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By definition, $x^1$=$x$ and $\sqrt[1]{x}$ = $x$

My own personal understanding of this rule is just rote memory, so if you were to ask me to explain why I wouldn't know what to say. For example does $9^1 = 9\cdot1$ ? Intuitively I would say yes because it gets the desired result, but somehow I doubt that's what it means.

Never mind the second definition with the first root. Absolutely no idea how that works (besides memorizing it for calculations of course).

Any explanation/proof appreciated. Thanks.

dps1212
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  • I have never seen the notation $\sqrt [1] x$ before. As to $x^1$, since $x^n= \underbrace{x\times \cdots \times x}_{\text {n times}}$, for any natural number $n$, we get the desired result just by letting $n=1$. Otherwise said, a good way to define $x^n$, for $n\in \mathbb N$, is to recursively define $x^1=1$ and $x^{n+1}=x\times x^n$ for $n≥1$. – lulu May 07 '20 at 18:45

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I think both are true by definition, since $\sqrt[1]{x}=x^{\dfrac{1}{1}}=x^{1}$ and: (from Wikipedia) enter image description here

BinyaminR
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