# Imaginary fun

Discussion in 'Math' started by Mark44, Apr 15, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
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.

3. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
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.

4. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
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.

5. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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

6. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
But arctan(-3/4) is exactly equal to arctan(3/-4)

7. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
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.

If two numbers are equal, their square roots are equal:
Code ( (Unknown Language)):
1.
2.   4 = 4
3. sqrt(4) = sqrt(4)
4.  -2 = 2
5.
See the problem? The square root of a number is not unique.

8. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
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

9. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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).

I've looked through the threads in the math forum, but apparently missed it.

For the third line above, I get 2 = 2.

I think the problem lies elsewhere...
Mark

10. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
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)

11. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
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.

Nov 26, 2007
626
1
Bingo
Mark

13. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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) = $\sqrt{x}$.

14. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You mean a parabola?

15. ### antimatter New Member

Apr 16, 2008
3
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half of one, f(x) = - $\sqrt{x}$ give you the other half.

a parabola, opening in the +x axis

16. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
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
Code ( (Unknown Language)):
1.
2. x = sqrt(4)
3.
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.

17. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
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 .

18. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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

19. ### recca02 Senior Member

Apr 2, 2007
1,211
0
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 .

Try this too

20. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
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.