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

• Jun 3rd 2010, 09:02 PM
chinchu
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
• Jun 3rd 2010, 09:08 PM
harish21
Quote:

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?
• Jun 3rd 2010, 09:10 PM
Prove It
Quote:

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 $a = b$ then $a - b = 0$.

You can't divide by $0$.
• Jun 3rd 2010, 09:16 PM
undefined
With division by 0 allowed, I can offer a much shorter proof.

1 * 0 = 2 * 0

1 = 2

Q.E.D. :D
• Jun 3rd 2010, 09:37 PM
Prove It
Quote:

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. :D

Except it's NEVER allowed. (Hi)
• Jun 4th 2010, 12:03 AM
chinchu
Thanks harish....
Thanks harish.......

Quote:

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?

• Jun 4th 2010, 12:40 AM
Prove It
Quote:

Originally Posted by chinchu
Thanks harish.......

(Wait) And what am I? Chopped liver?
• Jun 4th 2010, 12:03 PM
Mukilab
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
• Jun 4th 2010, 12:37 PM
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
• Jun 4th 2010, 12:56 PM
undefined
Quote:

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.

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