Question

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

Answer 1

Following are the answer to this question:

Explanation:

Given value:

`\to x > 0`

`\to S= { n\varepsilon N : x^n \leq 0 } \\\\ s \\eq \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 \\`