Chapter 12.4, Problem 24AYU

To determine

To prove: The given statement $0 < x < 1$, then $0 < x^n < 1$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 24AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the series $0 < x < 1$, then $0 < x^n < 1$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

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Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$0 < x < 1$, then $0 < x^n < 1$   -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is $x^1 = x$ which is obvious from the statement. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $0 < x < 1$, then $0 < x^k < 1$   -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that $0 < x < 1$, then $0 < x^{k + 1} < 1$

Consider $x^{k + 1} = x^k. x^1$

In here, $0 < x^k < 1$ and $0 < x < 1$. The two numbers which are between 0 and 1, then their product must be between 0 and 1.

Hence $0 < x^{k + 1} < 1$.

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.