So I have this problem I don`t quite know how to prove completely with $P(k)>P(k+1)$ implication: $x$ is a real number with the property that $x + \frac1x$ is an integer, then prove that $x^n + \frac{1}{x^n}$ is an integer, $n \in \mathbb N$. Help?
Asked
Active
Viewed 47 times
0
-
How do you use induction on real numbers? – TheRandomGuy Feb 28 '16 at 06:10
-
The induction is on $n$ and not on $x$. $x$ is fixed. @DhruvSomani – SchrodingersCat Feb 28 '16 at 06:17
-
See http://math.stackexchange.com/questions/936479/proving-that-frac-phi4001-phi200-is-an-integer – lab bhattacharjee Feb 28 '16 at 07:42
1 Answers
1
The induction comes up easily:
First of all, as given, we have that $(x+\frac{1}{x})$ is an integer. So $P(1)$ is true.
Now, using strong induction, assume that $P(1), \ldots ,P(k)$ is true i.e. $(x+\frac{1}{x}), \ldots ,(x^k+\frac{1}{x^k})$ are all integers.
Thus, we can say that $$x^{k+1}+\frac{1}{x^{k+1}}=\left(x+\frac{1}{x}\right)\left(x^k+\frac{1}{x^k}\right)-\left(x\cdot\frac{1}{x}\right)\left(x^{k-1}+\frac{1}{x^{k-1}}\right)$$ $$=\color{blue}{\text{an integer}}$$
So $P(k+1)$ is true and the statement given is proved to be true by strong induction.
SchrodingersCat
- 24,472