Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element
Jump to navigation
Jump to search
This article is complete as far as it goes, but it could do with expansion. In particular: we want to prove that for n at least 2, the inequality is strict. The only essential change for that is the base case for induction. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Expand}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Theorem
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
- $\circ^n x \ge x$
Proof
The proof proceeds by induction:
If $n = 1$, then $\circ^n x = x$.
So:
- $\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
- $\circ^n x > 1$
By Product of Positive Element and Element Greater than One:
- $x \circ \paren {\circ^n x} > x$
Hence:
- $\circ^{n + 1} x \ge x$
$\blacksquare$