Base 7 expansion of $\frac{31}{32}$?

Question

I want to find the base $7$ expansion of this fraction $\left(\frac{31}{32}\right)_{10}$.

I normally do this by long division in the desired base. Which is no problem for fractions with single digit denominators and numerators like $\frac{5}{6}$ and things like that, but when it comes to $2$ digits this method is useless, because I don't know my $32$ times table in base $7$.

Is there an easier way to do this or will I just have to bite the bullet? I would preferably like to keep doing the long division way.

Answer 1

I thought, since I didn't see one, I would present an instance of long division in native base $7$:

$$ \require{enclose} \begin{array}{r} 0.6531\phantom{6} \\[-3pt] 44 \enclose{longdiv}{43.00000}\kern-.2ex \\[-3pt] \underline{36.0}\phantom{0000} \\[-3pt] 3.40\phantom{000} \\[-3pt] \underline{3.16}\phantom{000} \\[-3pt] 210\phantom{00} \\[-3pt] \underline{165}\phantom{00} \\[-3pt] 120\phantom{0} \\[-3pt] \underline{44}\phantom{0} \\[-3pt] 430 \end{array} $$

At this point in the calculations, it becomes evident that the result is the repeating septimal value $0.\overline{6531}_7$.

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Here's another way. We recognize that this is a rational number, and therefore must have a septimal representation that is either terminating or repeating. Since $32 = 2^5$ is not a power of $7$, it must be repeating. Therefore, we seek a power of $7$ that is one more than a multiple of $32$; that is,

$$ 32k = 7^d-1 $$

and then the repeating decimal has a period of $d$ digits, which is the septimal representation of $31k$. There are more systematic ways of doing this, but the numbers here are small enough to do by inspection: We see that $7^4-1 = 2400 = 32 \times 75$, and $31 \times 75 = 2325 = 6531_7$, and that's our repeating period.

Answer 2

You do it by multiplication and "peeling".

For a proper fraction start at the "decimal" point. To move one spot to the right multiply by the base, 7, and render the result as a mixed fraction. The whole number part is tge first base-7 "digit" after the "decimal" point. The remaining fraction is then used in the next step of multiplication and peeling to get the second base-7 digit and remainder that is used for the third decimal point, etc. Eventually the fractions will repeat and you can then identify the repeating block.

And so:

$(31/32)×7=(217/32)=6+(25/32)$, thus 0.6 ... .

$(25/32)×7=(175/32)=5+(15/32)$, thus 0.65 ... .

$(15/32)×7=(105/32)=3+(9/32)$, thus 0.653 ... .

After one more round your remainder fraction is back to 31/32, thus (leaving the reader to find the last "digit" of the block)

$0.653x653x653x ..._7$.

Answer 3

Note that $32\gt 31$

Now $$31\times 7=217=6\times 32+25$$ and $$25\times 7 = 5\times 32 +15$$ and $$15\times 7 =3\times 32 +9$$ and $$9\times 7=1\times 32+31$$

So base $7$ you get $\frac {31}{32}=0.6531 \dots$ and since it is rational it will be a recurring sequence.

Answer 4

Note that \begin{align} \frac{31}{32}\times 7 & = 6 + \frac{25}{32} \\ \frac{25}{32}\times 7 & = 5 + \frac{15}{32} \\ \frac{15}{32}\times 7 & = 3 + \frac{9}{32} \\ \frac{9}{32}\times 7 & = 1 + \frac{31}{32} \\ \vdots \quad & = \quad \vdots \\ \end{align}

Answer 5

Note that $32$ divides $7^4-1$. We then have: $$ \begin{aligned} \frac {31}{32} &= \frac {31\cdot 75}{32\cdot 75} = \frac {2325}{7^4-1} \\ &= 6531_7\cdot \left(\frac 1{7^4}+\frac 1{7^8}+\frac 1{7^{12}}+\dots\right) \\ &= 6531_7\cdot 0,\ 0001\ 0001\ 0001\ \dots_7 \\ &= 0,\ 6531\ 6531\ 6531\ \dots_7 \ . \end{aligned} $$

Answer 6

This is just a little exercise in thinking in base seven:

$${1\over11}=.0606060\ldots$$

so

$${1\over22}=.0303030\ldots$$

so

$${1\over44}=.01350135\ldots$$

so

$${43\over44}=1-{1\over44}=.65316531\ldots$$

The trickiest step is doing the long division of $.0303030\ldots$ by $2$, noting that $3=2\cdot1+1$, which leaves $10=2\cdot3+1$ for the next digit, after which $13=2\cdot5$.