Proofs/explanations of root rules

jpanknin

Junior Member
Joined
Jan 8, 2020
Messages
108
I'm looking for a really intuitive explanation of why the radical rules work. I've gone through this process with exponent rules (like those below) and they make complete intuitive sense to me after working them out. I've also been able to find a lot of resources online about exponent rules (many provided by people on this forum - thank you), but not much at all about radicals/roots.

These make sense:

[math](\frac{a}{b})^2 = (\frac{a}{b})*(\frac{a}{b}) = \frac{a*a}{b*b} = \frac{a^2}{b^2}[/math]
or

[math](ab)^2 = (ab)*(ab) = (a*a*b*b) = (a*a)(b*b) = a^2b^2[/math]
But can't find much on ones such as:

[math]\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}[/math]
or

[math]\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]a}{\sqrt[n]b}[/math]
My book gives the rules, but not explanations for WHY these things are true. Any explanations or even links to good outside sources would be tremendously appreciated.
 
The proofs of what you are asking for actually requires a lot of mathematical background to understand them.
Basically at the level of basic algebra you should understand why the rules work for integer exponents and accept them for non-integer exponents.
 
I'm looking for a really intuitive explanation of why the radical rules work. I've gone through this process with exponent rules (like those below) and they make complete intuitive sense to me after working them out. I've also been able to find a lot of resources online about exponent rules (many provided by people on this forum - thank you), but not much at all about radicals/roots.

These make sense:

[math](\frac{a}{b})^2 = (\frac{a}{b})*(\frac{a}{b}) = \frac{a*a}{b*b} = \frac{a^2}{b^2}[/math]
or

[math](ab)^2 = (ab)*(ab) = (a*a*b*b) = (a*a)(b*b) = a^2b^2[/math]
But can't find much on ones such as:

[math]\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}[/math]
or

[math]\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]a}{\sqrt[n]b}[/math]
My book gives the rules, but not explanations for WHY these things are true. Any explanations or even links to good outside sources would be tremendously appreciated.
If we agree that a) [imath](ab)^n = a^n b^n[/imath], and if we define [imath]a^{1/n}[/imath] as inverse of [imath]a^n[/imath], i.e. [imath](a^{1/n})^n = a[/imath] then [imath](a^{1/n} b^{1/n})^n = ab[/imath] and thus [imath]a^{1/n} b^{1/n} = (ab)^{1/n}[/imath].
 
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