Prove inequalities relating a real number x to integer powers of x

Prove that for $n \geq 2$ ,

$x^n> x \qquad \text{if } x > 1$

and

$x^n < x \qquad \text{if } 0 < x < 1.$

Proof #1. Assume $x > 1$ . Then, if $n = 2$ we have

$x > 1 \ \implies \ x \cdot x > 1 \cdot x \ \implies \ x^2 > x.$

Where $x > 1$ implies $x > 0$ so multiplying both sides by $x$ preserves the inequality. So, the statement is true for $n =2$ . Assume then that the statement is true for some $n = k \in \mathbb{Z}_{\geq 2}$ . Then,

$\begin{align*} x^k > x &\implies& x^k \cdot x &> x\cdot x \\ &\implies& x^{k+1} &> x^2 \\ &\implies& x^{k+1} &> x & (\text{Since } x^2 > x.) \end{align*}$

Thus, the statement holds for $k+1$ ; and hence, for all $n \in \mathbb{Z}_{> 0}. \qquad \blacksquare$

Proof #2. Assume $0 < x < 1$ . Then, for the case $n=2$ we have

$x < 1 \ \implies \ x^2 < x$

since $0 < x < 1$ still allows us to preserve the inequality after multiplying both sides by $x$ . So, the inequality holds for the case $n=2$ . Assume then that it holds for some $n = k \in \mathbb{Z}_{\geq 2}$ . Then,

$\begin{align*} x^k < x &\implies& x^k \cdot x & < x\cdot x \\ &\implies& x^{k+1} &< x^2 \\ &\implies& x^{k+1} &< x &(\text{Since } x^2 < x). \end{align*}$

Thus, the inequality holds for $k+1$ ; and hence, for all $k \in \mathbb{Z}_{\geq 2}. \qquad \blacksquare$

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Prove inequalities relating a real number x to integer powers of x

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