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

`\frac17 = \frac18 + \frac1{8^2} + \frac1{8^3} + \cdots`

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

`\frac17 \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

`\frac17 = \frac2{16} + (2^2)/(16^2) + (2^3)/(16^3) + (2^4)/(16^4) + (2^5)/(16^5) + (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.

`(2^n)/(16^n) = (2^4*2^(n-4))/(16^n) = (2^(n-4))/(16^(n-1))`

That is,

`(2^4)/(16^4) = \frac1{16^3}`

`(2^5)/(16^5) = \frac2{16^4}`

`(2^6)/(16^6) = \frac4{16^5}`

etc. so that

`\frac17 = \frac2{16} + \frac4{16^2} + \frac9{16^3} + \frac2{16^4} + \frac4{16^5} + \frac9{16^6} + \cdots`

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

`\frac17 \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}.