Krdytkyu

I know how to convert from $769_{10}$ to base 8 that is $1401_8$ through dividing and remainder method but what is the method for $769.513_{10}$ to convert to octal?I know that it is to separate the integer part and the decimal place part that is $1401_8$ $bullet$ (answer from converting $0.513_{10}$ to octal).I just can't seem to work out the $0.513_{10}$ part to octal.

First of all, note that $lfloor8times.513rfloor=4$; that will your first digit.

On the other hand, $.513-4times8^{-1}=.513-.5=.013$. And $lfloor8^2times.013rfloor=0$: that will be your second digit.

Now, $lfloor8^3times.013rfloor=6$. So, your third digit will be $6$. And so on…

Actually, this process will never stop, since $.513=dfrac{513}{1,000}$ and this fraction is irreducible. In particular, it is not of the type $dfrac a{8^b}$, with $ainmathbb Z$ and $binmathbb{Z}_+$.

Hint:

$$.513=dfrac{513}{1000}=dfrac{a_1}8+dfrac{a_2}{8^2}+cdots$$ where $0le a_ile7$

$$implies513=125a_1+cdots$$

$implies a_1=leftlfloordfrac{513}{125}rightrfloor=?$

$513-500=dfrac{1000a_2}{8^2}+dfrac{1000a_3}{8^3}+cdots$

$implies a_1=leftlfloordfrac{(513-500)8^2}{1000}rightrfloor=?$

Can you take it from here?

So the question is to write
$$
x=0.513=0.513_{(10)}=frac{513}{1000}
$$
to octal. (If no base is written, numbers are considered in the usual decimal system. So $0.513$ is decimal, same for the numerator $513$, and denominator $1000$, decimal numbers.)

It is known that the result is a periodic number, there may be a part after the comma place which is "atypical", but then we repeat the period. First,
$$
frac 1{1000}=frac 18cdotfrac 1{125} .
$$
We have $frac {513}{125}=4+frac{13}{125}$. It will be enough to get the octal representation of $13/125$.

We now search for a number of the shape $8^k-1$ which is divisible by $125$.
We know (Euler, Euler indicator function) that $k=phi(125)=100$ does it, but it is a good idea to get the minimal $k$. Well, as it happens here, it is $100$. So we expect a period of this length! It is natural that i will use a computer to get it. Now
$$
begin{aligned}
frac 1{125}
&=
frac{(8^{100}-1)/125}{8^{100}-1}
=
Ncdotfrac{1}{8^{100}-1}text{ with }N=(8^{100}-1)/125inBbb N ,
\
&=
Ncdotfrac{1}{8^{100}}cdotfrac{1}{1-left(frac 18right)^{100}}
\
&=
Ncdot qcdotfrac{1}{1-q}text{ with } q=left(frac 18right)^{100} ,
\
&=
N(q+q^2+q^3+q^4+dots) .
end{aligned}
$$
The last number is simple, as follows: $q$ is the octal number $0,000dots1$, then $q^2$ is $0,000dots0 000dots1 $, and so on, the sum $q+q^2+dots$ is the periodic number
$0.(000dots1)$, where the parenthesis means "period". An then we multiply by $N$, thus $N$ "becomes the period", well, adjusted to $100$ digits as a "p hone number". In our case, we need instead $13/125$, so we use $13cdot N$. This is using sage

sage: 13*(8^100-1)/125

211851741538786552971918351594575328749352712941257370066158606732855655175387017443073327

in a decimal writing. Octal representation of it, again using sage:

sage: ''.join( [ str(digit) for digit in ZZ(13*(8^100-1)/125).digits(base=8)[::-1] ] )

'651767635544264162540203044672274324773716662132071260101422335136152375747331055034530040611156457'

So the final answer is
$$
begin{aligned}
0.513
&
=frac 18cdotfrac {513}{125}
=frac 18cdotleft(4+frac {13}{125}right)
\
&=
frac 18cdot
(4_{(8)}+
0.(065176763554426416254020304467227432
\&qquadqquad
477371666213207126010142233513615237574
\&qquadqquadqquad
7331055034530040611156457)_{(8)}
)
\
&=
frac 18cdot
4.(065176763554426416254020304467227432
\&qquadqquad
477371666213207126010142233513615237574
\&qquadqquadqquad
7331055034530040611156457)_{(8)}
\
&=
0.4(0651767635544264162540203044672274
\&qquadqquad
32477371666213207126010142233513615237
\&qquadqquadqquad
5747331055034530040611156457)_{(8)}
end{aligned}
$$
A small check for the first octal places after the comma:

sage: a - 4/8 - 6/8^3 - 5/8^4 - 1/8^5 - 7/8^6 - 6/8^7

61/131072000

sage: _ < 1/8^7

True