Math Induction?

Prove by math induction that if x > 1, then x^n > 1.

Relevance
• 9 months ago

You didn't state the domain on n. I can already find a counter example with negative values of n.

For example:

x = 2, n = -1

x^n = 2^(-1) = 1/2 < 1

Perhaps you meant to show this for the natural numbers, n = {1, 2, 3, ...}?

BASE CASE:

n = 1, then x^1 = x and we are told that x > 1, so x^1 > 1

INDUCTION STEP:

Assume it is true that x^k > 1.

When we multiply this by x, we have:

x^(k+1) = x^k * x

Multiplying a positive value by a number larger than 1 will always make it bigger. Since x^k > 1, x^(k+1) must also be bigger than 1.

We've shown it for the base case of n=1. Then we've shown that if it is true for k, it is true for k+1. So by the induction step, we can say it is true for all natural numbers n = {1, 2, 3, ...}

Q.E.D.

• JOHN
Lv 7
9 months ago

Suppose result is true for n = m.

Then x^m > 1

Since x > 1 > 0, x ^ (m + 1) = x(x^m) > x > 1

So result is true for n = m

implies result is true for m + 1

Result is true for n = 1 since x > 1

So by induction result is true for all n.