$711.7$ base $10$ to base $2$

Question

I am attempting to solve the following two questions. The first question I believe I have done the whole number calculation correctly but I am confused how to do the decimal portion. The second question I am not sure about.

$1)$ Convert $711.7$ in base $10$ to base $2$

My solution.

Looking at the whole number to the left of the decimal $711$

$$(711)_{10} = 2^9 + 2^7 + 2^6 +2^2 + 2^0 \\ =1\cdot2^9 + 0\cdot 2^8 + 1\cdot 2^7 + 1 \cdot2^6 + 0\cdot 2^5 + 0\cdot 2^4 + 0\cdot 2^3 + 1 \cdot 2^2 + 0\cdot 2^2 + 1\cdot 2^0 \\= (1011000101)_2$$

Looking at the decimal part:

$$(0.7)_{10} = 2^{-1} + 2^{-2} + 2^{-4} + 2^{-7}+...$$

This is getting closer to $0.7$ but it doesn't seem to ever actually equal it. So I how do I determine this? How do I show that it is repeating ?

My other question is can this answered be stored exactly on a computer?

0.7 does not translate to an even binary fraction. It's a repeating one, as you've found. So no, it won't fit into 53 or any other finite number of bits of precision. — Mad Physicist, Sep 17 '18 at 20:14
A fraction has a finite radix representation in a particular base if it is a factor of some power of the base. In this case, we have $7/10$; it will have finite representations in any base that's a multiple of 10, and it will not have such a thing in base 2. The only other option for fractions is repeating digits; in this case, it's $0.1\overline{0110}$. — Dan Uznanski, Sep 17 '18 at 20:27
@DanUznanski how is it $0.10110$ repeating? could you show this? — user123, Sep 17 '18 at 20:29
This is the same question but with a different real number (5.1). If you submitted the same question with 294.6 instead of 711.7, would that be a "different" question? — David G. Stork, Sep 17 '18 at 20:35
no but I am just trying to understand how I can show it is repeating, anyways thanks — user123, Sep 17 '18 at 20:37
@user123. You missed the overline in $0.1\overline{0110}$. It means the last 4 digits repeat forever. — Mad Physicist, Sep 17 '18 at 20:41

score 2 · Accepted Answer · answered Sep 17 '18 at 20:51

Although a duplicate, i try to give the insight. It will not be short, but this is my everyday problem.

As a parallel, let us consider in base ten a periodic fraction. I will take $8/13$ which has the periodic decimal representation $0.\ 615384\ 615384\ 615384\ \dots=0.(615384)$ as i learned it in school. Now the reason for this is the fact that the first power of ten which is $1$ modulo $13$ is $10^6$, i.e. $999999/13=76923$ is the first $99\dots 9$ number divisible by $13$, the period comes from $999999/13\cdot 8 = 615384$ and the representation is shown by the long line $$ \begin{aligned} \frac 8{13} &= \frac{615384}{999999} \\ &=615384\cdot\frac 1{10^6}\cdot\frac 1{1-\frac 1{10^6}} =615384\cdot\frac 1{10^6}\cdot\left(1+\frac 1{10^6}+\frac 1{10^{12}}+\dots\right) \\ &=615384\cdot 0.\ 000001\ 000001\ 000001\ \dots \\ &=0.\ 615384\ 615384\ 615384\ \dots\ . \end{aligned} $$ (There is a little more dancing around when the denominator is not relatively prime to the basis ten.) Sorry for this, but in our case we have a similar computation.

Same computational idea, an other basis.

$$ \begin{aligned} 0.7 &=\frac 7{10}=\frac 12\cdot \frac 75 =\frac 12\cdot\left(1+ \frac 25\right) \\ &\text{ and the first $2$-power which is $1$ modulo $5$ is $16=2^4$, so...} \\ &=\frac 12\cdot\left(1+ \frac 6{16-1}\right) \\ &=\frac 12\cdot\left(1+ 6\cdot\frac 1{2^4}{1-\frac 1{2^4}}\right) \\ &=\frac 12\cdot\left(1+ 6\cdot\left( \frac 1{2^4} + \frac 1{2^8} + \frac 1{2^{12}} + \dots \right)\right) \\ &=\frac 12\cdot\left(1+ \color{blue}{110}_{\color{red}{2}}\times 0.\ 0001\ 0001\ 0001\ \dots_{\color{red}{2}} \right) \\ &=\frac 12\cdot\left(1+ 0.\ 0\color{blue}{110}\ 0\color{blue}{110}\ 0\color{blue}{110}\ \dots_{\color{red}{2}} \right) \\ &=\frac 12\cdot1.\ 0\color{blue}{110}\ 0\color{blue}{110}\ 0\color{blue}{110}\ \dots_{\color{red}{2}} \\ &=0.1\ 0\color{blue}{110}\ 0\color{blue}{110}\ 0\color{blue}{110}\ \dots_{\color{red}{2}}\ . \end{aligned} $$ Computer check:

sage: sum( [ 2.^-k for k in [1, 3,4, 7,8, 11,12, 15,16]] )
0.699996948242188
sage: sum( [ 2.^-k for k in [1, 3,4, 7,8, 11,12, 15,16, 19,20, 23,24, 27,28]] )
0.699999999254942

Note: The number can be stored exactly as a fraction, but not in binary representation.

Note: Add $711$ in binary for the final answer.

$711.7$ base $10$ to base $2$

1 Answers1