Marilyn vos Savant: -1x-1=+1, instead of -1

  • #1
Marilyn vos Savant, who claims to be in the Guinness Hall of Fame for “World’s Highest IQ.”,
has written the book The World’s Most Famous Math Problem, and here is a direct quote from
the book, page 61:

The square root of +1 is a real number because +1 × +1 = +1; however, the square root of -1 is imaginary because -1 times -1 would also equal +1, instead of -1. This appears to be a contradiction. Yet it is accepted, and imaginary numbers are used routinely. But how can we justify using them to prove a contradiction?

Marilyn's book was reviewed by American Mathematical Monthly 102 (1995) 470-473, and the above quote was given as an example of author's misunderstanding and mangling the notion of proof by contradiction. Wikipedia writes that Marilyn is said to misunderstand imaginary numbers.

It seems to me that she is not taken seriously because she is not a mathematician but rather
a layperson. I see also the same contradiction, why should i^2 = -1 if a negative number times a
negative numbers is equal to a positive number? Are the mathematicians themselves making
a mistake and a contradiction? Or perhaps it is alright to break the rules if you can do
it so cleverly that no-one, especially the laymen, notices it. The best way is to
define new rules that apply to the contradiction and call the new numbers imaginary numbers
which obey their own rules. That way a negative number times a negative number can be made equal to a negative number, so -1 times -1 would be equal to -1 according to the new rules,
because these are just the imaginary numbers obeying the new rules. Maybe it does not matter
that these rules may violate the old rules, who cares?
 

Answers and Replies

  • #2
Office_Shredder
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I is not a negative number, so there's no contradiction
 
  • #3
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The square root of +1 is a real number because +1 × +1 = +1; however, the square root of -1 is imaginary because -1 times -1 would also equal +1, instead of -1. This appears to be a contradiction. Yet it is accepted, and imaginary numbers are used routinely. But how can we justify using them to prove a contradiction?
There is no contradiction.

Say we want to solve the equation X * X + 1 = 0. In the 1800s we would use quaternions, which do work but are cumbersome.
http://en.wikipedia.org/wiki/Quaternion#Square_roots_of_.E2.88.921

'Imaginary numbers' are just another way of solving the equation, not more or less. Solving an equation using 'Imaginary numbers' can also be done using 'quaternions', its just more cumbersome.

You need to think in terms of what problems people were trying to solve in the 1800s, and how they solved them.
 
  • #4
I is not a negative number, so there's no contradiction
If i is not a negative number, but a positive one ( also the same as +i), then -i is a negative number, and

(-i)x(-i) = (+i)x(+i) = i^2 = -1

so we have the result: a negative number times a negative number equals a negative number.
And yes, there appears to be a violation of the old rules.
 
  • #5
chiro
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The imaginary numbers are required to complete algebra.

Gauss did the necessary work to fill in the blanks with his various fundamental theorems of algebra, and basically unified a lot of the work that previous mathematicians like Cardano were doing but at a much more abstract level.

The imaginary numbers link all forms of geometry (negative, zero, positive curvature), periodicity/frequency/wave-stuff and exponential growth, and give us completely new insights to algebra, analysis, and geometry.

Periodicity is very critical: one can look at the complex numbers and look at how periodicity in many contexts is applied. At one level you get normal complex, at another the quaternions, then the octonions and so on. With Grassmanns work you get interior, exterior, and geometric products which all help generalize geometric algebras in terms of length, and angle.

You also generalize a lot of other functions like logs, exponentials, trig functions and link all the hyperbolic functions to the non-hyperbolic ones.

Periodicity though is really the big thing since it is a basic analytic foot-print of nature. Our universe is periodic as is our languages and even our number systems. Our base 10 number system has a periodic nature to it where each number goes from 0 to 9 and then repeats back again in particular sequence. Our planets, atoms, galaxies, and other physical entities also have periodic properties to them and things interact in a periodic manner.

We model everything with periodic models, and eventually we are going to get complex numbers in there somewhere that represent this.

There are many more, but the above are the more obvious ones.
 
  • #6
pbuk
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If i is not a negative number, but a positive one ( also the same as +i), then -i is a negative number,
i is neither negative nor posititve, and the same applies to -i; these terms only have meanings for real numbers similar to the way "odd" and "even" only have meanings for integers, not numbers like 1.5.

(-i)x(-i) = (+i)x(+i) = i^2 = -1
Well it is true that -i2 = -1, but your "proof" is not convincing. You need to write (-i)x(-i) = ((-1)x(i))x((-1)x(i)) = ((-1)x(-1))x((i)x(i))= (1)x(-1) = -1.

there appears to be a violation of the old rules.
The rules that apply to multiplication of complex numbers (including imaginary numbers like i) are consistent with (i.e. do not violate) the rules that apply to multiplication of real numbers. -1 x -1 = 1 is still true.
 
  • #7
Office_Shredder
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Ally, what you have in fact proven is that it is impossible to put an algebraic total ordering on the complex numbers (i.e. pick a set of numbers and declare them to be positive and expect them to satisfy the things we expect positive and negative numbers to satisfy)
 
  • #8
pwsnafu
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If i is not a negative number, but a positive one ( also the same as +i), then -i is a negative number, and

(-i)x(-i) = (+i)x(+i) = i^2 = -1

so we have the result: a negative number times a negative number equals a negative number.
And yes, there appears to be a violation of the old rules.
Zero is neither positive nor negative. Why do you think i is must be positive or negative?
 
  • #9
Well it is true that (-i)2 = -1, but your "proof" is not convincing. You need to write (-i)x(-i) = ((-1)x(i))x((-1)x(i)) = ((-1)x(-1))x((i)x(i))= (1)x(-1) = -1.
There is the result (+i)^2 = (-i)^2 = ((-1)x(-1))x((i)x(i)) = (i)x(i) = (1)x(-1) = -1.
→ i= 1 and i= -1.

It could be that there are the values i=1 and i=-1 besides the usual i=√-1,
although this may sound a strange idea. At least the usual multiplication rules are then
valid, and the original problem of multiplying two negative numbers and getting a negative
result is solved.
 
  • #10
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ay we want to solve the equation X * X + 1 = 0. In the 1800s we would use quaternions, which do work but are cumbersome.

'Imaginary numbers' are just another way of solving the equation, not more or less. Solving an equation using 'Imaginary numbers' can also be done using 'quaternions', its just more cumbersome.

You need to think in terms of what problems people were trying to solve in the 1800s, and how they solved them.
That's backwards. Imaginary numbers and complex numbers predate quaternions by quite a bit. Hamilton specifically invented the quaternions as an extension to the complex numbers.


And yes, there appears to be a violation of the old rules.
Of course. The same happens with negative numbers. It is no longer valid to use √(x2)=x when one allows x to be negative.


Note that the labels negative and imaginary (and also irrational) are rather pejorative. This is no accident. It took mathematicians a while to admit that expressions such as x+1=0 and x2+1=0 were worthy of anything but derision.
 
  • #11
D H
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With regard to "intelligence", I'd much rather be half as "intelligent" as was Richard Feynman with his supposedly mundane IQ of 125 than equally as "intelligent" as is Ms. Vos Savant with her supposedly superhuman IQ of 228. IQ tests do measure something, but the only thing that people agree on is that it they test the ability to take IQ tests.
 
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  • #12
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shows just because an obnoxious person has a high IQ doesn't make them good at math.

i remember she tried to discredit Andrew Wiles Last Theorem Solution using her outragous logic.

Ignoring her is the best solution.
 
  • #13
pwsnafu
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Note that the labels negative and imaginary (and also irrational) are rather pejorative. This is no accident. It took mathematicians a while to admit that expressions such as x+1=0 and x2+1=0 were worthy of anything but derision.
Heck, we see this even today with paradoxical sets.
 
  • #14
D H
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Heck, we see this even today with paradoxical sets.
Paradoxically, the axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?
 
  • #15
I did a little research about the possibility of i being multivalued, because above is
a formula where we can calculate that i=+1 and i=-1 besides the well known i=√-1 .
This means that +1 = -1, and we seem to have made a mathematical fallacy if
we trust Wikipedia which says that:

+1=√1 =[itex]\sqrt{}(-1)\dot{}(-1)[/itex] =√(-1)[itex]\dot{}[/itex]√(-1) = i [itex]\dot{}[/itex] i = -1

according to Wikipedia this is an invalid proof and the fallacy is that the rule [itex]\sqrt{}x\dot{}y[/itex] =√x [itex]\dot{}[/itex] √y is generally valid only if both x and y are positive, which is not the case here.

However, the credibility of Wikipedia can be questioned because there are other sources who claim the opposite, for example Wolfram MathWorld says
that " +1 is always an n'th root of unity, but -1 is such a root only if n is even " , direct quote.
We have found a proof that 1 has a second root of unity equal to -1:
√1 = -1
another second root of unity is +1 but it is obvious that √1 = 1.

To me it seems strange if no-one thought about i being multivalued, it seems to be the only
way to solve the problem how a negative number times a negative number equal to a negative
number.
 
  • #16
arildno
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AllyScientific:
This is a non-issue, based on your, and von Savant's severe misunderstanding of what complex numbers are, and what it means to "multiply" complex numbers, relative to "multiplying" what we call real numbers.

If we look at how Hamilton defined "complex numbers", each element (i.e, complex number) is a pair of real numbers (a,b), where we assign a specific "multiplication rule" for two complex numbers (a,b) and (c,d):

(a,b)*(c,d)=(ac-bd,ad+bd), where ac, bd and so on are standard "multiplication between reals".

Note that in the case a=c=0, b=d=1, we have (0,1)*(0,1)=(-1,0), whereas for b=d=0, we have (a,0)*(c,0)=(ac,0)

The latter case shows how we may regard "multiplication of real numbers" as subsumed into "multiplication of complex numbers" (we may identify the COMPLEX number (a,0) with the REAL number "a"), and the first case shows how i=(0,1) fulfills the identy i^2=-1, where "^" is multiplication between complex numbers, and "-1" is short hand identification of the complex number (-1,0) with the real number -1.
 
  • #17
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I did a little research about the possibility of i being multivalued, because above is
a formula where we can calculate that i=+1 and i=-1 besides the well known i=√-1 .
This means that +1 = -1, and we seem to have made a mathematical fallacy if
we trust Wikipedia which says that:

+1=√1 =[itex]\sqrt{}(-1)\dot{}(-1)[/itex] =√(-1)[itex]\dot{}[/itex]√(-1) = i [itex]\dot{}[/itex] i = -1

according to Wikipedia this is an invalid proof and the fallacy is that the rule [itex]\sqrt{}x\dot{}y[/itex] =√x [itex]\dot{}[/itex] √y is generally valid only if both x and y are positive, which is not the case here.
That is correct.

Any mathematical system that admits even one stinkin' contradiction is an invalid system. Armed with that single contradiction, one can simultaneously prove and disprove every statement made by that system. The system either has to be fixed somehow so that it no longer admits that contradiction or the entire system has to be tossed.

The complex numbers have been around for quite a while now. There are no contradictions with the complex numbers. That proof is invalid.

Many of those "standard rules" for exponentiation aren't so standard at all. They apply to the positive numbers only. For example, consider x=x2/2=(x2)1/2. In short, x=√(x2). This is not true even for the negative numbers, let alone complex numbers. We've helped a vast number of students who have made this basic error somewhere in their work.


However, the credibility of Wikipedia can be questioned because there are other sources who claim the opposite, for example Wolfram MathWorld says
that " +1 is always an n'th root of unity, but -1 is such a root only if n is even " , direct quote.
Huh? How do you see this as being a claim of the opposite?
 
  • #18
arildno
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"Armed with that single contradiction, one can simultaneously prove and disprove every statement made by that system."

Really?

Suppose a system contains axioms A, B, C and not-A

Proofs solely dependent upon axioms B and C should not be affected in their validity due to the inherent system contradiction in that both A and not-A have been accepted as true.
 
  • #20
[insult deleted]

It seems to me that you did not even read what I said. I have presented you a proof
that:
√1 = -1
and the proof is based on the concept of the roots of unity. Do your homework and learn
what are the roots of unity. You will find that a second root of unity means the same as
the square root of 1. And it has two values +1 and -1, on the complex plane these are
lying on the real axis, on opposite sides of the unit circle described by the Euler's formula
e^i[itex]\alpha[/itex] = cos[itex]\alpha[/itex] + i [itex]\dot{}[/itex]sin[itex]\alpha[/itex]

If you continue to trust Wikipedia as a source of scientific information, that is your choice.
I have proved that Wikipedia is not credible source to prove that √1 = -1 is a mathematical
fallacy.
 
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  • #21
That is correct.

Any mathematical system that admits even one stinkin' contradiction is an invalid system. Armed with that single contradiction, one can simultaneously prove and disprove every statement made by that system. The system either has to be fixed somehow so that it no longer admits that contradiction or the entire system has to be tossed.

The complex numbers have been around for quite a while now. There are no contradictions with the complex numbers. That proof is invalid.

Many of those "standard rules" for exponentiation aren't so standard at all. They apply to the positive numbers only. For example, consider x=x2/2=(x2)1/2. In short, x=√(x2). This is not true even for the negative numbers, let alone complex numbers. We've helped a vast number of students who have made this basic error somewhere in their work.



Huh? How do you see this as being a claim of the opposite?
I must ask you the same question as about ZeroPivot, did you read what I wrote?
I have given a proof that Wikipedia is wrong about proving that √1 = -1 is a mathematical
fallacy, while you assume that Wikipedia is correct. Do you know what are the roots of unity?
Especially what is the second root of unity?
Also I provided a direct quote from one example source found on the net (there are many other
similar sources, I just picked one):

Wolfram MathWorld says
that " +1 is always an n'th root of unity, but -1 is such a root only if n is even ".

In the case of n=2, we have the second root of unity, that is square root of 1, and so √1 = -1. Another second root of unity is of course +1 because √1 = +1.
 
  • #22
D H
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ZeroPivot, read your private messages.
 
  • #23
D H
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I have given a proof that Wikipedia is wrong about proving that √1 = -1 is a mathematical fallacy, while you assume that Wikipedia is correct.
Ally, you misread that wikipedia article on imaginary numbers, http://en.wikipedia.org/wiki/Imaginary_number#Multiplication_of_square_roots. I don't know how you misconstrued that as a proof that √1=-1.

This is a standard ploy that attempts to show that 1=-1, which is a contradiction. Note the bold "attempts". There are lots and lots of such so called proofs that 1=-1. You don't even need the complex numbers. All that's needed is division by zero, or using ##x=\sqrt{x^2}##.

That wikipedia article is not using this ploy to show that the complex numbers admit a contradiction. The article instead shows the flaw in that ploy. Perhaps it would have been better for the authors of that wikipedia article to not discuss that ploy at all.

Needless to say, the complex numbers do not admit a contradiction. At least as far as is known, they don't.
 
  • #24
Curious3141
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I must ask you the same question as about ZeroPivot, did you read what I wrote?
I have given a proof that Wikipedia is wrong about proving that √1 = -1 is a mathematical
fallacy, while you assume that Wikipedia is correct. Do you know what are the roots of unity?
Especially what is the second root of unity?
Also I provided a direct quote from one example source found on the net (there are many other
similar sources, I just picked one):

Wolfram MathWorld says
that " +1 is always an n'th root of unity, but -1 is such a root only if n is even ".

In the case of n=2, we have the second root of unity, that is square root of 1, and so √1 = -1. Another second root of unity is of course +1 because √1 = +1.
You don't seem to understand that (by universally understood convention), the square root symbol (##\sqrt{}##) always refers to the principal square root. When the argument is a non-negative real number, this is the unique non-negative square root.

So ##\sqrt{1} = 1## is a perfectly correct statement. It is incorrect to say ##\sqrt{1} = -1## or even ##\sqrt{1} = \pm 1##.

It *is* correct to say that ##x^2 = 1## has two solutions, ##-1## and ##1##. It is correct to call these the two square roots of unity. However, if you want to use the square-root symbol to represent the two square roots of unity, you have to do it as ##\sqrt{1}## and ##-\sqrt{1}##, because, as I said, the square root symbol always refers to the non-negative root when the argument is a non-negative real.

The square root symbol can be applied also to negative real numbers, and even imaginary and complex numbers. However, in this case, the usual "laws" such as distribution of the square root often fail. If you try to do them, you might run into fallacies, as the Wikipedia article shows. The Wiki is correct on this point. When you apply the square root symbol to anything other than a non-negative real, the result is still considered the principal square root, but it can no longer be said to be "non-negative" since this has no meaning when applied to imaginary and complex numbers. Instead a different set of rules applies to decide which is the principal root.
 
  • #25
arildno
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Ally:
It is understandable that the phrase "roots of unity" confuses you.

This phrase is directly derived from the usage of "roots" in the expressions like: "the roots of a polynomial P(x)"

The "roots" here mean the set of x-values so that P(x)=0

The "n-roots of unity" means those numbers that are roots to the polynomial x^n-1
 

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