SImpe question about square root function

[Ali Asadullah]

Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.

 

[CRGreathouse]

No..

 

[Mark44]

Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.

If a >= 0 and b >=0, $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$

 

[Xitami]

If a <= 0 and b <=0, $$\sqrt{ab} = -\sqrt{a}\sqrt{b}$$

 

[Mark44]

If a <= 0 and b <=0, $$\sqrt{ab} = -\sqrt{a}\sqrt{b}$$

No, that's not true at all.
$$\sqrt{(-2)(-8)} = \sqrt{16} = 4 = \sqrt{2}\sqrt{8}$$

 

[Mentallic]

And finally for the last case, a<0 b$\geq$0

$$\sqrt{ab}=\sqrt{-a}\sqrt{b}=i\sqrt{a}\sqrt{b}$$

 

[Char. Limit]

And finally for the last case, a<0 b$\geq$0

$$\sqrt{ab}=\sqrt{-a}\sqrt{b}=i\sqrt{a}\sqrt{b}$$

Actually, if only one of a and b are positive, we don't know if the answer is multiplied by i or by -i, I believe. That non-property is used to show the fallacy in a certain 1=2 argument.

 

[Martin Rattigan]

Has this thread something to do with the recent release of Alice in Wonderland?

Mark44: $\sqrt{(-2)(-8)}=\sqrt{2}\sqrt{8}$ (true) $=(\sqrt{-2}\sqrt{-8})/i^2=-\sqrt{-2}\sqrt{-8}$ (as in Xitami's formula)

Or setting $b=1,\sqrt{a}\neq 0$:

Mentallic: $1=i$

Char Limit: $1=\pm i$

 

[Mentallic]

Actually, if only one of a and b are positive, we don't know if the answer is multiplied by i or by -i, I believe. That non-property is used to show the fallacy in a certain 1=2 argument.

Of course I felt I can venture into this forbidden territory of complex numbers without any issues... I was wrong.

Mark44: $\sqrt{(-2)(-8)}=\sqrt{2}\sqrt{8}$ (true) $=(\sqrt{-2}\sqrt{-8})/i^2=-\sqrt{-2}\sqrt{-8}$ (as in Xitami's formula)

I'll have some fun here to try and explain where Martin's second expression could've arisen from.

a>0, b>0

$$\sqrt{ab}=\sqrt{(-a)(-b)}=\sqrt{-a}\sqrt{-b}$$

$$=\frac{i^2}{i^2}\sqrt{-a}\sqrt{-b}$$

$$=\frac{-1}{i^2}\left( i\sqrt{a}\right)\left( i\sqrt{b}\right)$$

$$=\frac{-1}{i^2}(i^2)\sqrt{a}\sqrt{b}$$

$$=-\sqrt{a}\sqrt{b}$$

as required... I suppose

Disclaimer: The first line, while mathematically sound, isn't allowed to be used in this square root function.

Or setting $b=1,\sqrt{a}\neq 0$:

Mentallic: $1=i$

Char Limit: $1=\pm i$

I'm not sure what you're getting at here. We were assuming a<0 so $\sqrt{a}\neq 0$ is a given.

 

[Martin Rattigan]

I'll have some fun here to try and explain where Martin's second expression could've arisen from.

If by second expression you mean $\sqrt{2}\sqrt{8}$ it was copied from Mark44's post (as was the first expression $\sqrt{(-2)(-8)}$, but for some reason itex doesn't extend the square root sign - it should have read $\sqrt{(-2)(-8)}$).

The third expression comes from $\sqrt{-2}=i\sqrt{2},\sqrt{-8}=i\sqrt{8}$ and substitution in the second expression.

The fourth expression comes from replacing $i^2$ by $-1$ and looks very much like what Xitami said and Mark44's post is claiming to be wrong.

I'm not sure what you're getting at here. We were assuming a<0 so $\sqrt{a}\neq 0$ is a given.

What I'm getting at is you're claiming $\sqrt{ab}=i\sqrt{a}\sqrt{b}$ when $a<0,b\geq 0$. Substituing $b=1$ (which is $\geq 0$) gives, for $a<0$

$\sqrt{a.1}=i\sqrt{a}\sqrt{1}$

or $\sqrt{a}=i\sqrt{a}$

and after division by $\sqrt{a}$ which is possible because, as you point out $a<0$ therefore $\sqrt{a}\neq 0$

$1=i$

 

[Mark44]

If by second expression you mean $\sqrt{2}\sqrt{8}$ it was copied from Mark44's post (as was the first expression $\sqrt{(-2)(-8)}$, but for some reason itex doesn't extend the square root sign - it should have read $\sqrt{(-2)(-8)}$).

The "some reason" is that you switched from [ tex] tags to [ itex] tags.

The third expression comes from $\sqrt{-2}=i\sqrt{2},\sqrt{-8}=i\sqrt{8}$ and substitution in the second expression.

The fourth expression comes from replacing $i^2$ by $-1$ and looks very much like what Xitami said and Mark44's post is claiming to be wrong.



What I'm getting at is you're claiming $\sqrt{ab}=i\sqrt{a}\sqrt{b}$ when $a<0,b\geq 0$. Substituing $b=1$ (which is $\geq 0$) gives, for $a<0$

$\sqrt{a.1}=i\sqrt{a}\sqrt{1}$

or $\sqrt{a}=i\sqrt{a}$

and after division by $\sqrt{a}$ which is possible because, as you point out $a<0$ therefore $\sqrt{a}\neq 0$

$1=i$

 

[Mentallic]

What I'm getting at is you're claiming $\sqrt{ab}=i\sqrt{a}\sqrt{b}$ when $a<0,b\geq 0$. Substituing $b=1$ (which is $\geq 0$) gives, for $a<0$

$\sqrt{a.1}=i\sqrt{a}\sqrt{1}$

or $\sqrt{a}=i\sqrt{a}$

and after division by $\sqrt{a}$ which is possible because, as you point out $a<0$ therefore $\sqrt{a}\neq 0$

$1=i$

but for a<0, $$\sqrt{a}=i\sqrt{-a}$$

 

[Mark44]

And finally for the last case, a<0 b$\geq$0

$$\sqrt{ab}=\sqrt{-a}\sqrt{b}=i\sqrt{a}\sqrt{b}$$

This isn't right. Neither equality is valid under the given conditions.

Since the possiblility of b being zero doesn't add anything useful, let me assume that b > 0 for the rest of the discussion, and that a < 0 as before.

$$\sqrt{ab} \neq \sqrt{-a}\sqrt{b}$$
In the expression on the left, ab < 0 so its square root is imaginary. In the expression on the right, both radicands are positive, so each square root is real and positive, making the product real and positive.

$$\sqrt{-a}\sqrt{b} \neq i\sqrt{a}\sqrt{b}$$
For reasons already given, the left side above is real and positive. The right side above is also real, but negative. The reason for this is that
$$i\sqrt{a}\sqrt{b} = i^2 \sqrt{-a}\sqrt{b}= -\sqrt{-a}\sqrt{b}$$

It might be helpful to show this with numbers instead of variables, replacing a with -2 and b with 8.
$$i\sqrt{-2}\sqrt{8} = i^2 \sqrt{2}\sqrt{8}= -\sqrt{2}\sqrt{8}$$

I think we have succeeded in going well beyond what the OP asked, and might have further confused him/her with a lot of extraneous information. I hope we are done here.

 

[Martin Rattigan]

but for a<0, $$\sqrt{a}=i\sqrt{-a}$$

I might very well agree with that. You can divide by $i\sqrt{-a}$ and prove $1=i$ that way if you like.

But if you want to leave the rabbit hole you'll need to change your original equation.

 

[Martin Rattigan]

The "some reason" is that you switched from [ tex] tags to [ itex] tags.

I thought that was more or less what I said.

I think we have succeeded in going well beyond what the OP asked, and might have further confused him/her with a lot of extraneous information. I hope we are done here.

I don't think there was any confusion up to and including Xitami's first (correct) post, but it went downhill from there. All that was required at that point was a correct formula for $\sqrt{ab}$ for real $a,b$ of opposite sign.

 

[Char. Limit]

I think that what we have here is notation error...

How about this then?

For positive real a and b...

$$\sqrt{(-a)(b)}=\sqrt{-a}\sqrt{b}=i\sqrt{ab}$$

Of course, a similar argument cannot be done for -a and -b because the square root function can't be "split" with two negative radicands. No, it can't.

 

[Martin Rattigan]

I think that what we have here is notation error...

How about this then?

For positive real a and b...

$$\sqrt{(-a)(b)}=\sqrt{-a}\sqrt{b}=i\sqrt{ab}$$

Of course, a similar argument cannot be done for -a and -b because the square root function can't be "split" with two negative radicands. No, it can't.

If you want to avoid confusion it's probably not a good idea to change the meaning of $a$ and $b$ from those in the original question (repeated under).

Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.

For real $a$ and $b$, $(ab)^{0.5}=a^{0.5}b^{0.5}$ unless both $a$ and $b$ are negative, in which case $(ab)^{0.5}=-a^{0.5}b^{0.5}$.

 

[Char. Limit]

No, if a and b are both negative, than the result is invalid. And I already proved the result if either a or b is negative.

Note that because of the discontinuous nature of the square root function in the complex plane, the law √zw = √z√w is in general not true. (Equivalently, the problem occurs because of the freedom in the choice of branch. The chosen branch may or may not yield the equality; in fact, the choice of branch for the square root need not contain the value of √z√w at all, leading to the equality's failure. A similar problem appears with the complex logarithm and the relation log z + log w = log(zw).) Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that −1*=*1:

So if both a and b are negative, the result is invalid (actually multivalued).

If either a or b is negative, my result follows above.

 

[Martin Rattigan]

From the original question:

... when a and b either both are negative or one of them is negative.

From your post:

... For positive real a and b...

What I was suggesting is that this is likely to lead to confusion.

No, if a and b are both negative, than the result is invalid.

By "the result" here, do you mean the result given in the original question or the result I gave in answer?

If the former then I would agree with you. That is the import of the minus sign in the last expression in my answer.

If the latter, can you give a specific pair of real values for a and b where you think the answer is incorrect?

With the usual definition of $z^w$ for complex $z,w$ viz. $z^w=e^{w Log(z)}$ the Log is taken to be the principal log, so the function is single valued, so I think your remarks about the result being multivalued are mistaken if by "the result" in this case you mean one of (ab)^0.5, a^0.5 or b^0.5.

 

[Char. Limit]

I mean your result.

$$\sqrt{(-2)(-8)}=\sqrt{16}=4$$

I only used multiplication and the square root of a positive number. If a and b are negative, the answer is a real, positive number, and your result is invalid. Assuming we don't take the principal root (always possible)...

$$\sqrt{(-2)(-8)}=\sqrt{16}=4 , -4$$

Which is multivalued. So if the principal root gives you a real, positive number, and the general root gives you multiple values, you won't receive a singlevalued negative real number from your operations.

(Note: I did post the wrong Wikipedia section... They are talking about complex numbers, not negative numbers. Sorry for that mixup.)

 

[Martin Rattigan]

I mean your result.

$$\sqrt{(-2)(-8)}=\sqrt{16}=4$$

I only used multiplication and the square root of a positive number. If a and b are negative, the answer is a real, positive number, and your result is invalid.

This is the same example that Mark44 previously used. As I said regarding that example:

... Mark44: $\sqrt{(-2)(-8)}=\sqrt{2}\sqrt{8}$ (true) $=(\sqrt{-2}\sqrt{-8})/i^2=-\sqrt{-2}\sqrt{-8}$ (as in Xitami's formula)

the result agrees with Xitami's formula, which matches my answer for $a,b$ both negative. So that is not a counter-example.

Assuming we don't take the principal root (always possible)...

$$\sqrt{(-2)(-8)}=\sqrt{16}=4 , -4$$

Which is multivalued.

Both notations $z^{0.5}$ and $\sqrt{z}$ are defined as the principal root (precisely to avoid multiple values). It's always possible for you not to take the principal root, but in that case you're not talking about the problem.

So I think the one line answer I included a few posts back was all that was really required for this thread.

PS. Neither OP nor I said anything about the expressions on either side being real. I would say it was clear that OP was well aware that they would not be.

 

[Char. Limit]

You're assuming that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if both a and b are negative. This isn't true. After all, here's another equation using your same logic (it's your logic, right?)

$$-1=i^2=i \times i=\sqrt{-1}\sqrt{-1}=\sqrt{-1 \times -1}=\sqrt{1}=1$$

And that's why $\sqrt{ab}$ does not equal $\sqrt{a}\sqrt{b}$ for negative a and b. Because if it did, -1 would equal 1.

Edit: Here's another place that shows the same logic, but with division. The relevant material is near the bottom.

http://www.math.toronto.edu/mathnet/falseProofs/guess11.html

 

[Martin Rattigan]

Absolutely not. If you read what I wrote a little more carefully it says that $\sqrt{ab}=-\sqrt{a}\sqrt{b}$ if both $a$ and $b$ are negative. This gives you:

$-1=i^2=i\times i=\sqrt{-1}\sqrt{-1}=-\sqrt{-1\times -1}=-\sqrt{1}=-1$

No surprise there.

 

[Char. Limit]

Ok, then tell me why every single calculator I've used on this iPod (six at last count) gave me real and positive numbers for every square root of the product of two negative numbers that I put in? According to you, those numbers should be negative. Also, shouldn't math be consistent? So why can't I multiply the numbers in the radicand to produce a positive product, when I can with ANY other number under that root sign? In other words, why does multiplying the numbers in the radicand and then taking the square root (all allowed formal operations) produce a positive number, but your method produces a negative number?

 

[Mentallic]

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

How did you come up with this?

The way I see it, and why I agree with Char.Limit is that $$\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$$ from $$\sqrt{a}\sqrt{b}=\sqrt{ab}$$

The problem arises from the order the process is undertaken. Repeating what wikipedia said that Char.Limit posted,

(Equivalently, the problem occurs because of the freedom in the choice of branch.

$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$$

or

$$1=\sqrt{1}=\sqrt{(-1)(-1)}$$ (now at this point we multiply under the brackets first, to go back to the previous equality, in which case we will get 1 again).

 

[jackmell]

. . . what a quagmire. Roots are multi-valued. $\sqrt{1}=\pm 1$. And if you carry this multivalued notion throughout the process, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for any values of a and b. For example,

$\sqrt{(-2)(4)}=\sqrt{-2}\sqrt{4}=\sqrt{-1}\sqrt{2}\sqrt{4}=\pm i 2\sqrt{2}$

$\sqrt{(-2)(-4)}=\sqrt{-2}\sqrt{-4}=\sqrt{-1}\sqrt{2}\sqrt{-1}\sqrt{4}=\pm i 2(\pm i)\sqrt{2}=\pm 2\sqrt{2}$

even:

$\sqrt{(2)(4)}=\sqrt{2}\sqrt{4}=(\pm 2)(\pm \sqrt{2})=\pm 2\sqrt{2}$

you've really got to get a handle on this concept because this is just the lowly square root with just two values. What happens when we up the root or consider the log function or inverse trig functions which are really infinitely-valued (in the complex sense)? For example, can I:

$\sqrt[3]{(-4)(2)}=\sqrt[3]{-4}\sqrt[3]{2}$?
and just carry-over the multiple-valued $\sqrt[3]{-1}$ throughout the steps and write:
$\sqrt[3]{(-4)(2)}=\sqrt[3]{-4}\sqrt[3]{2}=\sqrt[3]{-1}\sqrt[3]{4}\sqrt[3]{2}$
with the understanding that we interpret the cube-root of -1 as actually three values and $\sqrt[3]{2}$ and $\sqrt[3]{4}$ as the positive cube-roots.

 

[Martin Rattigan]

Curiouser and curiouser ...

Ok, then tell me why every single calculator I've used on this iPod (six at last count) gave me real and positive numbers for every square root of the product of two negative numbers that I put in? According to you, those numbers should be negative. Also, shouldn't math be consistent? So why can't I multiply the numbers in the radicand to produce a positive product, when I can with ANY other number under that root sign? In other words, why does multiplying the numbers in the radicand and then taking the square root (all allowed formal operations) produce a positive number, but your method produces a negative number?

I would ask you again to read my answer carefully.

According to that answer, when $a$ and $b$ are both negative $\sqrt{ab}$ is positive.

Specifically:

$\sqrt{ab}=-\sqrt{a}\sqrt{b}=-i\sqrt{-a}.i\sqrt{-b}=-i^2\sqrt{-a}\sqrt{-b}=-(-1)\sqrt{-a}\sqrt{-b}=+\sqrt{-a}\sqrt{-b}$

and the two square roots on the right are positive by definition.

When you ask if you can multiply two numbers in a radicand the answer is, "of course". What you can't do is assume the universal applicability of $\sqrt{ab}=\sqrt{a}\sqrt{b}$ with the standard definition of "$\sqrt{}$ ".

 

[Martin Rattigan]

. . . what a quagmire. Roots are multi-valued. $\sqrt{1}=\pm 1$. ...

Quagmire indeed. But defining roots to be in a Cheshire cat like state between one value and another will only make it more treacherous.

According to any respectable definition $\sqrt{1}=1$. The solutions of $x^2=1$ are $\pm 1$.

 

[Martin Rattigan]

How did you come up with this?

Directly from the answer I left for OP with a=-1 and b=-1.

The way I see it, and why I agree with Char.Limit is that $$\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$$ from $$\sqrt{a}\sqrt{b}=\sqrt{ab}$$

As Char.Limit himself was at pains to point out (correctly), the identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ that you quote is invalid. You can use the formula I posted for real numbers.

The problem arises from the order the process is undertaken. Repeating what wikipedia said that Char.Limit posted,

$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$$

But I would guess that Wikipaedia makes it clear that this is a fallacy. 1 and -1 really are not the same with their generally accepted meanings.

 

[jackmell]

You're looking only at the shadow when you take $\sqrt{1}=1$ when in reality, the (complex-analytic) function consists of two twisted surfaces in 3D space. Most (all?) of the functions you deal with in math are like that except in real calculus, we only concern ourselves with their cross-sections along one or more real axes which are often times, real-valued. Same way in differential equations. What I'd like to get across is if students would somehow begin to understand this more "global" definition of the functions in terms of their complex-analytic counterparts, then this quagmire would become less so. However I suspect some teachers would argue that just the real analysis is difficult to learn and injecting the complex part too soon would only empty the classrooms.

 

[Mark44]

. . . what a quagmire. Roots are multi-valued. $\sqrt{1}=\pm 1$. And if you carry this multivalued notion throughout the process, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for any values of a and b.

This thread seems to have diverged from the context in which the OP asked his question, which I believe was the context of the square root function defined on the reals, and not as a function defined on complex numbers, with branches and all that.

Assuming that a domain of the real numbers is the correct context for his question, roots are not multi-valued, and when we talk about the square root, cube root, etc. of a number we are talking about the principal square (cube, etc.) root of that number.

The OP's question was in essence whether the following equation is true for a and b both negative or a and b opposite in sign.
$$\sqrt{ab} = \sqrt{a}\sqrt{b}$$
Every college algebra/precalculus textbook I have ever seen places restrictions on a and b, namely, restricting them to a >= 0 and b >= 0. In this thread we have determined that the equation above is not true if a < 0 and b < 0. Kitami's response early in this thread shows a different formulation that can be shown.

For example,

$\sqrt{(-2)(4)}=\sqrt{-2}\sqrt{4}=\sqrt{-1}\sqrt{2}\sqrt{4}=\pm i 2\sqrt{2}$$\sqrt{(-2)(-4)}=\sqrt{-2}\sqrt{-4}=\sqrt{-1}\sqrt{2}\sqrt{-1}\sqrt{4}=\pm i 2(\pm i)\sqrt{2}=\pm 2\sqrt{2}$

even:

$\sqrt{(2)(4)}=\sqrt{2}\sqrt{4}=(\pm 2)(\pm \sqrt{2})=\pm 2\sqrt{2}$

I have never seen a single textbook that made a claim similar to the above. Can you name a textbook that does so?

you've really got to get a handle on this concept because this is just the lowly square root with just two values. What happens when we up the root or consider the log function or inverse trig functions which are really infinitely-valued (in the complex sense)? For example, can I:

$\sqrt[3]{(-4)(2)}=\sqrt[3]{-4}\sqrt[3]{2}$?
and just carry-over the multiple-valued $\sqrt[3]{-1}$ throughout the steps and write:
$\sqrt[3]{(-4)(2)}=\sqrt[3]{-4}\sqrt[3]{2}=\sqrt[3]{-1}\sqrt[3]{4}\sqrt[3]{2}$

Cube roots and odd roots in general are different from even roots such as the square root. The principal cube root of any real number is a real number. Same for 5th, 7th, and higher odd roots.

with the understanding that we interpret the cube-root of -1 as actually three values and $\sqrt[3]{2}$ and $\sqrt[3]{4}$ as the positive cube-roots.

The principal cube root of -1 is -1. There are two other cube roots of -1, but they are complex. Similarly, the principal cube root of 2 is a positive real number, as is the principal cube root of 4. The other two cube roots of these numbers or not negative, as I infer from what you wrote: they are complex.

I would ask you again to read my answer carefully.

You have said this several times, but given that your responses make up about a third of the ones in this thread, it would be helpful if you specified which post you are referring to.

 

[Char. Limit]

Not to mention the simple fact that to get your answer, Martin, you are using the property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for negative a and b, where it doesn't apply.

Let's assume positive a and b, because the notation gets confusing otherwise (with -a being positive and all that, too confusing). Here, as I see it, is your reasoning.

$$\sqrt{(-a)(-b)}=\sqrt{-a}\sqrt{-b}=i\sqrt{a}i\sqrt{b}=i^2\sqrt{ab}=-\sqrt{ab}$$

The problem is in the first step, where you wrongly apply the distributive property of square roots where it doesn't belong.

 

[Martin Rattigan]

Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.

For real $a$ and $b$, $(ab)^{0.5}=a^{0.5}b^{0.5}$ unless both $a$ and $b$ are negative, in which case $(ab)^{0.5}=-a^{0.5}b^{0.5}$.

You have said this several times, but given that your responses make up about a third of the ones in this thread, it would be helpful if you specified which post you are referring to.

The second quote above is the post to which I was referring. It is almost just a repetition of Xitami's post, which I believe is correct. (You posted that Xitami's post is incorrect, but my first post was intended to point out, among other things, that it's your response rather than Xitami's post that was wrong.)

I think my post above is all that should have been required for this thread.

 

[Martin Rattigan]

Not to mention the simple fact that to get your answer, Martin, you are using the property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for negative a and b, where it doesn't apply.

I don't believe so. Where?

Let's assume positive a and b, because the notation gets confusing otherwise (with -a being positive and all that, too confusing).

If the question specifies one or both of a and b are negative, and you want to refer to positive values, it would be much less confusing to introduce new variables, say c and d for the positive values to which you wish to refer.

Here, as I see it, is your reasoning.

$$\sqrt{(-a)(-b)}=\sqrt{-a}\sqrt{-b}=i\sqrt{a}i\sqrt{b}=i^2\sqrt{ab}=-\sqrt{ab}$$

The problem is in the first step, where you wrongly apply the distributive property of square roots where it doesn't belong.

No, that is not my reasoning. As you point out the first step is incorrect, but it's not my step.

 

[Mentallic]

No, that is not my reasoning. As you point out the first step is incorrect, but it's not my step.

Ok, fine. Please explain how you get this:

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

I have yet to see any variations of the distributive properties for square roots in any textbook that says,

For $a\geq 0, b\geq 0$ then $$\sqrt{a}\sqrt{b}=\sqrt{ab}$$

However $$\sqrt{-a}\sqrt{-b}=-\sqrt{a}\sqrt{b}=-\sqrt{ab}$$