(0) = 2(0)
0 = 0 //Hence, possible//
First of all, you need to solve the values on one side(say, LHS) and then may compare it with the resultant value of the other side(RHS) for further solving.
And so in this case, you need to subtract 1 from 1 on either sides first, multiply the difference with 1 and 2 respectively and then consider the resultant "0"s for equaling.
Hence, possible !
1x1 - 1x1 = (1+1)(1-1) [:-a^2- b^2= (a-b)(a+b)]
1(1-1) =(1+1)(1-1)
1(0) = 2(0)
0 = 0 //Hence, possible//
First of all, you need to solve the values on one side(say, LHS) and then may compare it with the resultant value of the other side(RHS) for further solving.
And so in this case, you need to subtract 1 from 1 on either sides first, multiply the difference with 1 and 2 respectively and then consider the resultant "0"s for equaling.
Hence, possible !
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