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My question is simple. I'm teaching in a college (pre-calculus course) and I'm asking myself why high school books say $\sqrt{xy}=\sqrt x\sqrt y$. This is false (counterexample is $x=y=-1$).

How can I state this rule in a more general and correct manner?

Thanks

Rafael Chavez
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    Any book that does not say that the equality is true only for $x,y>0$ is wrong. – 5xum Sep 21 '15 at 11:40
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    I assume that those books mention that $x,y \ge 0$. – Hetebrij Sep 21 '15 at 11:41
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    @5xum In fact, we need only one of the variables being positive, right? – Rafael Chavez Sep 21 '15 at 11:42
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    On the other hand, in a high-school context (where only nonnegative reals have square roots), the equality is true whenever both sides are defined. – hmakholm left over Monica Sep 21 '15 at 11:42
  • And with only the assumption that $xy\ge 0$ they should say $\sqrt{xy}=\sqrt{|x|}\sqrt{|y|}$. Anyway, as a teacher you can explain to your students that what's in the books is wrong unless $x,y\ge 0$. – Scientifica Sep 21 '15 at 11:42
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    @RafaelChavez Since the square root function is usually only defined for positive numbers, we need both of them to be positive. Remember, $i=\sqrt{-1}$ is not true. It is only true that $i^2=-1$, but since $-i$ is also a "square root" of $-1$, you don't know which one to choose. – 5xum Sep 21 '15 at 11:44
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    To sum up, because, as every mathematical statement, this one comes with some hypothesis about the objects it applies to (in the present case, that $x$ and $y$ are two nonnegative real numbers). – Did Sep 21 '15 at 11:45
  • @5xum so even in a more general context (complex analysis context) I can't define $\sqrt {-1}$. Can I say that? – Rafael Chavez Sep 21 '15 at 11:48
  • @RafaelChavez Not without some complication, no. Because the square root function, no matter how you define it, will have discontinuities. Look up "branch cut". – 5xum Sep 21 '15 at 11:49

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$\sqrt{x}:\mathbb{R}_{\ge0}\to\mathbb{R}_{\ge0}$ is defined as the unique $y\in\mathbb{R}_{\ge0}$ so that $y^2=x$ where $\mathbb{R}_{\ge0}=\left\{x\in\mathbb{R}:x\ge0\right\}$.

For this function, and this should be the function in your book, $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

Unfortunately, after complex numbers are introduced, sometimes writers define an inverse of $z\to z^2$ on $\mathbb{C}$ minus some branch cut and call it $\sqrt{z}$. However, for this function, it is not the case that $\sqrt{zw}=\sqrt{z}\sqrt{w}$.

robjohn
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  • However, it is clear that $\sqrt{z} \sqrt{w} = \pm \sqrt{zw}$. So $\sqrt{z} \sqrt{w}$ is some square root of $zw$, but not always the "chosen one" – spin Sep 21 '15 at 12:24
  • do you want to say that some writers do not define it as $\sqrt z$? – Nikolay Gromov Sep 21 '15 at 12:51
  • @spin: There is no canonical definition for $\sqrt{z}$ on $\mathbb{C}$; but certainly, for any plausible definition of a function that calls itself $\sqrt{z}$ one would have $\sqrt{zw}=\sqrt{z}\sqrt{w}$ for some $z$ and $w$ and $\sqrt{zw}=-\sqrt{z}\sqrt{w}$ for others. This is simply because $z^2-w^2=(z+w)(z-w)$ still has the same factorization over $\mathbb{C}[z,w]$. – robjohn Sep 21 '15 at 14:19
  • @NikolayGromov: some don't. Often, writers will leave most of the complication to the $\log$ function, defining its branches and cuts. Then they define $z^{1/2}=e^{\frac12\log(z)}$. I personally don't like to use $\sqrt{z}$ for complex arguments since it has such specific meaning on $\mathbb{R}_{\ge0}$ and students learn formulas like $\sqrt{xy}=\sqrt{x}\sqrt{y}$. – robjohn Sep 21 '15 at 14:22
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Almost every mathematical statement may become wrong/absurd without its proper context.

For instance, the apparently harmless $$ (xy)^2 = x^2 y^2 $$ might not hold if $x$ and $y$ are two matrices or two differential operators.

So the "right way" is just to be clear about the context, just like in everyday's life.

Jack D'Aurizio
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In general this assumes $\arg(x) + \arg(y) \leq \pi || \arg(x) + \arg(x) \geq 3\pi || \arg(x) \leq \pi {\;\rm and\;} \arg(y) > \pi || \arg(x) > \pi {\;\rm and\;} \arg(y) \leq \pi $

Hope I am not missing any particular case. I think someone should once and forever give the most general domail of the applicability of the $\sqrt{ x y}=\sqrt{x}\sqrt{y}$ formula! Please don't down-vote, that's not fare...

Nikolay Gromov
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Corrected:

If $x$ and $y$ are allowed to be complex, your proposed counterexample is no such thing:

When $x = -1$ and $y = -1$,

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

and

$$\sqrt x\sqrt y = \sqrt {-1} \sqrt {-1} = (\pm i) (\pm i) = \pm (-1) = \pm 1 $$

So the high-school text books are more correct than they seem to be.

The above interprets $ \sqrt x $ as a multivalued function. If we take $\sqrt x$ as the principle value, the contention is, as you say, false.

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