Root is Commutative
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Theorem
Let $x \in \R_{> 0}$ be a (strictly) positive real number.
Let $a$ and $b$ be nonzero integers.
Then:
- $\sqrt [a] {\sqrt [b] x} = \sqrt [b] {\sqrt [a] x}$
Proof
Let $y = \sqrt [a] {\sqrt [b] x}$.
Then
| \(\ds \sqrt [a] {\sqrt [b] x}\) | \(=\) | \(\ds y\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [b] x\) | \(=\) | \(\ds y^a\) | Definition of Root of Number | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \paren {y^a}^b\) | Definition of Root of Number | ||||||||||
| \(\ds \) | \(=\) | \(\ds y^{a b}\) | Product of Indices of Real Number | |||||||||||
| \(\ds \) | \(=\) | \(\ds y^{b a}\) | Integer Multiplication is Commutative | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {y^b}^a\) | Product of Indices of Real Number | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [a] x\) | \(=\) | \(\ds y^b\) | Existence and Uniqueness of Positive Root of Positive Real Number | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [b] {\sqrt [a] x}\) | \(=\) | \(\ds y\) | Existence and Uniqueness of Positive Root of Positive Real Number | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [b] {\sqrt [a] x}\) | \(=\) | \(\ds \sqrt [a] {\sqrt [b] x}\) | Definition of $y$ |
$\blacksquare$