# Distributional property of $n$-th root extraction respect to the multiplication

I have seen this exercise in a old book of the 2011. It is this with $$a,b\in\Bbb R$$.

$$\large \sqrt{a^5b^2}=\sqrt{a^5}\cdot \sqrt{b^2}=\sqrt a \cdot \sqrt{|b|}. \tag 1$$

For my opinion it is wrong infact must be:

$$\large \sqrt{a^5b^2}=\sqrt{a^5}\cdot \sqrt{b^2}=\sqrt{|a|} \cdot \sqrt{b}. \tag 2$$ Roots with odd index are defined everywhere. The absolute value does not go on $$b$$ but not on $$a$$ because the square root must make sense.

• Have you considered a relatively simple case like $a = 1$ and $b = -1$? Sep 21 at 7:51
• With the $\sqrt{1} = -1$ case, take both sides to the $5$'th power instead to get $\sqrt{1}=-1$. Note I don't want to basically answer your question within the comments. Instead, these basic examples are just to get you thinking about the consequences of your suggestion, and to consider whether or not it makes sense. As such, I'm not going to comment any further here about them. Sep 21 at 8:03
• In order for the initial expression to make sense we must have $a\geq 0$ and $b$ can be arbitrary, both positive or negative. The first transformation takes all this into account. Your transformation does not. Sep 21 at 8:14
• Because $\sqrt{b^2}$ is not equal to $\sqrt{b}$. For example, they are not equal when $b=-1$. However $\sqrt{b^2}=\sqrt{|b|}$ always. It is the same reason why $\sqrt{b^2}=|b|$ and not just $b$. We treat $\sqrt{}$ as a function defined in some standard way we all agreed on, for example for positive inputs it returns the positive root if there are two to choose from etc. Sep 21 at 11:25
• @JohnOmielan I have undestood my mistake. Thank you very much. Sep 21 at 11:45

The correct steps are those given in $$(1)$$

$$\sqrt{a^5b^2}=\sqrt{a^5}\cdot \sqrt{b^2}=\sqrt a \cdot \sqrt{|b|}$$

instead of the $$(2)$$ for the following reasons.

Remembering that $$a,b\in\Bbb R$$ we consider $$\sqrt{a^5b^2}$$. Since we have a root with an even index $$10$$ it necessarily needs to be $$a\geq 0$$. For $$b$$ into the $$\sqrt{\vphantom{b}\ }$$, $$b$$ being squared we not have any problem. This fact it is important to be able to apply distributional property of $$n$$th root extraction respect to the multiplication. We have

$$0\leq \sqrt{a^5b^2}=\sqrt{a^5}\cdot \sqrt{b^2}$$

Being $$a\geq 0$$ then $$\sqrt{a^5}=\sqrt{a}$$ and no absolute value is required. For $$b\in\Bbb R$$ we have

$$\sqrt{b^2}\geq 0$$ and must be $$\sqrt{|b|}\ge 0$$ necessarily with absolute value in order to have concordance of the sign of the two radicals $$\sqrt{b^2}$$ and $$\sqrt{|b|}$$, where the first radical is positive or null. If I did not put the absolute value at $$b$$, and wrote $$\sqrt{b}$$, this radical can be positive, negative or null and we would have the contradiction

$$0\leq \sqrt{b^2}=\sqrt{b}<0$$

Another example can be $$\sqrt{(x-1)^2}$$ where the existence field is defined for each $$\Bbb R$$, Being $$\sqrt{(x-1)^2}\geq 0$$ then by applying the invariant property we must $$\sqrt{(x-1)^2}=\sqrt{|x-1|}\geq 0$$, absolute value is necessary because the concordance of the initial sign must be preserved.

The invariantive property like other properties of radicals can be applied to simplify radicals if the base of the radical is positive or null; if it is negative, sign concordance could be lost. For example $$\sqrt{(-2)^6}\neq \sqrt{(-2)^3}$$, in fact the first radical is positive while the second is negative. Instead $$\sqrt{(-2)^3}=\sqrt{-2}$$ because in this case the concordance of the sign is preserved, in fact although the base is negative, the exponent remains odd, preserving the sign of the base. If the radical has a negative base and in the simplification its exponent changes from even to odd, it is necessary to put the absolute value. Example $$\sqrt{(-2)^6}=\sqrt{|-2^3|}\geq 0$$. If the radical is literal the same procedure is followed: whenever studying the sign of the radical we find that the base may be negative, if the exponent of the radical changes from even to odd, we put the modulus to ensure concordance of the sign $$\sqrt{x^6}=\sqrt{|x|^3}\geq 0$$.

Thank you very much for the comments of the users.