Search results
Results From The WOW.Com Content Network
Use octal digits as hexadecimal ones and then just add the numbers you got with weights of 8n 8 n, working in hexadecimal. For example: consider a =3478 a = 347 8. To get a hexadecimal representation without resorting to binary or decimal, you can use the fact that 8 <16 8 <16, i.e. just take the digits as they are.
A simpler way is to go through binary (base 2) instead of base 10. 0x1A03 = 0001 1010 0000 0011. Now group the bits in bunches of 3 starting from the right. 0 001 101 000 000 011. This gives. 0 1 5 0 0 3. Which is your octal representation. Share.
The reason this works is because 8 8 is a power of 2 2. In particular 8 =23 8 = 2 3, which is why we split into blocks of 3 3. Each block of 3 3 in binary corresponds to a unique number in octal, and vice-versa. Say I have a binary number. x = anan−1 ⋯a2a1a0.a−1a−2 ⋯a−m. x = a n a n − 1 ⋯ a 2 a 1 a 0. a − 1 a − 2 ⋯ a − m.
12 bits so 2^12 - 1. Decimal -> 4095. Hexadecimal -> FFF. Octal -> 777. Binary -> 111111111111. You can see in the BInary representation, all 12 bits are filled to the brim with 1s. Thus this would logically be the last address. Share. Cite.
2. I'm trying to learn on how to convert a negative decimal number (with fraction) to binary, octal and hexadecimal. So, base 10: -89.3125. Here is what I did to convert this to binary: -89= 1011001 (positive) // 0100110 (invert) // +1 (add one) = 0100111. I then represent the number in paper as (1)0100111.
A very similar thing happens from binary to octal because eight is an exact power of two. So we could write the binary numeral 1101000111010 1101000111010 as an octal numeral in the form 1, 101, 000, 111, 010 1, 101, 000, 111, 010, where each grouping of three bits takes up one octal place. However, because we only need eight symbols for octal ...
One strategy is to just convert to decimal and back, but a much better strategy is to remember that in hexadecimal each digit corresponds to four binary digits ($2^4=16$) and in octal each digit is three bits ($2^3=8$).
I've heard of many methods. Some people divide the decimal by 16 and find the remainder. Others convert the decimal into binary and then convert that binary into hexadecimal. One I am quite curious of is a method where the decimal is converted into octal, and the octal to binary, and the binary to hexadecimal. Is this an efficient method at all?
Then octal & hexadecimal follow just by grouping the digits in groups of 3 or 4 respectively. To convert to binary, repeatedly divide by 2 and collect the remainders (which must all be either 0 or 1) until you reach zero. The binary representation is then given by the remainders, starting with the last first. For example, take 23 23.
Octal (base-8) and hexadecimal (base-16) numbers are a reasonable compromise between the binary (base-2) system computers use and decimal (base-10) system most humans use. Computers aren't good at multiple symbols, thus base 2 (where you only have 2 symbols) is suitable for them while longer strings ,numbers with more digits, are less of a problem.