Show that $\exists x \in \Bbb{R}$ such that $2^x< y< 2^{x+\epsilon}$.

Question

Let $y>1$ and $\epsilon > 0$. Show that $\exists x \in \Bbb{R}$ such that $2^x< y< 2^{x+\epsilon}$. Build the logarithm on base 2.

So, I tried so many things to start, but nothing works. :( I just can use sequences proprieties and things like that.

Answer 1

Let $\epsilon>0$ and $y>1$

For $x=\log_2{(y)}-\frac{\epsilon}{2}$

$x<\log_2{(y)} \Rightarrow 2^x<y$

Also $\log_2{y}=x+\frac{\epsilon}{2}<x+\epsilon$

Thus $y<2^{x+\epsilon}$

Answer 2

Consider the sequence $(x_n)_{n \in \mathbb{N}}$ given by $x_n = \frac{\epsilon (n-1)}{2}$ (with the convention that $\mathbb{N} = \{1,2,\cdots\}$. By assumption, $2^{x_1} = 1 < y$, and since the sequence increases without bound there is a largest $n$ such that $2^{x_n} < y$.

Can you complete the proof by showing that $x = x_n$ has the desired property $2^x < y < 2^{x+\epsilon}$?

Answer 3

It suffices to solve the equation

$$y={2^x+2^{x+\epsilon}\over2}$$

for $x$. The value that you get from this is

$$x=\log_2\left(2y\over1+2^\epsilon\right)$$

Note, this is not the only value that satisfies the inequalities $2^x\lt y\lt2^{x+\epsilon}$, it's just one such value.