Prove that for $a,b,x \in \mathbb R_+$ and $a,b \ge 1$, then $a<b \implies a^x < b^x$.

Question

I have shown that this is true for $x \in \mathbb Q$. We define $a^x = \sup \{x^p : p \in \mathbb Q, p < x\}$. I let $S(a) = \{a^p: p \in \mathbb Q, p < x\}$. And so by definition. $a^x = \sup S(a), b^x = \sup S(b)$.

Suppose $a<b$.

Suppose $a^x > b^x$. Then, $\sup S(a) > \sup S(b)$. There must exist a $a^p, p \in \mathbb Q, p < x$ such that $a^p > \sup S(b)$. Else, $\sup S(b)$ would be an upper bound for $S(a)$ that is less than $\sup S(a)$. However, $b^p \in S(b)$ by definition of $S(b)$. But as the proposition is true for rational $x$, then $b^p > a^p$, which contradicts $a^p$ being an upper bound. Hence, $a^x \le b^x$.

From here, I can't find a way to derive a contradiction from $a^x = b^x$.

Answer 1

May I suggest the following approach: it would first be very helpful to establish the algebraic relation $(uv)^x=u^x v^x$, for any $u, v>0$ and $x \in \mathbb{R}$. This can be easily done from the definition, exploiting the fact that:

• for instance $u^x=\displaystyle\lim_{n \to \infty}u^{p_n}$ for any say strictly increasing sequence of rationals $p \in \mathbb{Q}^{\mathbb{N}}$ such that $p_n \xrightarrow[]{n \to \infty} x$ (fact which should not be too difficult to prove) and that:
• the relation $(uv)^q=u^qv^q$ is indeed valid for any rational exponent $q \in \mathbb{Q}$.

Once you have achieved this, you can write $b^x=\left(\frac{b}{a}\right)^xa^x$ and in order to prove your inequality it will suffice to show that for any $u>1$ and $x>0$ one has $u^x>1$. This should be rather straightforward from the definition.

Answer 2

The other answer is the way to go, but here is another approach, assuming you know that $b^{x+y}=b^xb^y$. And it has the benefit of giving a from scratch construction of a logarithm, together with nice applications of Bernoulli's inequality.

Toward a contradiction, it suffices to prove that if $b>1,\ y>0$ then there is a unique $x$ such that $y=b^x$, for then we will have $y=b^x=a^x=y'$ with $y\neq y'$ by uniqueness.

This proof is exercise $1.7$ in Baby Rudin:

First prove by induction the Bernoulli inequality: $b^n-1\ge b(n-1)$ for any interger $n$. This is easy.

Then, prove that if $t>1$ and $ n>\frac{b-1}{t-1}$ then $b^{1/n}<t$. This is not hard either, for in this case we have $n(t-1)>b-1\ge n(b^{1/n}-1)$ where the second inequality arises from subsituting $b^{1/n}$ for $b$ in Bernoulli's inequlality.

Setting $t=yb^{-w}$, it follows that if $w$ is such that $y>b^w$ then $y>b^{w+1/n}$ if $n$ is large enough. Similarly, if $y<b^w$ then $y<b^{w-1/n}$ for sufficiently large $n$.

Now define $x=\sup\{w:b^w<y\}$ and show by considering cases that $y=b^x$:

if $y>b^x$ then $x+1/n\in \{w:b^w<y\} $ for large $n$ and so $x$ cannot be an upper bound.

if $y<b^x$ then $x-1/n$ is an upper bound for $\{w:b^w<y\}$ for large $n$ and so $x$ cannot be the least upper bound.

Therefore, $y=b^x$

Uniqueness is a calculation: if $x'>x$ then $y=b^{x'}=b^{x+x'-x}=b^{x'-x}b^{x'}>b^{x'}=y$ (you should verify the last inequality!)