SImpe question about square root function

[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}

 

[jgens]

Suppose that a&lt;0 and b&lt;0, in which case ab&gt;0 so \sqrt{ab} exists and is real valued. Now since \sqrt{a} = i\sqrt{|a|} and likewise \sqrt{b} = i\sqrt{|b|}, it follows immediately that \sqrt{a} \sqrt{b} = - \sqrt{|a|} \sqrt{|b|} = - \sqrt{|a||b|}. Because |a||b| = ab, using the last equality, we have \sqrt{a} \sqrt{b} = - \sqrt{ab} as desired.

This is the same argument that Martin Rattigan has been pushing throughout the whole thread.

 

[Martin Rattigan]

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}

That's probably because the textbooks you're looking at don't define \sqrt{x} for negative x.

I used the definition \sqrt{x}=e^{\frac{1}{2}Log(x)} where Log(x) is the principal log of x. You can find this definition in e.g. Mathematical Analysis, Apostol, Addison Wesley, Chapter 1, and in any other analysis book that covers complex analysis.

For positive real numbers x it gives \sqrt{x}=y where y is the positive real number s.t. y^2=x. For negative real numbers x it gives i\sqrt{-x}.

It is then a simple matter to work out for yourself that for real a,b if a\geq 0 or b\geq 0 then \sqrt{ab}=\sqrt{a}\sqrt{b} otherwise \sqrt{ab}=-\sqrt{a}\sqrt{b}.

Please just try it.

 

[Mentallic]

Like I've already said, it depends on the order of operations.

Let a=b, then you can either look at it as \sqrt{a^2}=|a| or \sqrt{a}\sqrt{a}=\left(\sqrt{a}\right)^2=a. You're using the second, I've used the first. This is the discrepancy in dealing with the distributive law when it comes to negative numbers, which you've seemed to ignore and start dealing with definitions of complex numbers instead. This is another topic entirely.

You can't disagree that there is a discrepancy can you? Just because you've shown -1=-1 as you've expected, it has been clearly shown that using the property \sqrt{a}\sqrt{b}=\sqrt{ab} causes problems for a,b<0.

 

[jgens]

You can't disagree that there is a discrepancy can you? Just because you've shown -1=-1 as you've expected, it has been clearly shown that using the property \sqrt{a}\sqrt{b}=\sqrt{ab} causes problems for a,b<0.

No one here is claiming that if a&lt;0 and b&lt;0, then \sqrt{ab} = \sqrt{a} \sqrt{b}, so I struggle to understand your point.

Edit: And to clear up any confusion, could you point out where (in this thread) someone has made that claim recently. Further, if you dispute the proof that I've given that \sqrt{ab} = - \sqrt{a} \sqrt{b} if a,b&lt;0, then please point out the fallacious steps.

 

[Mentallic]

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}

Depending on the order of operations, one can say \sqrt{(-2)(-8)}=\sqrt{16}=4

or, incorrectly, \sqrt{(-2)(-8)}=\sqrt{-2}\sqrt{-8}=i^2\sqrt{2}\sqrt{8}=-4

It is then a simple matter to work out for yourself that for real a,b if a\geq 0 or b\geq 0 then \sqrt{ab}=\sqrt{a}\sqrt{b} otherwise \sqrt{ab}=-\sqrt{a}\sqrt{b}.

Now for Martin to come to his result, while it seems as though he used complex numbers to do so, it's the same as undertaking the second order of operation as I've shown.

 

[jgens]

Unless I'm misunderstanding what you mean by the "second order of operations," you're not reading Martin's formula correctly. Here's what Martin's formula says:

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

which is entirely true.

 

[Char. Limit]

Wait, the formula still gives the same answer?

See, this is why I hate confusing notation. I look at a and b and say that they're positive, because there's no negative sign. I do this subconsciously. So I looked at his formula and got a negative number at the end, and a positive number at the beginning. But see, here's why I don't like confusing notation like a or b being negative.

So, defining u=-a and v=-b, for a<0 and b<0, u and v are positive, and...

First, I think it can be proven that ab=uv under the conditions given. Then...

\sqrt{ab}=\sqrt{uv}=\sqrt{u}\sqrt{v}=\sqrt{-a}\sqrt{-b}=i\sqrt{a}i\sqrt{b}=i^2\sqrt{a}\sqrt{b}=-\sqrt{a}\sqrt{b}

Was that your reasoning? I used u and v as far as I could to avoid strange notation... And after I switched back, it didn't matter if a or b was negative or not, the root was already split into two positive radicands, u and v.

 

[Martin Rattigan]

Correct.

But note there is an implicit step omitted between

\dots=\sqrt{-a}\sqrt{-b}=i\sqrt{a}i\sqrt{b}=\dots

(which doesn't invalidate the proof).

For positive r, \sqrt{-r}=i\sqrt{r}, but for negative r, which a and b are here, \sqrt{-r}=-i\sqrt{r}.

So with the insertion of the omitted step it would read

\dots=\sqrt{-a}\sqrt{-b}=(-i\sqrt{a})(-i\sqrt{b})=i\sqrt{a}.i\sqrt{b}=\dots

 

[jgens]

Char. Limit: The notation isn't confusing; in fact, it's abundantly clear what's what. While you do run into problems if you mindlessly throw numbers into the equation, that doesn't necessarily mean that the problems are with the notation.

 

[Char. Limit]

Char. Limit: The notation isn't confusing; in fact, it's abundantly clear what's what. While you do run into problems if you mindlessly throw numbers into the equation, that doesn't necessarily mean that the problems are with the notation.

If it wasn't confusing, then why is it that I couldn't get what he was talking about until I created new notation, u=-a and v=-b, that were positive for negative a and b?

I didn't mindlessly throw numbers into the equation; in fact, I only inserted actual numbers once to prove a counterexample, and I was done in by the fact that I expect constants to be positive unless shown with a negative sign. By the way, can you name a textbook that says "b is a negative real number"? No, they always define the constants as either positive or arbitrary, and work with negative numbers using -b.

I'm not mindless.

 

[jgens]

I would call putting numbers into an equation while neglecting negative signs mindless. It's all clearly there, so unless you omit them somehow, you weren't doing what the equation says (and it's an extremely simple one at that). Call that what you like.

And to be quite honest, I don't really care what you have or have not seen in an algebra text. Unless you can dispute the proof I gave in post 36 (and please quote the specific lines you find troublesome) the fomula stands.