Math Help - Prove:

1. Prove:

For any integer $n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1

2. Originally Posted by newtoinequality
For any integer $n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
Hi

It can be done by induction
The property is true for n=2 (16 = 2^4)

Suppose that it is true for n. This means that there exists p integer such that $10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

You have to demonstrate that there exists q integer such that $10^{n} \leq 2^q < 2 \cdot 10^{n}$

3. Originally Posted by running-gag
Hi

It can be done by induction
The property is true for n=2 (16 = 2^4)

Suppose that it is true for n. This means that there exists p integer such that $10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

You have to demonstrate that there exists q integer such that $10^{n} \leq 2^q < 2 \cdot 10^{n}$
You mean $10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?

4. Originally Posted by newtoinequality
For any integer $n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
First, show existence. If the statement were false, then there would exist $m,\,n$ such that $2^m<10^{n-1}$ and $2^{m+1}\ge2\cdot10^{n-1}.$ But the former gives $2^{m+1}<2\cdot10^{n-1}$ which is a contradiction of the latter.

Hence for each $n$ there must be $m$ such that $10^{n-1}\le2^m<2\cdot10^{n-1},$ i.e.

$2^m\ <\ 2\cdot10^{n-1}\quad\ldots\ \fbox1$

$2^m\ \ge\ 10^{n-1}\quad\ldots\ \fbox2$

To prove uniqueness, observe that $2^{m-1}<10^{n-1}$ (from $\fbox1)$ and $2^{m+1}\ge2\cdot10^{n-1}$ (from $\fbox2).$ The result follows since if $m'\ne m$ then either $2^{m'}\le2^{m-1}$ or $2^{m'}\ge2^{m+1}$ and so $2^{m'}$ cannot be between $10^{n-1}$ and $2\cdot10^{n-1}.$

5. Originally Posted by HallsofIvy
You mean $10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?
No

For instance $10^3 \leq 2^{10} < 2 \cdot 10^3$