What's wrong with my proof of i?

Question

Ok.... i^2 = -1, so I rule that i = -1^.5. So, i^2 = -1^.5 * -1^.5 should equal -1, being i squared. But, I can simplify that to i^2 = (-1*-1)^.5, and then to i^2 = 1^.5, which simplifies to i^2 = 1, and i = 1.

Ok.... i^2 = -1, so I rule that i = -1^.5. So, i^2 = -1^.5 * -1^.5 should equal -1, being i squared. But, I can simplify that to i^2 = (-1*-1)^.5, and then to i^2 = 1^.5, which simplifies to i^2 = 1, and i = 1.

f_m - but what about the property that states

a^y * b^y = ab^y

? Or did I remember that wrong?

Thanks, bandf. That's pretty much the best answer out of all the people I've asked this, even from mathforum.org =)

Answer 1

Your mistake is when try to go from:
i² = -1^.5 * -1^.5
to
i² = (-1 * -1)^.5

When you have two numbers with the same base, you add the exponents, so this would be changed to
i² = -1^(0.5 + 0.5) = -1^1 = -1

So you are back to i² = -1.

The rule you quote a^y * b^y = (ab)^y is *only* true if both a and b are greater than or equal to zero. -1 is definitely *not* greater than or equal to zero.

A second mistake is when you go from i² = 1^.5. The square root of 1 isn't just 1, it can also be -1, which would return you to i² = -1

Wikipedia has copy of this same invalid "proof" (shown a little more clearly) with an explanation of why it is invalid. See the link below.

Answer 2

Simple. Some rules for real numbers fail for complex numbers. Here's another.

Every real number is either zero, positive, or negative by the trichotomy. But i fails: We know i is not zero, but suppose i is positive. So i > 0. Multiply both sides by i: i^2 > 0i, which simplifies to -1 > 0, which is false. So i is negative, or i < 0. Multiplying by i, however, gives us i^2 > 0i because when we multiply by a negative, the inequality sign changes direction. Thus, -1 > 0, which is wrong. Thus, i is neither positive, negative, nor zero.

Similarly, that rule you applied going from sqrt(-1) * sqrt(-1) = sqrt(1) doesn't work for complex numbers in general.

Answer 3

1. Your proof is very messy and hard to read. That, in itself, is an error.

2. You cannot take the square root of a number and assume the answer will be positive. Once you do that, the proof is no longer valid.

Answer 4

Your error is in the step i^2 = (-1*-1)^.5

In general, the following holds
x^a * x^b = x^(a+b)

So, (-1^.5) * (-1^.5) = -1(.5 + .5) = -1^1 = -1

What you were stating, incorrectly is that

x^a * x^a = 2*x^a

Answer 5

how do you go from -1 to (-1*-1)^.5? that doesnt make sense