On changing the base of $x<1$ from $10$ to everything else

Question

Assume that we have a number $x$ which is less than $1$ and is written in base 10.

For example, if we want to write $x$ in base $4$, The algorithm says that each time we multiply $x$ by $4$. Then $4x$ has an integer part and the rest is the decimal part. We write the decimal part and then again multiply the decimal part by $4$. The process continues until we reach a number having $0$ as its fractional part.

The question is:

Why do we just multiply the fractional part? What's the logic of omitting the digits and then multiplying the fractional part?

Note: My question may seem too easy. But I'm trying to understand the algorithm. So, an explanation of what we're doing here would be great.

Thanks in advance.

Why do you stop at a 0 digit? You should either stop when the remainder (fractional part) is 0, or a specified number of places has been reached. By your method, 0.1(10) converts to 0(4). — Weather Vane, Oct 06 '17 at 08:46
@WeatherVane Not at the 0 digit. When the part which is after $.$ becomes 00. Because $0 \times 4$ is $0$. — Arman Malekzadeh, Oct 06 '17 at 08:57
Clearly, your example stops when the first 0 digit pops out. — Weather Vane, Oct 06 '17 at 08:57
@WeatherVane I mean, for example, if i reach $2.0$ then i stop. It it wrong? — Arman Malekzadeh, Oct 06 '17 at 08:59

gammatester · Accepted Answer · 2017-10-06T09:20:02.903

Your arithmetic is wrong, the correct computation is as follows:

$0.74 \times 4 = 2.96$
$0.96 \times 4 = 3.84$
$0.84 \times 4 = 3.36$
$0.36 \times 4 = 1.44$
$0.44 \times 4 = 1.76$
$0.76 \times 4 = 3.04$
$0.04 \times 4 = 0.16$
$0.16 \times 4 = 0.64$
$0.64 \times 4 = 2.56$

So $\implies (0.74)_{10}=(0.233113002\dots)_4,$ which can be verified with Wolfram Alpha.

You better say 'we use the fractional part', and this is the same as when you compute the base 10 representation of a positive number $0 \le x \lt 1.$

Without complications you want to compute the coefficients $a_k$ in base $b$ $$x=\sum_{k>0}{a_k b^{-k}},$$
repeated multiplication with $b$ and taking the fractional part gives $a_1, a_2 \dots$

You start with $xb = a_1 + \sum_{k>0}{a_{k+1} b^{-k}},$ where the sum is $\lt 1$, now repeat the proceess until the new sum is zero.

That was just an example! excuse me for that :) The question is something else! Why just multiplying the fractional part? What's the logic of it? Why not $2.96 \times 4 $? Why $0.96 \times 4 $? — Arman Malekzadeh, Oct 06 '17 at 08:59

On changing the base of $x<1$ from $10$ to everything else

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