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I understand that $i = \sqrt{-1}$, but it seems as if $i^2$ would be:

$$\sqrt{-1}\sqrt{-1} = \sqrt{1} = \pm 1$$

Why is this not the case?

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Because $\sqrt{1}=1$, not $\pm 1$. – JohnD Feb 4 '13 at 20:06
By definition, $\sqrt{1} > 0$ and so $\sqrt{1} \neq -1$. If $x^2=1$ then $x = \pm\sqrt{1} = \pm 1$. – Fly by Night Feb 4 '13 at 20:06
Then why is √1 by definition greater than 0? √25, for instance is ±5, no? – Smovies Feb 4 '13 at 20:09
No. The solution to the equation $x^2=25$ is $x=\pm 5$ because $(5)^2=(-5)^2=25$. By definition, the symbol $\sqrt{25}$ means the positive number whose square is 25. That's why you get $\pm \sqrt{25}$ in solutions, one is positive: namely $\sqrt{25}$ and one is negative: namely $-\sqrt{25}$. – Fly by Night Feb 4 '13 at 20:12
This was a double-whammy of misconceptions :) – rschwieb Feb 4 '13 at 20:22
up vote 13 down vote accepted

If I understand correctly, you're thinking is as follows: $$\ \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = \pm 1. $$

I'll address the primary question here: in writing this, you are assuming the "rule" that tells us: $$ \sqrt{ab} = \sqrt{a}\sqrt{b} $$ also applies to $a, b$ when both are negative. However:

$\sqrt{ab} = \sqrt a\sqrt b\;$ if and only if $a$ and $b$ are not both negative real numbers.

The usual laws of exponents (and roots) do not apply to complex imaginary numbers, nor, in the case of the square root, do they apply to negative numbers of any sort.

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It seems that if either a OR b were negative, though, the usual rule would apply, as in: (√5)(√-3) = √-15 ? – Smovies Feb 4 '13 at 20:11
Yes, then you could write $\sqrt{-15}$ as $i\sqrt {15}$, or rather, $\sqrt 5 \sqrt {-3} = i\sqrt 5 \sqrt 3 = i\sqrt{15}$ – amWhy Feb 4 '13 at 20:12
@amWhy Not exactly. "However: √ab =√a√b if and only if a,b are non-negative real numbers." We just showed above that √ab = √a√b when a OR b is negative, no? As in: (√a)(√-b) = √-ab – Smovies Feb 4 '13 at 20:30
As Smovies points out, the "if and only if" part as stated is false. For real $a,b$, the rule $\sqrt{ab}=\sqrt{a}\sqrt{b}$ will be true if and only if $a,b$ are not both negative, i.e. if $a$ or (not and) $b$ is nonnegative. – anon Feb 4 '13 at 21:16
In order to ask meaningfully whether $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true, one has to have definitions fixed for what $\sqrt{ab}$, $\sqrt{a}$, and $\sqrt{b}$ mean. Only in the case where $a\geq 0$ and $b\geq 0$ is there a fairly universal canonical definition of $\sqrt a$ and $\sqrt b$. When extending the notation $\sqrt{a}$ by defining particular branches of the square root on subsets of the complex plane, whether or not such identities hold will depend on the branches chosen and the particular $a$,$b$ chosen. It will hold sometimes when neither $a$ nor $b$ is real. – Jonas Meyer Feb 5 '13 at 2:15

The rule that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b$ are positive.

To add some more context: these rules hold only for positive numbers for a few reasons, but in general, the existence of complex numbers is something necessitated by an inadequacy of the reals to solve certain types of problems.

In a way, we might look at $i^2 = -1$ as a definitive property of $i$ (that is, a property that holds because we define it to be so), rather than a consequential property.

Complex arithmetic is essentially synthetic: it's something we made up. It happens that the rules we chose are very useful and have exceptionally nice properties, so we stick with them.

But in the end, we define $(a,b)(c,d) = (ac-bd,ad+bc)$. This is exactly equivalent to defining $i^2 = (0,1)(0,1) = (-1,0)$.

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Isn't defining property less odd? – Mariano Suárez-Alvarez Feb 4 '13 at 20:13
@MarianoSuárez-Alvarez That's perhaps a question for English.SE. – Emily Feb 4 '13 at 20:15
Not really: I have never seem definitive used in that way in the many, many years I have spent reading math, and always seen defining. I have seen definitive used that way in other contexts, though. – Mariano Suárez-Alvarez Feb 4 '13 at 20:17
The participle as an adjective is a crude option when a perfectly suitable---and much more pleasant sounding---alternative is available, regardless of discipline. ;) – Emily Feb 4 '13 at 20:19
In fact, I would say that this at least confusing. I do not care that much really. – Mariano Suárez-Alvarez Feb 4 '13 at 20:28

It seems like you are thinking that $$\ \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1. $$ As others have already mentioned, in doing this, you are assuming that $$ \sqrt{ab} = \sqrt{a}\sqrt{b} $$ but this rule only holds for non-negative real numbers.

One way to think about $i$ is simply as something whose square is $-1$. You can then defined the complex numbers as all linear combinations $a + bi$, where $a$ and $b$ are real numbers. Now when you start to manipulate/use algebra with these complex numbers you just remember that $i^2 = -1$ (by definition). Likewise some math books will simply define that for a positive real number $a$ $\sqrt{-a}$ is $i\sqrt{a}$.

The morale of all this is that when you move on to something "new" in math, you have to be very careful because the "usual" rules don't always apply.

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Ah, that's clearer. I undertand it, for sure, taking i^2 = -1 as a first principle. It just seems inelegant to have to that and I wanted to understand it from the perspective of the first principle i = √-1 – Smovies Feb 4 '13 at 20:15

A main misconception here is that $\sqrt{a^2}=\pm a$. That is not true, but $\sqrt{a^2}=|a|$ is true for any real number $a$.

What you are confusing it with is solving the equation $x^2=c$. Some students are taught to "take the square root of both sides, but to avoid dropping a solution, use $\pm$":


It's important not to confuse these two things: finding the square root of a number, and solving an equation of the form $x^2=c$.

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$\sqrt{a^2}=|a|$ is true when $a$ is real and $\sqrt{}:[0,\infty)\to[0,\infty)$. (I thought it might be worth mentioning because there are also nonreal numbers in the question.) – Jonas Meyer Feb 5 '13 at 2:09
@JonasMeyer That's worth adding: thanks. – rschwieb Feb 5 '13 at 12:50

It may help to avoid using the $\sqrt{}$ notation, but rather exponents. Remember that $\sqrt a$ is actually $a^{1/2}$.

Your argument is then this. Why cannot we do: $(-1^{1/2})(-1^{1/2}) = (-1\cdot -1)^{1/2} = 1^{1/2} = 1$?

But how the product of powers of the same base actually works is this: $x^ax^b = x^{a+b}$.

So, in other words, $-1^{\frac{1}{2} + \frac{1}{2}} = -1^1 = -1$.

We cannot simply take things out from under the exponent and multiply them out!

In the case that $a = b$, it reduces to $x^ax^a = x^{2a}$.

So then the pattern you're trying to follow looks like this, with exponents: $x^ax^a = x^{2a} = (x^2)^a = (xx)^a$. Now, that is actually correct depending on the nature of the $x$ and $a$, but not when $x$ is negative and $a$ is fractional.

The problem is that the exponent law $x^{(mn)} = (x^m)^n$ does not hold in general. It does not hold when $x$ is negative and $m$ or $n$ are fractional.

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if you square the square root of 'something', the result is equal to the 'something' i.e. the (square root of two) squared is equal to two

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