A formal-logic formula for decimal to binary conversion
How can one calculate that decimal 2 is binary 10 or that decimal 10 is binary 1010?
Instead of memorizing decimal to binary conversion tables, what formal-logic formula can one use to do generally any decimal to binary calculation?
3 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
I am giving Python code.
# Function to convert decimal number
# to binary using recursion
def DecimalToBinary(num):
if num >= 1:
DecimalToBinary(num // 2)
print(num % 2, end = '')
# Driver Code
if __name__ == '__main__':
# decimal value
dec_val = 24
# Calling function
DecimalToBinary(dec_val)
I hope it is understandable. Keep diving until you get zero. And, note every remainders. Remainders is binary.
Note : sometime some computers use 0 before binary. E.g. binary of 85 is 1010101 but, some computer will write 01010101.
I think it happens for encoding. Any computer can decode both and, return same value.
0 comment threads
The formal formula for base conversion of a non-negative number is $$x = \left\lfloor \frac{x}{b} \right\rfloor b + (x \bmod b)$$ For binary, $b=2$.
0 comment threads
If you have a non-negative integer $n$, and the base $b$ representation of $n$ is a sequence of digits, numbered from the right as $d_k\dots d_2d_1d_0$, then $d_i = \left\lfloor \frac{n}{b^i} \right\rfloor \bmod b$.
For example, decimal 10 is binary 1010 because:
$$ d_3 = \left\lfloor \frac{10}{2^3} \right\rfloor \bmod 2 = 1 \bmod 2 = 1 $$ $$ d_2 = \left\lfloor \frac{10}{2^2} \right\rfloor \bmod 2 = 2 \bmod 2 = 0 $$ $$ d_1 = \left\lfloor \frac{10}{2^1} \right\rfloor \bmod 2 = 5 \bmod 2 = 1 $$ $$ d_0 = \left\lfloor \frac{10}{2^0} \right\rfloor \bmod 2 = 10 \bmod 2 = 0 $$
(For all $i > 3$, $d_i = 0$, because $10 < 2^i$.)
This can be computed more efficiently using a loop or recursion, but this is the more succinct mathematical description.
By the way, extending this formula to non-integers is straightforward—just continue the digit numbering past the point as $d_k\dots d_2d_1d_0.d_{-1}d_{-2}\dots$, and the formula is still valid, modulo the usual philosophical concerns about infinite sequences of digits.
0 comment threads
0 comment threads