Real numbers: powers inequality Trying to prove the following inequality from textbook.


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*Let $x>1$ be a real number, and let $q>0$ be a rational number. Then $x^q>1$.

*Let $p<q$ be a rationals numbers, and let $x>1$ be a real number. Show that $x^p<x^q$.

 A: It depends on your premises. I will work with the followings definitions, supposing you know the exponentiation for integers and rationals:

Definition 1. Let $x>0$ be a positive real, and let $n\ge 1$ be a positive integer. We define $x^{1/n}:=\sup\{y\in\mathbf R:y\ge 0\text{ and }y^n\le x\}$.
Definition 2. Let $x>0$ be a positive real, and let $q$ be a rational number such that $q=a/b$ for some integer $a$ and positive integer $b$. We define $x^q:=(x^{1/b})^a$.
Lemma 3. Let $x>0$ and $y>0$ be real numbers, and let $n\ge1$ be a integer. We have $y=x^{1/n}$ if and only if $y^n=x$.
Lemma 4. Let $x>0$ and $y>0$ be real numbers, and let $n\ge1$ be a integer. We have $x>y$ if and only if $x^{1/n}>y^{1/n}$.
Lemma 5. Let $x>0$ be a positive real, and let $n\ge 1$ be a positive integer. Then $x^{1/n}$ is a positive real number.

Remark. In order to understand the proofs, you can prove the above lemmas from definitions and rational exponentiation. Any questions please ask.

Exercise 6. Let $x$ be a real number, and let $q>0$ be a positive rational. If $x>1$, then $x^q > 1$.
Proof. Suppose $x>1$ be a positive real number, and let $q$ be a rational such that $q=a/b$ for some integer $a$ and positive integer $b$. By Lemma 4, we have $x^{1/b}=1^{1/b}$. Using the Lemma 3, we can to obtain $1^{1/b}=1$, so $x^{1/b}>1$. Now, using our knowledge about rational exponentiation, we have $(x^{1/b})^a>1^a=1$, i.e., $x^q>1$, by Definition 2.

Exercise 7. Let $x>1$ be a positive real number, and let $p<q$ be rational numbers in the sense of Definition 2. Show that $x^p<x^q$.
Proof. Suppose for sake of contradiction that $x^p\ge x^q$. By Lemma 5, we can obtain $x^p/x^q\ge1$ (since $x^q$ is no zero), so we have $x^{p-q}\ge1$ (you can prove $x^p/x^q=x^{p-q}\ge1$ by Definition 1 and rational exponentiation). Since $p<q$ by hypothesis, we have $x^{-r}\ge1$ for some negative rational $-r:=p-q$. Suppose $-r:=(-a)/b$ (since $b\ge0$ by Definition 2). Then we have $x^{-r}=x^{(-a)/b}=(x^{1/b})^{-a}\ge1$ by Definition 2. Using the rational exponentiation knowledge, we have $(x^{1/b})^{-a}=1/((x^{1/b})^a)=1/x^r\ge1$ by Definition 2. By Lemma 5, we can obtain $x^r>1$, but from $1/x^r\ge1$ we obtain $x^r\le1$ (using Lemma 4 and rational exponentiation), a contradiction.
A: (i) Suppose $\alpha^r < 1$, then
$$
(\alpha^r)^{1/r} < 1^{1/r} \Rightarrow \alpha < 1.
$$
Since $\alpha > 1$ we have a contradiction. Therefore, $\alpha^r > 1$.
(ii) According to the previous item we have that
$$
\alpha^{b-a} > 1
$$
once that $b-a>0$. Thus
$$
\alpha^a < (\alpha^a)(\alpha^{b-a}) = \alpha^{a+b-a} = \alpha^b.
$$
