How can I prove $2^x > x$ and $x \in \mathbb R$ I can prove this for the Natural numbers with induction. But I don't know how to do this for the real numbers $\Bbb R$.
They only thing I come up is proving 3 parts $x \leq 0$, $0 \lt x \lt 1$ and $x \geq 1$.
Please help, thanks in advance!
 A: Just define the function $f(x) = 2^x - x$. Notice that $f'(x)= \log (2)2^x-1$. Let's find where is this derivative equal to zero:
$$\log(2)2^x-1=0\implies 2^x=\frac{1}{\log(2)}\implies x=\log_2\left(\frac{1}{\log(2)}\right).$$
Now notice that this value is between $0$ and $1$ (because $\frac{1}{\log(2)}<2$ because $\log(2)>1/2$), and in our case, $f'(0)=\log (2)-1<0$, and $f'(1)=\log(2)2-1>0$, so we can conclude $x_0=\log_2\left(\frac{1}{\log(2)}\right)$ is an absolute minimum of $f$ in $\mathbb R$.
Finally, notice that evaluating $f$ in its minimum gives us a positive value, so necessarily the function is positive in every point:
$$f(x_0)=\frac{1}{\log(2)}-\log_2\left(\frac{1}{\log(2)}\right)>0,$$
because $\frac{1}{\log(2)}>1$ and $x_0\in(0,1)$ as we said before, and this proves then that
$$f(x)=2^x-x>0, \forall x\in\mathbb R,$$
hence $2^x>x$, $\forall x\in\mathbb R$.
A: For $x \leq 0$, the inequality is trivial. Thus we focus on $x > 0$. By taking $\log$, it suffices to prove that $x\log(2) > \log(x)$, or equivalently, that $f(x) = x \log(2) - \log(x) > 0$ for all $x > 0$. Note that $f(x) \to \infty$ as $x \to 0$ and $f(x) \to \infty$ as $x \to \infty$. Since $f$ is continuous, this implies $f$ attains a minimum at some point $x_0 \in (0, \infty)$. By Fermat's principle, $x_0$ satisfies $f'(x_0) = 0$. Solving for $x_0$ gives $x_0 = \frac{1}{\log(2)}$. We have $f(x_0) = 1 - \log(\frac{1}{\log(2)}) \approx 0.633 > 0$. Thus $f(x) > f(x_0) > 0$ for all $x \in (0, \infty)$.
A: For $x\le 0$, this is clearly true since $2^x$ is positive. For $x>0$, you can try with Bernoulli's inequality. That requires an integer exponent, so you can instead use $\lfloor x \rfloor$:
$$\begin{align}2^x &\ge 2^{\lfloor x\rfloor} \\
&=(1+1)^{\lfloor x\rfloor} \\ &\ge 1+\lfloor x\rfloor \\ &> x  \end{align}$$ Note the last line uses $\lfloor x\rfloor>x-1$.
A: If you know some basic properties of $2^x$ (positive, monotone increasing, $2^{a+b} = 2^a2^b$, $2^0=1$, $2^1=2$, and $2^2=4$) then this result follows pretty simply from the case when $x$ is a natural number.
Since $2^x > 0$ for all $x$, we have, in particular, that $2^x > x$ for all $x < 0$.
For $0 \leq x < 1$, we have $2^x \geq 2^0 = 1 > x.$
For $1 \leq x < 2$, we have $2^x \geq 2^1 = 2 > x.$
For $2 \leq x < 4$, we have $2^x \geq 2^2 = 4 > x.$
For $x \geq 4$, we have $$2^x \geq 2^{\lfloor x\rfloor} = 2(2^{\lfloor x\rfloor - 1}) > 2(\lfloor x\rfloor - 1) = \lfloor x\rfloor + (\lfloor x\rfloor -2) \geq \lfloor x \rfloor + 2 > x$$
In the last case, I use $2^{\lfloor x\rfloor - 1} > \lfloor x\rfloor - 1$, which is where I've used the result $2^n > n$ for natural numbers $n$.
