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

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.

• 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. – Don Thousand Mar 23 at 19:42
• @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 $x<(1+x)^n$. – Clayton Mar 23 at 19:58
• @Clayton True! I just thought that the casework wasn't hard and it's pretty intuitive. – Don Thousand Mar 23 at 20:00
• @DonThousand Part c is not "just algebra". It's rather analysis. – amsmath Mar 23 at 20:31
• @amsmath I didn't really know what to call it. Whatever it is, it follows pretty directly from the rest. – Don Thousand Mar 23 at 20:31

For a) Consider $$x \geq 0$$, then for the Newton's Binomial Theorem $$$$(x + 1)^n = \sum_{i = 0}^{n}\binom{n}{i} x^{n-1}1^i=\sum_{i = 0}^{n - 2}\binom{n}{i} x^{n - 1} + nx + 1 > x.$$$$ For b) Just consider the clearly continuous function $$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$$.