Thread: Algebra showing 2 = 1. Where's the error?

Can anyone explain how this is possible?



1. Let a and b be equal non-zero quantities

a = b



2. Multiply through by a (^ : is raised to)

a^2 = ab



3. Subtract b^2

a^2 - b^2 = ab - b^2



4. Factor both sides

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



5. Divide out (a - b)

a + b = b



6. Observing that a = b

b + b = b



7. Combine like terms on the left

2b = b



8. Divide by the non-zero b

2 = 1

Originally Posted by chinchu

Can anyone explain how this is possible?



1. Let a and b be equal non-zero quantities

a = b



2. Multiply through by a (^ : is raised to)

a^2 = ab



3. Subtract b^2

a^2 - b^2 = ab - b^2



4. Factor both sides

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



5. Divide out (a - b)

a + b = b



6. Observing that a = b

b + b = b



7. Combine like terms on the left

2b = b



8. Divide by the non-zero b

2 = 1

Very old trick which would amuse idiots only.

look at step 4.. If a=b, then (a-b) = 0, and in step 5, you are dividing by 0. What does your algebra class tell you about divding by 0?

Originally Posted by chinchu

Can anyone explain how this is possible?



1. Let a and b be equal non-zero quantities

a = b



2. Multiply through by a (^ : is raised to)

a^2 = ab



3. Subtract b^2

a^2 - b^2 = ab - b^2



4. Factor both sides

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



5. Divide out (a - b)

a + b = b



6. Observing that a = b

b + b = b



7. Combine like terms on the left

2b = b



8. Divide by the non-zero b

2 = 1

If $\displaystyle a = b$ then $\displaystyle a - b = 0$.

You can't divide by $\displaystyle 0$.

With division by 0 allowed, I can offer a much shorter proof.

1 * 0 = 2 * 0

1 = 2

Q.E.D.

Originally Posted by undefined

With division by 0 allowed, I can offer a much shorter proof.

1 * 0 = 2 * 0

1 = 2

Q.E.D.

Except it's NEVER allowed.

Thanks harish.......

Originally Posted by harish21

Very old trick which would amuse idiots only.

look at step 4.. If a=b, then (a-b) = 0, and in step 5, you are dividing by 0. What does your algebra class tell you about divding by 0?

Originally Posted by chinchu

Thanks harish.......

And what am I? Chopped liver?

Heh this brings back memories of 'you just divided by zero' posters, often featuring someone's head exploding behind a book or a giant whirlpool in the ground :P

This reminds me of a math teacher I had who had dozens of "proofs" that 1 = -1, or 0 = 2, or some such. Whenever he made a mistake working a problem on the board he would try to show why his mistake wasn't really a mistake using another such "proof." Here are two - enjoy!

1. 1 ^2 = -1 ^2
2. Take the square root of both sides: sqrt (1^2) = sqrt (-1^2)
3. since sqrt(a^2)= sqrt(a): 1 = -1
QED

And another:
1. 1 = -1^2
2. log(1) = log (-1^2) = 2 x log(-1), from log(a^n) = n log(a)
3. log(1) = 0, so 0 = 2 x log(-1)
4. Divide through by 2: 0 = log(-1)
5. Since 0 = log(1): log(1) = log(-1)
6. If log(a) = log(b) then a = b, so: 1 = -1. QED

Originally Posted by ebaines

This reminds me of a math teacher I had who had dozens of "proofs" that 1 = -1, or 0 = 2, or some such. Whenever he made a mistake working a problem on the board he would try to show why his mistake wasn't really a mistake using another such "proof." Here are two - enjoy!

1. 1 ^2 = -1 ^2
2. Take the square root of both sides: sqrt (1^2) = sqrt (-1^2)
3. since sqrt(a^2)= sqrt(a): 1 = -1
QED

And another:
1. 1 = -1^2
2. log(1) = log (-1^2) = 2 x log(-1), from log(a^n) = n log(a)
3. log(1) = 0, so 0 = 2 x log(-1)
4. Divide through by 2: 0 = log(-1)
5. Since 0 = log(1): log(1) = log(-1)
6. If log(a) = log(b) then a = b, so: 1 = -1. QED

Might as well use expressions like this then.

$\displaystyle \frac{\log (-(\sqrt{-1})^2-1)}{0}$