jamteddy83 | Student

2=1, cannot be proven. 

Both "proofs" above, divides by a 0.   A number can not be divided by 0 because in essence it's like trying to take a group and divide into a group of nothing.  A group already exist, to make it vanish will be subtraction not division.

a-a, from "wakeuprj" means a number minus itself so that is 0.

a-b, from "samantha96" is 0 because the assumption was made that a=b meaning a and b are the same number.  Using additive inverse, subtract b from both sides a-b=0.

2=1 is obviously wrong and inlogical, mathematicians have no problem to admit something is not true.

wakeuprj | Student
(a^2-a^2)=(a^2-a^2) we can say this because both LHS and RHS are the same now, a^2-a^2 can be written as (a+a)(a-a) therefore, (a+a)(a-a)=(a^2-a^2) now, as (a^2-a^2) can be written as: a(a-a) so (a+a)(a-a)=a(a-a) now divide both sides by (a-a) we get a+a=a 2a=a now. divide both sides by a we get 2=1 HENCE PROVED THAT 2=1...
samantha96 | Student

a = b

multiply both sides by a

a^2 = a*b

subtract b^2 from both sides

a^2-b^2 = a*b-b^2

apply the distributive law to both sides

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

divide both sides by (a-b)

(a+b) = b

substitute all a's for b's (remember, if a = b you can do this)/

a+a = a

regroup the two a's in the left side, and rename it 2a

2a = a

divide both sides by a

2 = 1

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