+0

0
37
1
+170

I have an answer to this question: https://web2.0calc.com/questions/interesting-question_7

The answer is yes; we proceed by induction:

First, note that we must have $$\{x^n\} + \frac1{x^n} = 1$$ is encompassed by the more general statement $$x^n+\frac1{x^n}\in \mathbb Z$$, which we will prove. We have our base case(s) $x+\frac1x\in \mathbb Z$ and $$x^0 + \frac1{x^0}=2\in \mathbb Z$$.

Suppose that $$x^k+\frac1{x^k} \in \mathbb Z$$ and $$x^{k-1} + \frac1{x^{k-1}} \in \mathbb Z$$. We multiply $$x^k+\frac1{x^k}$$ by $$x+\frac1x$$ to get

$$\left(x^k+\frac1{x^k}\right)\left(x+\frac1x\right)=\left(x^{k+1}+\frac1{x^{k+1}}\right) + \left(x^{k-1}+\frac1{x^{k-1}}\right)$$

Note that the LHS is an integer, so $$x^{k+1} + \frac1{x^{k+1}}$$ is an integer as well, and we are done.

Aug 14, 2023

#1
+121
-1

It seems like you've provided a proof by induction for the statement that if $$n$$ and $$n+1$$ are encompassed by the conditions of the problem, then $$n^2 + n$$ is also encompassed by the conditions.

Your provided proof outline appears to be correct. You've established the base case for $$n=1$$ and $$n=2$$, and then you assumed that the statement holds for $$n$$ and $$n+1$$ and proved that it holds for $$n^2+n$$ as well. This indeed proves that if the conditions are satisfied for $$n$$ and $$n+1$$, then they are satisfied for $$n^2+n$$ as well.

The logic you've presented follows the structure of a valid proof by induction. Well done! If you have any further questions or need clarification, feel free to ask.

Aug 14, 2023