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-i Not a Negative Number

Date: 12/12/2001 at 05:59:17
From: Mark Lewis
Subject: Complex number and -i


My 13 year-old daughter has recently been introduced to complex 
numbers in her algebra class (I might add she's excited at the new 
world it has opened in the math realm; she loves math). Being curious, 
she used her new found knowledge to solve the cornerstone x^2 = -1 
problem. She correctly determined that the solutions for x are i and 
-i. Plugging them back into the equation she obtained the not uncommon 
result of 0 = 0 for i and 2 = 0 for -i. Obviously, she was perplexed 
by the result 2 = 0. 

I pointed out that the problem in her solution was that i^2 and -i^2 
are both -1, and that using this fact would result in the correct 
answer for both values of x. Nevertheless, she pointed out that this 
didn't make sense because x^2 = -i^2 = (-1)*(-1) = 1, ergo, 2 = 0. A 
compelling argument that I cannot adequately dispel. The best I could 
do was to go to an old algebra text that gives the following 
definition for i:

"The number i has the property that i^2 = -1. We write SQRT(-1) = i, 
and agree that the two square roots of -1 are i and -i."

This definition, from my daughter's viewpoint, conflicts with the 
sign rule that states a minus times a minus is a positive as well as 
the convention that -x = -1 * x.

I pointed out that clearly both i and -i must be -1 for the system to 
work. But, as to a better and more concise explaination, I have none. 
Searching the Internet and the Dr. Math forums hasn't helped either.

Could you help with a better mathematical explanation or motivation 
for why -i^2 is (must be) -1 and not 1, i.e., what is the motivation 
for and derivation of the proof?

Thank you.

Date: 12/12/2001 at 08:49:52
From: Doctor Peterson
Subject: Re: Complex number and -i

Hi, Mark.

I recall being excited by complex numbers, too - I'm glad your 
daughter has the same pleasure in this discovery! I hope I can help.

I'm afraid I don't follow everything you've said; what equation does 
she plug i and -i into to get 2 = 0? I don't see it anywhere. But 
there are several things I can clarify.

When you write -i^2, the rules for order of operations say that should 
be taken as -(i^2), not as (-i)^2, because exponents are done first. 
So you should always use parentheses to say what you mean. I don't 
know whether this is part of your problem, taking -i^2 sometimes as (-
i)^2 = -1, and sometimes as -(i^2) = 1; but that is an easy mistake to 
make. I don't understand why you would say -i^2 = (-1)*(-1); where did 
i disappear to?

Here is the proof you asked for:

    (i)^2 = -1
   (-i)^2 = (-1 * i)^2 = (-1)^2 * (i)^2 = -1

That is, both i and -i are square roots of -1. That's all there is to 

The "minus times minus" problem is a confusion about the meaning of 
"negative." A negative number is always REAL; the opposite of a 
positive (real) number is called negative. The negative of an 
arbitrary number can be anything: the opposite of -1 is +1, a positive 
number. So just because you can express a number as -x, you can't call 
it negative. In particular, -i is NOT a negative number. It is an 
imaginary number, and its square is positive. And there's nothing 
wrong with saying that -i = -1 * i. That's still true; negation is the 
same as multiplication by -1.

Please write back and tell me if there's anything I haven't covered. 
Here are a few pages in our archives that may help, which I found by 
searching for "-i":

   Imaginary (Complex) Numbers   

   Imaginary Numbers - History and Commentary   

   Multiplying Radicals of Negative Numbers   

- Doctor Peterson, The Math Forum   

Date: 12/13/2001 at 03:28:02
From: Nice Doctor
Subject: Re: Complex number and -i

Greetings Dr. Peterson,

Thank you for your timely response. My daughter has read your reply 
and now understands the mistake she made. Your demonstration of proof 
was right on target! Thanks again! (Daddy is a bit brighter too...and 
a bit embarrassed at missing an obvious error.)

In the original email I abbreviated my daughter results as I assumed 
them a common mistake. In the event that this thread gets posted in 
the forum, and for the benefit of others, I am including the 
computation sequence that was in error (with the solution via the 
binomial theorem abbreviated as it is obvious).

The problem was to solve

        x^2 = -1

rearranging to quadratic form

        x^2 + 1 = 0

applying the binomial theorem yields

        x = i  and  x = -i

Plugging these results back into the original quadratic -

for the case of x = i

        i^2 + 1 = 0

        -1 + 1 = 0        (by definition i^2 = -1)

        0 = 0

for the case of x = -i

       -(i^2) + 1 = 0

        -(-1) + 1 = 0        (by definition i^2 = -1)

   ( -1)*(-1) + 1 = 0        (by -x = -1 * x)

            1 + 1 = 0

                2 = 0

Obviously 2 = 0 is nonsense. The error occurs in the parenthesization 
at the outset of solving for the case of x = -i which seperates the 
minus sign from i. Instead of

       -(i^2) + 1 = 0        WRONG

should have been

       (-i)^2 + 1 = 0        CORRECT

The above error had the subtle effect of ignoring the fact that

        -x = -1 * x,

or equivalently,

       -ax = -a * x

where a = 1, and thus equivalently

       x^2 = (-ax)^2.

Therefore the -1 term was separated from exponentiation, resulting in 
a nonsense result.

Solving with the correct binding of the minus sign for the case of 
x = -i,

        (-1 * i)^2 + 1 = 0        (by -x = -1 * x)

        ((-1)^2 * (i)^2) + 1 = 0        (by ab^n = a^n * b^n)

        (1 * -1) + 1 = 0

        -1 + 1 = 0

        0 = 0

The correct result for x = -i.

The moral of the story is to be very careful when manipulating the 
minus sign.

Mark Lewis

Date: 12/13/2001 at 08:53:11
From: Doctor Peterson
Subject: Re: Complex number and -i

Hi, Mark.

I had guessed that you were referring to x^2 + 1 = 0, but since that 
wouldn't make sense unless my guess about the error in handling the 
sign was correct, I didn't want to make too many guesses at once!

There is a fairly well-known "false proof" of 1 = 2, depending on 
imaginary numbers, but based on a more subtle error (not just 
misreading of the order of operations, but an actual problem with 
complex numbers). You might like to look at it:

   False Proofs, Classic Fallacies   

Look in the links at the bottom for

   1 = 2: A Proof using Complex Numbers   

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Imaginary/Complex Numbers

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