i^2

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Portia
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i^2

Post by Portia »

If you have
\sqrt{xy} = \sqrt x \sqrt y

but apparently both x and y have to be greater than or equal to zero?

When a student I'm tutoring tried to solve i^2, he wrote (sq rt -1) * (sq rt -1)

This ≠ sq rt (-1 * -1), i,e, sq rt (1) = principal root = 1

One way to explain getting the correct answer is to think of squaring as the inversion operation of taking the square root; i.e., the "^2" and the radical "cancel each other out," leaving -1. But I am having a hard time explaining WHY his answer was wrong. That preceding ≠ sign must be the case, but he wrote = and it seems awfully convincing. :( Is this just a rule of evaluating square roots? He said his teacher had drilled in the fact that if x^2=36, x = 6, -6, which is true, but conflicts with this case in imaginary numbers.

Help? Why was he wrong? Why isn't i^2 positive AND negative 1?
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Defy V
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Re: i^2

Post by Defy V »

One square root of 1 is -1, I guess, which is i^2. And sqrt(36) = sqrt(-6*-6) = sqrt(-6)*sqrt(-6) = -6.
Neither of these are the principal roots, but I think that is one way to find the other square roots.
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Portia
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Re: i^2

Post by Portia »

Defy V wrote:One square root of 1 is -1, I guess, which is i^2. And sqrt(36) = sqrt(-6*-6) = sqrt(-6)*sqrt(-6) = -6.
Neither of these are the principal roots, but I think that is one way to find the other square roots.
The problem here is that if you put i^2=1 on a test, you'll be marked wrong. It's -1. I don't know the best way to explain that . . .
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Defy V
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Re: i^2

Post by Defy V »

Unless you want to get all complicated and throw some basic complex analysis at the kid (which I don't recommend), then the best reason I can find is given by Wikipedia:

For all non-negative real numbers x and y, \sqrt{xy} = \sqrt{x}\sqrt{y}

So the rules fall apart for negative numbers. Why that is, well, . . . imaginary numbers are weird. And that's the best reason I can think of at a high school level.

But yeah, it does seem like there should be a better reason besides "it's just not allowed."
NerdGirl
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Re: i^2

Post by NerdGirl »

The normal rules of manipulating square roots don't apply with square roots of negative numbers. If you try to use them, you run into contradictions (like saying that 1= -1). There are people on here with actual math degrees (like TheAnswerIs42), so one of them could probably give you a more detailed explanation of the reasons for that, I'm sure. And DefyV already pretty much said that, so I don't know that I'm adding much to the conversation here. But one thing that might be helpful is to teach him to turn square roots of negative numbers into i * the square root of the positive number. So if you had sqrt(-9), you can write it as i * sqrt(9) and then you end up with 3*i. When I was learning this stuff someone told me to do that and I think it was helpful.
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Indefinite Integral
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Re: i^2

Post by Indefinite Integral »

Defy V basically gave the answer I was thinking of (high school math teacher here). When doing any square root problem, we tend to automatically take the positive square root, however, there are actually TWO answers to any square root problem - a positive and a negative one.

You could say \sqrt{(-6)^2} = 6, and that is totally valid, but \sqrt{(-6)^2} = -6 as well. Usually we denote whether we want the positive or the negative square root, then you can only take one of them.

Hmm...actually come to think about it, I'm pretty sure it boils down to order of operations. In some cases (specifically cases when x is positive) \sqrt{(x)^2} = (\sqrt{x})^2 However, that is because there are no other operations occurring inside the parentheses created by the sqaure root operation. However, when you add the negative in, then switching the order in which you do the operations does make a difference. Basically, they were taught a "rule" that really only sometimes is possible. It's a shortcut which only works when there is nothing else interfering with it.

Does that make sense? It just came to me in a blinding flash, and it would be so awesome if it works the way I'm thinking it does.
"The pursuit of mathematics is a divine madness of the human spirit." ~ Alfred North Whitehead
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Portia
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Re: i^2

Post by Portia »

Indefinite Integral wrote:Defy V basically gave the answer I was thinking of (high school math teacher here). When doing any square root problem, we tend to automatically take the positive square root, however, there are actually TWO answers to any square root problem - a positive and a negative one.

You could say \sqrt{(-6)^2} = 6, and that is totally valid, but \sqrt{(-6)^2} = -6 as well. Usually we denote whether we want the positive or the negative square root, then you can only take one of them.

Hmm...actually come to think about it, I'm pretty sure it boils down to order of operations. In some cases (specifically cases when x is positive) \sqrt{(x)^2} = (\sqrt{x})^2 However, that is because there are no other operations occurring inside the parentheses created by the sqaure root operation. However, when you add the negative in, then switching the order in which you do the operations does make a difference. Basically, they were taught a "rule" that really only sometimes is possible. It's a shortcut which only works when there is nothing else interfering with it.

Does that make sense? It just came to me in a blinding flash, and it would be so awesome if it works the way I'm thinking it does.
Thanks, I think order of operations is the perfect way to think about it.

Please PM me and say where and what level you teach! I'm a new math tutor for 4th through 12th and would love any insight I can get.
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