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        riddles >> hard >> 1=0?  (Math skills needed)
        
(Message started by: godskook on Mar 7th, 2004, 5:46pm)
Title: 1=0?  (Math skills needed)
Post by godskook on Mar 7th, 2004, 5:46pm
I stumbled on this wierd equation just today so I don't know how accurate my algebra is.

(1/-1)1/2=1/i=(1/i)(i/i)=i/i2=-i

or

(1/-1)1/2=(-1)1/2=i

which means that i=-i  or 1=-1 or 1=0  ?

What is wrong with this?

<for those who don't know (-1)1/2=i>
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 8th, 2004, 2:30am
1 = [sqrt]1 = [sqrt](-1*-1) = [sqrt]-1*[sqrt]-1 = i*i = -1

of course [sqrt](x2) = [pm]x and not just x, and so [sqrt]1 = [pm]1
So there's not really a problem.
If you limit [sqrt]x to the principle root, I don't think you can count on [sqrt](a*b) = [sqrt]a * [sqrt]b any more..
Title: Re: 1=0?  (Math skills needed)
Post by Sir Col on Mar 8th, 2004, 6:24am
I don't think that's quite right, towr. Although it is true to say that given x2=1, x=[pm]1, but [sqrt]1=1; that is, [sqrt] is just a function.

I think the problem with the "proof", as you suggested, is that [sqrt](a/b)=[sqrt]a/[sqrt]b only holds for real numbers.
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 8th, 2004, 7:05am
I usually try to check my claims in another source, and I did so here as well..

http://mathworld.wolfram.com/SquareRoot.html
"Note that any positive real number has two square roots, one positive and one negative"
Title: Re: 1=0?  (Math skills needed)
Post by Earendil on Mar 8th, 2004, 7:43am
I think both towr and Sir Col have a piece of truth.
What happens is that the function square root is different in the Real domain and in the Complex domain.
The "Real" sqrt of k returns only the positive value of the equation x² = k and the "Complex" sqrt of k returns all values that satisfy the equation x² = k.

Example
(real)[sqrt]1 = 1
(complex)[sqrt]1 -> x² = 1 -> x = +- 1

So what you have found are both answers (x1,x2) to x² = -1 and that does not imply that x1 = x2. If you suppose that they were equal then you would also have to suppose that both the solutions you obtain by Basquara are also always equal, and that sure isn't true :P
Title: Re: 1=0?  (Math skills needed)
Post by Benoit_Mandelbrot on Mar 8th, 2004, 9:23am
I believe what's happening is that you're assigning i to a different value.  You're saying that i=1/i, which is untrue.  Therefore, you can't do what you dit.
Title: Re: 1=0?  (Math skills needed)
Post by Sir Col on Mar 8th, 2004, 9:29am
I know it is unlike you to ever post anything which you haven't check first, towr, which was why I posted it with maximum respect and a rather tentative, "I don't think that's quite right...". It was just that I recall being drilled with identifying and removing this "misconception" when I was training to be a teacher.

I must say, though, that the jury is still out for me. However,  I'm inclined to go with Earendil's explanation about the difference between a principle value in the real domain versus multiple values in the complex domain; confer nth root of unity.

I stumbled two interesting articles that address the question:

The Art and Science of Mathematical Sin (http://www.decisionsciences.org/DecisionLine/VOL28/28_3/ed.htm) (by H. Arsham, University of Baltimore; the relevant part is in the penultimate paragraph)
Square Roots in Complex Numbers (http://mathforum.org/library/drmath/view/52243.html) (Ask Dr. Math @ Math Forum)
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 8th, 2004, 10:03am
If [sqrt]x2 could only be x, then [sqrt] can't be the inverse of 2. But if it is the inverse then it can't be a function. And most people don't want it to not be a function.
But of course since [bbr] [subset] [bbc], if it is multivalued in [bbc], then it can be so for [bbr].
Title: Re: 1=0?  (Math skills needed)
Post by Earendil on Mar 8th, 2004, 12:40pm
Indeed it can be multiple-valued in R and it will still be called (complex)sqrt (This is only a matter of convention). While the (real)sqrt is single valued.
Anyway if any function is of the type f(x) = (t1,t2) you certainly can not imply that t1 = t2. I guess we all agree about that
Title: Re: 1=0?  (Math skills needed)
Post by godskook on Mar 8th, 2004, 7:32pm
As far as the answer to this problem goes, I talked with my math teacher.

The flaw in my algebra is that exponentials don't all distribute when negatives are involved.

i.e.   (ab)1/2=!a1/2b1/2
unless "a" and "b" are both positive.

Concerning the sub issue of how many answers there are for the square root function, there is only one.

i.e.   11/2=1   all the time.

What most of you are thinking about is that when one applies the square root function to an existing equation one picks up a +- sign which only stays until the all variables are known.

i.e.   x2=1
      (x2)1/2= +-((1)1/2)
      x= +-1

But  if x=3 and x=1+-(2)1/2
     then we choose the sign that makes the equation work.

p.s.  Earendil is your login name from the Silmarillion. If so are you related to Elros? LOL
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 9th, 2004, 5:22am

on 03/08/04 at 19:32:53, godskook wrote:
As far as the answer to this problem goes, I talked with my math teacher.
As a general rule, teachers don't know everything.


Quote:
Concerning the sub issue of how many answers there are for the square root function, there is only one.
Yes, if it is a function.
But it can't be an inverse relation to the square in that case..


Quote:
What most of you are thinking about is that when one applies the square root function to an existing equation one picks up a +- sign which only stays until the all variables are known.
No, that isn't what we are thinking of.. We don't apply the square root function, but we apply the relation, because we want the inverse to the square function.
Applying a function will always only give one answer, and never pick up a [pm]
Title: Re: 1=0?  (Math skills needed)
Post by Earendil on Mar 9th, 2004, 8:18am
Hum... what godskook said in the last post made me think that possibly some concepts are being wrongly understood. When you solve the equation x² = 4 usually people go directly to the obvious approach:
x² = 4
x = +-4

But there is a little thing missing in between the steps...

x² = 4
sqrt(x²) = sqrt(4)
|x| = 2
x = +-2
___

Back to the main theme, what godskook said was correct indeed... In the Real Domain you can't use the property

sqrt(b) = sqrt(-b) * sqrt(-1)

But that has nothing to do with the topic as the property used wasn't that one but

sqrt(b) = sqrtb(a*b/a)

...
The problem is that sqrt() isn't a funcion in the Complex Domain so it delivers 2 values always =þ
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 9th, 2004, 7:27pm
towr is right.  Just my 2 cents:

1) There is no such thing as a positive complex number!  The complex numbers have no ordering[in the sense you're used to], so to say that a proposition only applies to "positive complex numbers" is nonsense.
 (If you don't believe me - Is i "positive"?  Suppose it is.  Then you're going want -i to be "negative".  You'll also want two "negatives" to multiply to be a "positive", and two "positives" to multiply to be a "positive", etc..  But then,
i*i = -1, so -1 is "positive"?
(i)*(-i) = -1, so -1 is "negative"?
and so on...
)

2) Just think of the square root of a complex number as a _set_ and you shouldn't have these problems.  Thus, the initial equations simply state that both i and -i are square roots of -1.  What's wrong with that?  They both square to -1.  So they're both roots.  Here are some more true statements,

(1)1/4 = i, -i, 1, and -1.  That is, 1 has four fourth roots.  This also illustrates why you shouldn't use the symbol  [pm] when in the complex numbers (in the real number we would have said (1)1/4 = [pm]1, and we would have missed two values if we tried to infer the same thing holds in the complex numbers)
Title: Re: 1=0?  (Math skills needed)
Post by Icarus on Mar 9th, 2004, 8:41pm
Okay, I have now written and rejected two long posts trying to make my point without rambling. I can't seem to accomplish it, so I am just going to lay out what I want to address.

1) x2 = 4 [bigto] x = [pm]2 really is all there is to it. The rest is only elaboration.

2) There are two accepted usages for "square root", even for positive reals: It is acceptable to use "square root of A" as both an arbitrary reference to either solution of x2 = A and to use it to mean specifically the positive solution only. In the former, it is used, as towr has said, as a general relation. In the latter, it is used as that specific form of relation known as a "function".

3) When we move beyond the non-negative reals, we lose any conventional "functional" definition for "square root". Only the more general meaning remains as a convention.

4) We can choose for our own purposes to define a square root function on broader sets of numbers, but this definition is ours and not a universally accepted standard. So we must bear in mind that it needs to be explained to anyone we are talking to on the subject. And we need to be aware that just because a relation works for the positive reals does not mean it will also work with our extended definition.

5) The reason no convention exists for extending [surd]x or x1/2 beyond the non-negative reals as a function is that any particular definition we choose will prove to be a poor choice in certain situations. For this reason, it is preferred to let each person define the function as best fits the need of the problem at hand.

6) In addition to simply throwing out one of the values, like we do in the positive reals, there are two other ways of defining the square root as a function on broader sets. The first is to change the target space from [bbc] to [bbc] + [bbc] - the set of un-ordered pairs from [bbc], under which the square root consists of the set of both roots. The other means is to change the domain from [bbc] to the Riemann Surface generated by the polynomial x2. This consists of two copies of [bbc] pinned together at 0 and connected in such a way that circling 0 twice will take you from one to the other and then back again to where you started on the first. The same point on both copies is mapped to differing roots on each.

So it is not completely true that [surd] is not a function for complex numbers. But neither is it completely true that it is.
Title: Re: 1=0?  (Math skills needed)
Post by Icarus on Mar 9th, 2004, 8:51pm
I have to add a comment about this which you posted while I was attempting to put together something coherent.


on 03/09/04 at 19:27:59, kellys wrote:
(1)1/4 = i, -i, 1, and -1.  That is, 1 has four fourth roots.  This also illustrates why you shouldn't use the symbol  [pm] when in the complex numbers (in the real number we would have said (1)1/4 = [pm]1, and we would have missed two values if we tried to infer the same thing holds in the complex numbers)


No, this only provides example that a 4th degree polynomial (x4 - 1) has 4 roots. There is nothing wrong with using [pm] signs, such as x = [pm]1, [pm]i. It is not the sign that causes missed roots!
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 9th, 2004, 10:35pm
Well, as with any argument in mathematics, it's all about definitions.  I hope we can agree that the answer to godskook's question is that the equalities are perfectly valid, and are essentially the same as those in towr's initial response.

Of course [pm] is still a perfectly valid symbol, but the equations xn=1 have a different symmetry than in the reals, so I would avoid [pm] in this case.
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 10th, 2004, 12:36am
The equalities aren't perfectly valid, since they leave something out. Otherwise 1=-1 would be valid, and it's not.
You can't choose one of the two values on the one side, the other on the other side, and then claim they are equal. In otherwords, [sqrt]1={-1,1} doesn't imply -1=[sqrt]1=1
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 10th, 2004, 11:08am
Is either equality, taken on its own, valid?
Title: Re: 1=0?  (Math skills needed)
Post by Sir Col on Mar 10th, 2004, 11:39am
My understanding (and I still don't feel comfortable with it!) is that [sqrt] is a one-to-many function (called a multvalued function). Similar to how cosine is a many-to-one function: cos(0)=1 and cos(2[pi])=1, but this does not imply that 2[pi]=0. That is, [sqrt]1=1 and [sqrt]1=-1, but they are separate branches of the same value.

I found this related article:
http://mathworld.wolfram.com/MultivaluedFunction.html
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 10th, 2004, 11:52am

on 03/07/04 at 17:46:02, godskook wrote:
which means that i=-i  or 1=-1 or 1=0  ?

What is wrong with this?

Oh oops - when I said that godskook's equalities were valid I wasn't counting the conclusion.  Anyways, the point is that the first two equalities are valid, but the inference is not.
Title: Re: 1=0?  (Math skills needed)
Post by Icarus on Mar 10th, 2004, 7:52pm
No, godskook's first equation cannot be considered valid. It depends on the following two choices for definition of ()1/2: 11/2 = 1 (as opposed to -1), and (-1)1/2 = i (as opposed to -i). With these choices, the identity (a/b)1/2 = a1/2/b1/2 fails to hold. But the equation uses this false identity anyway. Note that the equation concludes that (-1)1/2 = -i, but uses (-1)1/2 = i at one stage in the calcuation.

Sir Col - "multi-valued functions" are not true functions. They are relations between two variables (all functions are such relations, but also have a uniqueness of value condition that "multi-valued functions" do not). They can be associated with actual functions by any of the means I mentioned, however.

A point of clarification or obfuscation - you decide which: most branches of mathematics would accept the word "function" used to describe the map [surd] : [bbc] [to] [bbc] + [bbc], where the latter is the set of all unordered pairs from [bbc]. Analysts, however, are more restrictive in their definitions. To the analyst, "function" means that the target space is either [bbr] or [bbc], or on rare occasion some other base field. Anything else, such as this "multi-valued" definition of [surd] is called a "map".
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 10th, 2004, 9:54pm
So (-1)1/2[ne]-i?
And I suppose that also means,
(1)1/3 = 1
But,
(1)1/3 [ne] -1/2 +  i ([sqrt]3)/2,
(1)1/3 [ne] -1/2 -  i ([sqrt]3)/2 ?
Also, what is (-i)1/3 and why did you choose that particular value?
Title: Re: 1=0?  (Math skills needed)
Post by Sir Col on Mar 11th, 2004, 12:47am
If you're right, kellys, then what I thought from the beginning is true...

Given z3=1, it is true that z=1, -(1/2)[pm][sqrt]3/2 are solutions.

But 11/3=3[sqrt]1=1.

If x1/2[equiv][sqrt]x, I agree that x2=1 [bigto] x=[pm][sqrt]1=[pm]1, but [sqrt]1=11/2=1, and has one positive value/root.

I am thoroughly confused!  ???
Title: Re: 1=0?  (Math skills needed)
Post by towr on Mar 11th, 2004, 3:02am
I think,
for [bbr] 11/3 = 1
for [bbc] 11/3 [in] {1, 1/2+i[sqrt]3/2, 1/2-i[sqrt]3/2}, it isn't equal to a one of them a priori

The confusion comes from what [sqrt] (or cuberoot etc) means
[sqrt] = {(x,a) | x [in] [bbr] [wedge] a2 = x [wedge] a > 0} (a functional relation, [forall]a,b,c: (a,b) [wedge] (a,c) [to] b=c)
or
[sqrt] = {(x,a) | x [in] [bbc] [wedge] a2 = x } (a mapping relation which isn't functional, [forall]a[ne]0[exists]b,c: (a,b) [wedge] (a,c) [wedge] b[ne]c)

What you get is [sqrt]x [in] {a | (x,a) [in] [sqrt]}, which in the one case has just one element, and in the other two.
Title: Re: 1=0?  (Math skills needed)
Post by Sir Col on Mar 11th, 2004, 4:48am
Nice explanation! It now makes perfect sense to me and I'll stick with that. Thanks, towr.
Title: Re: 1=0?  (Math skills needed)
Post by kellys on Mar 11th, 2004, 5:32pm
So the "nth root" problem isn't particular to the complex numbers, since the nth root is a natural definition in *any* field (or group).  I think the thing to tell early students of mathematics is that the = sign is constantly abused, and its meaning will often change.  For example, I would write,
a = [sqrt]i
To mean, "a is a square root of i," even though what would be more appropriate is, a [in] { z | z2=i }.

Of course, another way to make the  nth root a function in the complex number is to use z1/2 = exp(log(z) / 2), since by log we generally mean the function defined on the primary branch.  But that might not help.
Title: Re: 1=0?  (Math skills needed)
Post by Icarus on Mar 13th, 2004, 8:27am
No, it definitely would not help, since by log(z) of complex numbers we often do not mean the primary branch!

You could define all exponentials that way, by specifying the primary branch, but to do so means that you have to give up all the nice laws of handling exponents, since they would no longer hold in some cases, as godskook's example shows.

We generally find it preferable to keep the laws, but with the understanding that they hold only "by abuse of notation", in that the two expressions on either side of the equal signs may actually be conjugal roots rather than the same one.
Title: Re: 1=0?  (Math skills needed)
Post by Nigel_Parsons on Apr 10th, 2004, 3:43pm
This may appear elsewhere, but:

if A=B
 
Then  A*A=A*B (multiplying both by A)

Then A*A - B*B = A*B - B*B (subtract B^2 from both sides)

Then (A+B)*(A-B) =B*(A-B)  [left side difference of 2 sqrs, right side simplification!]

Then (A+B) = B  (divide both sides by (A-B)

Then (B+B) =B  (substitution, A=B line 1)

thus 2B=B  thence 2=1

Subtract 1 from each side & 1=0


QED!   :)
Title: Re: 1=0?  (Math skills needed)
Post by Benoit_Mandelbrot on Apr 13th, 2004, 6:06am
Yes, but from the original equation A=B, you can find that A-B=0.  You divided by zero.
Title: Re: 1=0?  (Math skills needed)
Post by towr on Apr 13th, 2004, 6:26am
that and 2B=B gives (B=0 [vee] 2=1) and since 2=1 [equiv] [perp] this simplifies to just B=0


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