# What is the logic behind the octal to decimal conversion using the expansion method?

I am just utterly confused regarding number systems and the internet articles/searches haven't helped me so far.

Consider this number $123_{10}$ in base-10. Now it is fairly easy to understand that this $123_{10}$ actually is a shorthand form which translates to: $$1*10^2+2*10^1+3*10^0$$ which is basically nothing but multiplying each digit with its place value.

But let's now consider a number $123_{8}$ which, as with the base-10, actually should translate to $$1*8^2+2*8^1+3*8^0$$

Now firstly, why does evaluating octal numbers like this yield its decimal equivalent? Evaluating any base number like this, why does it yield its decimal equivalent? Isn't it just supposed to be a way to expand the shorthand way that we normally write the numbers in.

Writing this, another thing comes to mind, when we expand base-8, we are multiplying it with exponents of 8, which isn't even a digit in octal system. And then again, the exponents? What number system they belong to?

Now firstly, why does evaluating octal numbers like this yield its decimal equivalent? Evaluating any base number like this, why does it yields its decimal equivalent?

Only because we are used to doing computations in decimal. Someone (or a society) adept at doing computations in hexadecimal could use the same expansion to convert $123_8$ to hexadecimal:

\begin{align} 1 \cdot 8^2 + 2 \cdot 8^1 + 3 \cdot 8^0 &= 1 \cdot 40 + 2 \cdot 8 + 3 \cdot 1 \\ &= 40 + 10 + 3 \\ &= 53 \end{align}

$53$ is the hexadecimal form of $123_8$. This works because all the computations, e.g., $2 \cdot 8 = 10$, are done in hexadecimal. The only reason you get the decimal form $83$ from this computation is because you do the computation in decimal.

Writing this, another thing comes to mind, when we expand base-8, we are multiplying it with exponents of 8, which isn't even a digit in octal system.

This is true! This is a good hint that the computation is actually being done in decimal, as I noted above. In octal, the $8$ would be replaced by $10$, and the computation would look like this:

\begin{align} 1 \cdot 10^2 + 2 \cdot 10^1 + 3 \cdot 10^0 &= 1 \cdot 100 + 2 \cdot 10 + 3 \cdot 1 \\ &= 100 + 20 + 3 \\ &= 123 \end{align}

This computation confirms the tautology that the octal $123$ has the form $123$ in octal.

And then again, the exponents? What number system they belong to?

All the writing systems describe the same numbers -- they are just different ways of writing them. The exponents belong to the common system of numbers. How you write them is up to you; usually you'll write them the same way you write the bases, so if your computation is in decimal, you'll write the exponents in decimal; if your computation is in hexadecimal, you'll write the exponents in hexadecimal; etc.

• Yes, I think @Y. Forman made a very good point in highlighting that the decimal system is only the system in which you perform your computations – Lucio Tanzini Jan 28 '18 at 18:32

I think you are confused by the distinction between representation and actual quantity.

When you convert from octal to decimal you are "decoding" the number in octal, and that is fairly easy to do if you remember the meaning of the positions of a digit in a number, now, you can represent the underlying quantity any way you want, and of course, if you decide to do your calculations on a decimal system you'll get an answer in the decimal system.

I try to better explain it. Start from a set of objects, that set has a quantity, now, if you want to represent it into a number you need to find a good way of representation that links every possible quantity of a set to one and only one "word". Now, since the possible quantities are infinite, it is unfeasible to give an arbitrary name to all the quantities, so you can come up with a rule that allow you to go from name to quantity and vice versa. The romans invented(?) the now obsolete letter-system with I,V,D,L,M etc... and that worked as well, yet not as efficiently as the modern decimal system.

• So from what I understand, I can expand any number in any number system in such a way that WOULD yield its binary equivalent? – Parker Queen Jan 28 '18 at 18:20
• Whay do you mean by binary equivalent? – Lucio Tanzini Jan 28 '18 at 18:28
• By binary equivalent, I meant that same number in binary system. – Parker Queen Jan 28 '18 at 18:29
• Then the answer is definitely yes – Lucio Tanzini Jan 28 '18 at 18:33
• Alright. Thanks a lot mate. While Forman cleared up my concerns pretty well, you sure did a great job at making me understand deeper. – Parker Queen Jan 28 '18 at 18:34