“Prove” 2 = 0. Can You Find The Mistake?




Posted March 23, 2017 By Presh Talwalkar

I received an email I wanted to share with you. Wolfgang came up with a false proof that 2 = 0. No one in his class, not even his teacher, could figure out the mistake. Can you?
I present the false proof and the mistake in a new video.

Here is the “proof” in text.
2 = 1 + 1
2 = 1 + √(1)
2 = 1 + √(-1 * -1)
2 = 1 + √(-1)√(-1)
2 = 1 + i(i)
2 = 1 + i2
2 = 1 + (-1)
2 = 0
Or keep reading for a text explanation.


.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
False Proof 2 = 0
2 = 1 + 1
2 = 1 + √(1)
2 = 1 + √(-1 * -1)
2 = 1 + √(-1)√(-1)
2 = 1 + i(i)
2 = 1 + i2
2 = 1 + (-1)
2 = 0
The mistake is between lines 3 and 4.
√(-1 * -1) ≠ √(-1)√(-1)
This is a misapplication of the product rule for square roots. The product rule is guaranteed to work only when both values are positive.
If xy ≥ 0, then
√(xy) = √(x)√(y)
When x = y = -1, the product rule may not apply, and as demonstrated, it is not a valid step because it leads to the conclusion that 2 = 0.
When you learn a property in math class, make sure to pay attention to the specific conditions when it applies. If you don’t you could end up with an absurd result like 2 = 0!
So how are we supposed to simplify a square root of a negative number? It is actually a mistake to use the product rule (which KhanAcademy teaches):
√(-52) = √(-1)√(52) = i √(52)
This is a mistake! You should not use the product rule unless both terms are positive–although in this case you do get the correct answer.
The correct way is that we define the square root of negative numbers as follows (see page 529 in here):
If b is a real number greater than 0 , then
√(-b) = i √b
So the correct way to find the answer is by definition:
√(-52) = i √(52)
You might think this is a nit-picking distinction as the KhanAcademy method gets to the correct answer. But remember that the process matters in math–it is not about getting the correct answer, it’s about getting the correct answer by the correct method.
Previous
Next Post »