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Let $A=[0,\infty)\subseteq \mathbb{R}$ , and $n \in \mathbb{N}$.

a) Prove that for any $x\in A$ we have $0 \leq x < (1+x)^n$

b) Use the Intermediate Value Theorem to deduce that for all $x\in A$ there exists $z\in A$ such that $z^n=x$

c) Prove that this $z$ is unique by proving that if $z, z' \in A$ and $z^n=(z')^n$, then $z=z'$

I'm thinking maybe induction for part a? For part b, we haven't done anything using IVT yet, just looked at the proof of it briefly so I am not completely sure how to implement it in a proof.

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  • $\begingroup$ For part a, there are two cases: Case 1: If $x\leq1$, you can easily show that $(1+x)^n>1$. Case 2: If $x>1$, then $x^n>x$... Part b is literally just the definition of IVT. Part c is just algebra. $\endgroup$ – Don Thousand Mar 23 at 19:42
  • $\begingroup$ @Don: you don’t have to split $a$ into cases. We always have $x<1+x$ and $1\leq1+x$. For the latter inequality, multiplying both sides by $1+x$ yields $1+x\leq(1+x)^2$, etc. Thus, for any $n\in N$, we have $x<(1+x)^n$. $\endgroup$ – Clayton Mar 23 at 19:58
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    $\begingroup$ @Clayton True! I just thought that the casework wasn't hard and it's pretty intuitive. $\endgroup$ – Don Thousand Mar 23 at 20:00
  • $\begingroup$ @DonThousand Part c is not "just algebra". It's rather analysis. $\endgroup$ – amsmath Mar 23 at 20:31
  • $\begingroup$ @amsmath I didn't really know what to call it. Whatever it is, it follows pretty directly from the rest. $\endgroup$ – Don Thousand Mar 23 at 20:31
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For a) Consider $x \geq 0$, then for the Newton's Binomial Theorem $$ \begin{equation}(x + 1)^n = \sum_{i = 0}^{n}\binom{n}{i} x^{n-1}1^i=\sum_{i = 0}^{n - 2}\binom{n}{i} x^{n - 1} + nx + 1 > x. \end{equation}$$ For b) Just consider the clearly continuous function $f:A \longrightarrow A$ with $f(z) = z^n$, then because of Intermediate Value Theorem $\forall x \in A \hspace{.2cm} \exists z \in A: \hspace{.2cm} f(z) = z^n = x$. For c) you can simply suppose that $x^n = y ^n \Leftrightarrow (x^n)^{1/n} = (y^n)^{1/n} \Leftrightarrow x = y$ with $x,y \in A$.

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