Imaginary fun

Thread Starter

Mark44

Joined Nov 26, 2007
626
I'm used to using i for \[\sqrt{-1}\], but I'll use j since that's what most people in this forum seem to be more comfortable with.

Each step below seems reasonable, but obviously there is a problem.

\[\frac{1}{-1}\] = \[\frac{-1}{1}\]

If two numbers are equal, their square roots are equal:
\[\sqrt{\frac{1}{-1}}\] = \[\sqrt{\frac{-1}{1}}\]

The square root of a quotient is the quotient of the square roots:
\[\frac{\sqrt{1}}{\sqrt{-1}}\] = \[\frac{\sqrt{-1}}{\sqrt{1}}\]

Simplifying:
\[\frac{1}{j}\] = \[\frac{j}{1}\]

Cross-mulitiplying:
1\[^{2}\] = j\[^{2}\]

Therefore: 1 = -1 :eek:
 

Wendy

Joined Mar 24, 2008
21,847
The cross multiplying is the faulty step. It does not follow...

It also helps to think of j as a sign, just like plus and minus.
 
Why is the cross multiplication a faulty step? Even if you don't cross multiply, 1/j = -j which means there's something quirky even before the cross multiplication.
 
It is messed up because of the way the signs are initially assigned... its kind of like arctan(-3/4) is not equal to arctan(3/-4) although the arguments have the same value.
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
The cross multiplying is the faulty step. It does not follow...
Well, I'll do it without cross-multiplying.
\[\frac{1}{j}\] = \[\frac{j}{1}\]

Instead of cross-multiplying (a truly nebulous concept), I'll multiply both sides by 1 * j, resulting in:
1 = j * j = j\[^{2}\]

So now I'm back to 1 = -1
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
It is messed up because of the way the signs are initially assigned... its kind of like arctan(-3/4) is not equal to arctan(3/-4) although the arguments have the same value.
But arctan(-3/4) is exactly equal to arctan(3/-4)
 

Papabravo

Joined Feb 24, 2006
12,384
One problem is that square root is actually a multivalued function and so not actually a function at all in the strictest sense. Since it is a 1 to many mapping it cannot be uniquely inverted and so your inconsistent results are not surprising. It takes a bit of mathematical sophistication to understand what is going on here. You are also not the first person to have initiated this particular thread which has appeared serveral times before on this or other forums.

Here is your false assumption
If two numbers are equal, their square roots are equal:
Rich (BB code):
  4 = 4
sqrt(4) = sqrt(4)
 -2 = 2
See the problem? The square root of a number is not unique.
 
Wow, i'm kinda an idiot haha. I was getting mixed up with having to add pi/2 or whatever to certain arctan values to get the value you're looking for because arctan is defined -pi/2<x<pi/2 and i was getting messed up with the reasoning somewhere haha.

Anyway, wiki has a section on this proof and why its faulty under both the article on square roots and in the article on the imaginary unit
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
One problem is that square root is actually a multivalued function.
I don't think so. While it's true that every positive real number has two square roots (and we can extend this to negative reals, also), an expression such as sqrt(4) evaluates to a single number, namely 2. If we want to refer to the negative square root of 4, we write -sqrt(4).

You are also not the first person to have initiated this particular thread which has appeared serveral times before on this or other forums.
I've looked through the threads in the math forum, but apparently missed it.

Here is your false assumption
If two numbers are equal, their square roots are equal:
Rich (BB code):
  4 = 4
sqrt(4) = sqrt(4)
 -2 = 2
See the problem? The square root of a number is not unique.
For the third line above, I get 2 = 2.

I think the problem lies elsewhere...
Mark
 
I think (from what i've read on wiki) that the problem is that some identities involving square roots are only applicable for real positive numbers. Wiki cites sqrt(a*b) = sqrt(a)*sqrt(b) as being only valid for positive real numbers. I would imagine there's some condition like this for sqrt(x/y) = sqrt(x)/sqrt(y)
 

Wendy

Joined Mar 24, 2008
21,847
1 does not equal j * j or j^2

-1 = j^2 = j * j

Another way of stating where the falicy is...

Both 2 and -2 are the square root of 4, but they do not equal each other. A square root can be more than one value.
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
I think (from what i've read on wiki) that the problem is that some identities involving square roots are only applicable for real positive numbers. Wiki cites sqrt(a*b) = sqrt(a)*sqrt(b) as being only valid for positive real numbers. I would imagine there's some condition like this for sqrt(x/y) = sqrt(x)/sqrt(y)
Bingo :)
Mark
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
Both 2 and -2 are the square root of 4, but they do not equal each other. A square root can be more than one value.
Both 2 and -2 are square roots of 4, but the principal square root of 4 is 2. A square root has only one value. As proof, look at the graph of the function f(x) = \[\sqrt{x}\].
 

Papabravo

Joined Feb 24, 2006
12,384
Both 2 and -2 are square roots of 4, but the principal square root of 4 is 2. A square root has only one value. As proof, look at the graph of the function f(x) = \[\sqrt{x}\].
I do know what I am talking about. The square root of each positive number greater than zero has two values and not one. The mapping is one to many and therefore is not a function in the strictest sense. I did not make a mistake on line three of my previous post, it was intentional because both 2 and -2 are solutions to the equation
Rich (BB code):
x = sqrt(4)
That is why I said it requires a degree of mathematical sophistication to understand what is going on. You either understand it or you don't.
 

Wendy

Joined Mar 24, 2008
21,847
If -2 and 2 are both the square root of 4, then they are both the correct answer. Anything else is opinion. :)

You can prove anything with a false hypothisis, such as -1 = 1 .
 

Thread Starter

Mark44

Joined Nov 26, 2007
626
The square root of each positive number greater than zero has two values and not one. The mapping is one to many and therefore is not a function in the strictest sense. I did not make a mistake on line three of my previous post, it was intentional because both 2 and -2 are solutions to the equation
Rich (BB code):
x = sqrt(4)
From the Wikipedia article (http://en.wikipedia.org/wiki/Square_root) on square roots:
Every non-negative real number x has a unique (emphasis mine) non-negative square root, called the principal square root and denoted with a radical symbol as √x. For example, the principal square root of 9 is 3, denoted √9 = 3, because 32 = 3 × 3 = 9. If otherwise unqualified (emphasis mine), "the square root" of a number refers to the principal square root: the square root of 2 is approximately 1.4142.​

Clearly, the equation x\[^{2}\] = 4 has two solutions for x: 2 and -2. If you solve this equation by taking the square root of each side you get \[\sqrt{4}\], or 2, on the right side, and \[\sqrt{x^{2}}\] on the left. This expression equals x if x > 0 and equals -x if x < 0.

So, from x\[^{2}\] = 4 you get x = 2 or -x = 2.
Mark
 

recca02

Joined Apr 2, 2007
1,211
If two numbers are equal, their square roots are equal:
\[\sqrt{\frac{1}{-1}}\] = \[\sqrt{\frac{-1}{1}}\]

The square root of a quotient is the quotient of the square roots:
\[\frac{\sqrt{1}}{\sqrt{-1}}\] = \[\frac{\sqrt{-1}}{\sqrt{1}}\]
I think the problem lies here.
AFAIK separation like that isn't valid for complex numbers.
Other explanation has already been given once. To prove that similar thing can happen with real numbers. Thus, I concur here with Mr. Bravo.

Edit: I don't recall much but, if square root of X is taken as a function only positive values are considered. Else it can not be considered as a function as it returns more than one value for a single value of X. But IMO it doesn't mean that the operation itself is invalid.

some similar 'complex'ities :D.


Try this too :p
 

Papabravo

Joined Feb 24, 2006
12,384
From the Wikipedia article (http://en.wikipedia.org/wiki/Square_root) on square roots:
Every non-negative real number x has a unique (emphasis mine) non-negative square root, called the principal square root and denoted with a radical symbol as √x. For example, the principal square root of 9 is 3, denoted √9 = 3, because 32 = 3 × 3 = 9. If otherwise unqualified (emphasis mine), "the square root" of a number refers to the principal square root: the square root of 2 is approximately 1.4142.
Clearly, the equation x\[^{2}\] = 4 has two solutions for x: 2 and -2. If you solve this equation by taking the square root of each side you get \[\sqrt{4}\], or 2, on the right side, and \[\sqrt{x^{2}}\] on the left. This expression equals x if x > 0 and equals -x if x < 0.

So, from x\[^{2}\] = 4 you get x = 2 or -x = 2.
Mark
I agree that principal square root is a function in the strict sense precisely because it is single valued and therefore invertable. One must always interpret the unqualified square root as being one of two possible values. If an operation has either of two values, then it is trivially easy to make nonsense equations like the one that started this thread. Otherwise any conclusion can be arrived at from a false premise.
 
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