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sqrt(x) = -1

conkermaniac

conkermaniac

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Is it possible to find some number x so that sqrt(x) = -1? i^4 may be a solution, but I'm not so sure...there's two ways of looking at it:

sqrt(i^4) = i^2 = -1

However...this is also true:

sqrt(i^4) = sqrt(1) = 1

So is i^4 a solution of the equation sqrt(x) = -1? :confused:
 
Nope. Any number multiplied by itself would either be negative*negative or positive*positive. As you no doubt learned in 5th grade, any of those would turn out positive. THus nothing sqrt[x]=-1 is NOT possible. However, sqrt[x]=+-1 is possible. IN FACT any square roots answer can be either positive or negative. You can't tell because they both equal the same thing.
 
Shizzle, it seems that you are going by the definition of principal root, but I heard that there is a different situation when one is dealing with complex numbers. How does De Moivre's theorem fit into all of this? I'm confused by your last statement, however. While x^2 = 1 can yield two solutions of x (+/- 1), sqrt(1) can only yield one answer, which is positive. :)

Carebear, I know that i^4 = 1 (in fact, I used that in my second example), but what does sqrt(i^4) equal then? Isn't there also a law that says sqrt(i^4) = ((i^4))^(1/2)? Using the laws of exponents, we know that that equals i^2, which is -1. :confused2
 
Originally posted by conkermaniac
Carebear, I know that i^4 = 1 (in fact, I used that in my second example), but what does sqrt(i^4) equal then? Isn't there also a law that says sqrt(i^4) = ((i^4))^(1/2)? Using the laws of exponents, we know that that equals i^2, which is -1. :confused2
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What's wrong with that? :confused:

sqrt(i^4) = i² = -1

i² * i² = i^4 = 1
-1 * -1 = 1

You're just replacing one number with something's that is equivalent with it.. I could say (x + 1)² - x(x + 2) is the sqrt(1) which is correct but just a different form
 
Conker. You have to do it the second way you have listed. For a very simple reason to. The order of operations. You have to do the work inside the parenthesis first, before you do the outside work.

Plus, you have to treat i like a number, not a variable. You can not simplify i^4 to i^2. That is like:

sqrt(-1^2) = -1

Thats false. -1 squared is 1 and the square root of 1 is 1.

i^4 IS 1. I is not a variable but rather an sqrt(-1), specifically.

Therefore your first method is an incorrect method. What you are doing there is really:

i^sqrt(4)
 
Damn, I can't believe people actually sit around and discuss this stuff :p.
 
Well. Actaully, looking at it, both of us are right, and wrong. Becasue when you get the square root(or even root) of something, it is plus OR minus. For instance:

sqrt(1) = +1 or -1

Both are fine actually.

So both the first and second answer are correct, although the method may be a bit suspect for the first one. In most cases, you should work as far as possible inside before you do the outside work, because with odd roots, you don't have the fortune of the answer being + or -.
 
Originally posted by shizzle
Nope. Any number multiplied by itself would either be negative*negative or positive*positive. As you no doubt learned in 5th grade, any of those would turn out positive. THus nothing sqrt[x]=-1 is NOT possible. However, sqrt[x]=+-1 is possible. IN FACT any square roots answer can be either positive or negative. You can't tell because they both equal the same thing.
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Owen, way to come up with the same thing as the second post in the thread!
 
Shizzle: I was correcting a statement I made beforehand. However, my main point in posting it was to show his answer was correct, but not the method, because +- is only in even roots and not odd roots. Lets say he wanted the third root of i^6.

His first method:
(i^6)^(1/3) = i^3 = -i

The correct method:
(i^6)^(1/3) = -1^(1/3) = -1

Take my second post as simply in conjunction with my first post and not as two seperate points of though. It wasn't an infringement on what you said, but rather correcting a mistake I made.

Besides, its not the first time a specific thought is posted twice by different people, nor will it be the last. Is there a specific reason for pointing it out?
 
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Originally posted by CareBear
actually the third root of i^6 would be i^2 :)
Both methods give the same result.
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You are right. Geez... Looks like I either need to brush up on my algebra 2 or learn how to do powers again. Oy!

Try to seem smart and I look dumb! Gosh! :eek:
 
Lets forget every bit of trash I have written thus far. Lets look at the original problem

sqrt(i^4)

To simplify it(at least on paper), lets change i to -1^(1/2) and use exponents instead of roots.

Therefore:

((-1)^(1/2))^4)^(1/2)

Simplifying it,
((-1)^(4/2))^(1/2)
(-1)^(4/4)
(-1)^1

So the answer comes out -1.

I guess the first method, despite what I have thought is the correct one after looking at it. I's are complex(hence their number system name). An i is more like a half negative number, which wouldn't perform under the same rules are real numbers do in our math system.

Working it out in exponent form, I would surmise that the first method is the correct way. Following the order of operations though, it is the second way.

I hated complex numbers and I always will.
 
Originally posted by Nick
Damn, I can't believe people actually sit around and discuss this stuff :p.
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Me too. Anything that requires me to think is a no-do.
 
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