CODEWITHSUNDEEP
decimalbitoctal
user3297557
user3297557

Reputation: 115

representing a number in the octal system

I am not looking for help with my homework. I just need someone to show me the direction to do it. I know the answer theoretically. I just stuck with idea of how to prove it mathematically. here is the question.

Representing a number in the octal system require, on the average, about 10 percent more characters than in the decimal system.

How can I prove this mathematically?

Upvotes: 0

Views: 184

Answers (2)

Chele
Chele

Reputation: 100

Suppose you wanted to represent a given number x in both systems. In the decimal system, this will take in the order of log10(x) digits. In the octal system, it will take in the order of log8(x) digits.

For any a and b, loga(b) can be written as logc(b)/logc(a) for a given c. In particular, let c=10. Therefore, log8(x) = log10(x)/log10(8) ~= 1.1 log10(x), which means log8(x) is about 1.1 times greater than log10(x) for any given x. Note that this result is exact aside from the rounding. What is not exact is approximating the number of digits by log10(x) and log8(x).

Upvotes: 1

galinette
galinette

Reputation: 9292

The approximative number of decimal digits required for representing a number is : log10(x), and the number of octal digits is : log8(x)

Which means that the average ratio is log8(x)/log10(x)

As log8(x) = ln(x)/ln(8) and log10(x) = ln(x)/ln(10)

The average ratio is ln(10)/ln(8) = 1.1073...

Of course this is not a 100% exact demonstration, a real demonstration would define exactly the number we are trying to find (such as the average number of digits for numbers between 0 and n when n goes to infinity, etc...) and would compute the exact number of digits (which is an integer) and not an approximation.

Upvotes: 1

Related Questions