How to prove that $a^x>0$

Question

How to prove that $a^x>0$ ( Simple proof ) ?

My Try : There must be no answer for $a^x \le0 $ How to prove ?

Answer 1

$a^x=e^{x\ln(a)}>0$ is positive since the image of exponential is $>0$.

Answer 2

Note that by definition $\forall m,n \in\mathbb{N}$

$$e^\frac{m}{n}=\frac{\stackrel{m \, times}{e\cdot e\cdot ...\cdot e} }{\stackrel{n \, times}{e\cdot e\cdot ...\cdot e}}>0$$

then since $\forall x$ and $\forall \epsilon >0$ $\, \exists q_1=\frac{m_1}{n_1},q_2=\frac{m_2}{n_2}\in\mathbb{Z}$ such that $q_1\leq x \le q_2$ with $|q_i-x|<\epsilon$ and

$$e^{q_1}\le e^x \le e^{q_2}\iff e^\frac{m_1}{n_1} \le e^x\le e^\frac{m_2}{n_2} $$

it follows that $e^x>0$.

Answer 3

Starting from the base that $a\in \mathbb{R}^+$ and $x\in \mathbb{Z}^+$, then is by definition:

$$\tag1 a^n = 1 \cdot \underbrace{{ a\cdot a\cdot a\dotsm a }}_{x} > 0 \implies a^x>0$$ This follows $$\tag2 \text{for } x \in \mathbb{Z}^- \implies a^{-x} = (a^{-1})^x = \left(\frac1a\right)^x = \frac1{a^x} >0\implies a^x>0$$

$$\tag3 \text{for }x \in \mathbb{R} \implies a^{\frac{x}{y}} = \sqrt[y]{a^x} = (\sqrt[y]{a})^x > 0\implies a^x>0$$

$$\tag4 \text{for } x = i \in \mathbb{C}\implies a^{i} = e^{i\ln(a)} > 0 \implies a^x >0$$

Answer 4

At least you should be familiar with the fact that $a^x>0$ if $a>0,x\geq 0$. And if $x<0$ then $a^x=1/a^{-x}>0$.

Next we deal with the familiar fact mentioned above. If $a=1$ the claim is true (obvious). If $a>1$ then $a^x>1^x=1>0$. And if $0<a<1$ then $a^x=1/(1/a)^x>0$.

Needless to say all the above argument works by assuming fundamental properties of exponential functions and these can be established using algebra if $x$ is rational.