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### -i Not a Negative Number

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

Greetings,

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.
```

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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
it.

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
http://mathforum.org/dr.math/problems/kyle2.26.98.html

Imaginary Numbers - History and Commentary
http://mathforum.org/dr.math/problems/engel9.4.97.html

http://mathforum.org/dr.math/problems/amy.7.12.00.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

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

Greetings Dr. Peterson,

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

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
http://mathforum.org/dr.math/faq/faq.false.proof.html

Look in the links at the bottom for

1 = 2: A Proof using Complex Numbers
http://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Imaginary/Complex Numbers

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