1
$\begingroup$

If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:

$\begin{align} 2^{-1} &= 0.5 \\ 2^{-2} &= 0.25 \\ 2^{-3} &= 0.125 \\ 2^{-4} &= 0.0625 \\ 2^{-5} &= 0.03125 \\ ... \end{align}$

It looks like it's as simple as saying $2^{-n} = 5^n \times 10^{-n}$, and when we calculate that out, it's correct:

$\begin{align} 5^1 \times 10^{-1} &= 5 \times 0.1 = 0.5 \\ 5^2 \times 10^{-2} &= 25 \times 0.01 = 0.25 \\ 5^3 \times 10^{-3} &= 125 \times 0.001 = 0.125 \\ 5^4 \times 10^{-4} &= 625 \times 0.0001 = 0.0625 \\ 5^5 \times 10^{-5} &= 3125 \times 0.00001 = 0.03125 \\ ... \end{align}$

I calculated this out for $n = [0, 10]$ and it works out, but I have no idea how to prove it fully.

$\endgroup$
1
  • 1
    $\begingroup$ The first thing you should do is to write out clearly exactly what it is you are trying to prove. Hint: try something like "for every positive integer $n$, the digits after the decimal point in $2^{-n}$ are. . ." $\endgroup$
    – David
    Jan 20, 2014 at 23:43

4 Answers 4

9
$\begingroup$

You want to prove that $$2^{-n} = 5^n \times 10^{-n}.$$ And

$$ 5^n10^{-n} = \frac{5^n}{10^n} = \left(\frac{5}{10}\right)^n = \left(\frac{1}{2}\right)^n = \frac{1}{2^{n}} = 2^{-n}. $$

$\endgroup$
0
1
$\begingroup$

$\dfrac{1}{2^x}= \dfrac{5^x\cdot2^x}{2^x\cdot10^x}=\dfrac{5^x}{10^x}$

$\endgroup$
0
1
$\begingroup$

$$10^x = (2\cdot 5)^x = 2^x\cdot 5^x$$

$\endgroup$
1
$\begingroup$

$\dfrac{1}{2} = \dfrac{5}{10}$ (or $\dfrac{2}{4}$, $\dfrac{3}{6}$, or $\dfrac{4}{8}$ for that matter)

Then just raise to the power of $n$ on both sides and use the fact that $a^{-n} = \dfrac{1}{a^n}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .