Question

Use the Well-Ordering Principle to prove that given a > 0, a^n > 0 for every positive integer n

Answer 1

Answer:

Following are the answer to this question:

Step-by-step explanation:

Given value:

$\to x > 0$

$\to S= { n\varepsilon N : x^n \leq 0 } \\\\ s \neq \phi \\\\ \to x^n \leq 0$

$\to x^{n-1} x\leq 0\\\\ \to x>0 = x^{n-1} \leq 0 \\\\\to n-1 \varepsilon s \\ \ \ _{where} \ \ n-1 < n \\\\\to s= \phi \\\\\to \hence x^n > 0$