# working in base 2 /binary

• Jan 9th 2008, 07:40 AM
yellow4321
working in base 2 /binary
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• Jan 9th 2008, 08:40 AM
topsquark
Quote:

Originally Posted by yellow4321
ok the base 2 representation of

571 = 2^9 + 2^5 + 2^4 + 2^3 + 2 + 1 = (1000111011)

but i dont understand how the numbers 2^9 for example correspond to the 1's and 0's. i know how to compute the powers i just dont see how from that i cant write it out in 1s and 0s.
thanks.

Look at the coefficients of the 2s.

An example in base 10 might help:
$\displaystyle 571 = 5 \cdot 10^2 + 7 \cdot 10^1 + 1 \cdot 10^0$

Similarly:
$\displaystyle 571 = 1 \cdot 2^9 + 0 \cdot 2^8 + 0 \cdot 2^7 + 0 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 \implies 1000111011$

Does this help?

-Dan
• Jan 9th 2008, 08:43 AM
Jhevon
Quote:

Originally Posted by topsquark
Look at the coefficients of the 2s.

An example in base 10 might help:
$\displaystyle 571 = 5 \cdot 10^2 + 7 \cdot 10^1 + 1 \cdot 10^0$

Similarly:
$\displaystyle 571 = 1 \cdot 2^9 + 0 \cdot 2^8 + 0 \cdot 2^7 + 0 \cdot 2^6 + 1 \cdot 2^5 + 1 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 \implies 1000111011$

Does this help?

-Dan

yes, that does help, i've never seen this method before. thanks :D

it kind of seems clumsy though. how would you know how to represent 571 as a combination of different powers of 2's quickly and easily. do you start with the highest power of 2 you can get without exceeding the number, and then just stumble your way through the lower powers?

EDIT: wait, i think i got it, lemme check

EDIT 2: yeah, i got it, it wasn't as tedious as i thought it would be. that's nice!
• Jan 9th 2008, 08:53 AM
Plato
Quote:

Originally Posted by yellow4321
571 = 2^9 + 2^5 + 2^4 + 2^3 + 2 + 1 = (1000111011) but i dont understand how the numbers 2^9 for example correspond to the 1's and 0's. i know how to compute the powers i just dont see how from that i cant write it out in 1s and 0s.

Note that we end up with a 10-bit string.
We have a 1 in each place where that is a power of two.
See that there is $\displaystyle 2^9$ but none of $\displaystyle 2^8 \; 2^7 \; 2^6$ but there is a $\displaystyle 2^5$.
Thus we see 10001… . That is, there is a 1 where there is a power of two.
From right to left, there are ones in the first two positions but a zero in the third position because there is no $\displaystyle 2^2$.
• Jan 9th 2008, 09:17 AM
topsquark
Quote:

Originally Posted by Jhevon
yes, that does help, i've never seen this method before. thanks :D

You're kidding me, right? I thought that technique was basic. How else would you convert a number to binary?

-Dan
• Jan 9th 2008, 09:28 AM
yellow4321
yeah i remember now! and the powers are just formed from the euclidean algorithm, many thanks.
• Jan 9th 2008, 09:33 AM
Jhevon
Quote:

Originally Posted by topsquark
You're kidding me, right? I thought that technique was basic. How else would you convert a number to binary?

-Dan

i was taught the method where you repeatedly divide the number by two. in doing so, you will always get a remainder of 0 or 1. the trick is to keep track of these remainders, at the end, the remainders give you the number in binary when you read them in reversed order.

that method (as well as your method) is shown here

look at the "Short division by 2 with remainder" section for the method i described
• Jan 9th 2008, 09:44 AM
topsquark
Quote:

Originally Posted by Jhevon
i was taught the method where you repeatedly divide the number by two. in doing so, you will always get a remainder of 0 or 1. the trick is to keep track of these remainders, at the end, the remainders give you the number in binary when you read them in reversed order.

that method (as well as your method) is shown here

look at the "Short division by 2 with remainder" section for the method i described

I don't know if is in actuality another method or not. I would simply say that it is a slightly different book-keeping device for the same method.

-Dan