SImpe question about square root function

[jgens]

Suppose that $a<0$ and $b<0$, in which case $ab>0$ 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 $a<0$ and $b<0$, 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 $a,b<0$, 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.