Question

What are the first 10 digits after the decimal point (technically the hexadecimal point...) when the fraction frac17 is written in base 16?

Answer 1

We happen to have

$\dfrac17 = \dfrac18 + \dfrac1{8^2} + \dfrac1{8^3} + \cdots$

which is to say, the base-8 representation of 1/7 is

$\dfrac17 \equiv 0.111\ldots_8$

This follows from the well-known result on geometric series,

$\displaystyle \sum_{n=1}^\infty ar^{n-1} = \frac a{1-r}$

if $|r|<1$. With $a=1$ and $r=\frac18$, we have

$\displaystyle \sum_{n=1}^\infty \frac1{8^{n-1}} = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac1{1-\frac18} = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac87 = 1 + \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots \\\\ \implies \frac17 = \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots$

Uniformly multiplying each term on the right by an appropriate power of 2, we have

$\dfrac17 = \dfrac2{16} + \dfrac{2^2}{16^2} + \dfrac{2^3}{16^3} + \dfrac{2^4}{16^4} + \dfrac{2^5}{16^5} + \dfrac{2^6}{16^6} + \cdots$

Now observe that for $n\ge4$, each numerator on the right side side will contain a factor of 16 that can be eliminated.

$\dfrac{2^n}{16^n} = \dfrac{2^4\times2^{n-4}}{16^n} = \dfrac{2^{n-4}}{16^{n-1}}$

That is,

$\dfrac{2^4}{16^4} = \dfrac1{16^3}$

$\dfrac{2^5}{16^5} = \dfrac2{16^4}$

$\dfrac{2^6}{16^6} = \dfrac4{16^5}$

etc. so that

$\dfrac17 = \dfrac2{16} + \dfrac4{16^2} + \dfrac9{16^3} + \dfrac2{16^4} + \dfrac4{16^5} + \dfrac9{16^6} + \cdots$

and thus the base-16 representation of 1/7 is

$\dfrac17 \equiv 0.249249249\ldots_{16}$

and the first 10 digits after the (hexa)decimal point are {2, 4, 9, 2, 4, 9, 2, 4, 9, 2}.