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

24 Views Asked by At

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.

1

There are 1 best solutions below

2
On BEST ANSWER

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.