This is my first question here so sorry if I'm messing up the format, still trying to get the hang of this.
Given is: $a\in\mathbf{R}$ with $a\ge1$ and $A_n=\sqrt[n]{a}$. and you can infer that $a = (A_n)^n$
I need to prove a bunch of things about the sequence $A_n = \sqrt[n]{a}$ ($n$ th root of $a$), but I'm still struggling with the first question still... For the first part I need to show that $A_n≥1$ for all positive integers.
I think I need to show this by induction. I've tested for $n = 1$. For assumption: Let $n = k$, then $\sqrt[k]{a} \ge 1$ is true for at least one positive integer $k$. Then I think what I need to prove is that also $\sqrt[k+1]{a} \ge 1$.
I thought the best thing to do would be to start with my assumption and to get there: $\sqrt[k]{a}\ge 1$. Multiply by $\sqrt[k]{a}$.
$$(\sqrt[k]{a})\cdot\sqrt[k]{a} \ge 1\cdot\sqrt[k]{a}$$
$$\sqrt[k+1]{a} \ge \sqrt[k]{a}$$
Can I argue here that since $\sqrt[k]{a} \ge 1$ (Per assumption) then by transitivity is also $\sqrt[k+1]{a}\ge 1$?
That's where I got, and I'm not even sure if half the stuff I did is correct, or if I could even do it... I don't really understand the rules for $n$ th roots that well to be honest and we just recently started with induction, so I still lack practice with it, especially for inequalities.
Hope someone can point me to the right direction.
Thanks!
$\sqrt[k]{a}$ is the same thing as $a^{\frac{1}{k}}$, so $\sqrt[k]{a} \cdot \sqrt[k]{a} = a^{\frac{1}{k}} \cdot a^{\frac{1}{k}} = a^{\frac{1}{k}+\frac{1}{k}} = a^{\frac{2}{k}}$.
A hint on where to go would be to start at your assumptions and recall what $1^\frac{1}{x}$ equals. Also, you don't necessarily need induction to prove this. If you are studying induction right now, it's likely that a later part of the question is where induction becomes important.