Noobie question ( bits )

[ichihgo]

So each digit gets it own 01011 ... from the binary code 

 

If I want to create say 64 digit how bit will that be !? and how to calculate it !? ( starting from 0  )

[givingtnt]

what can you explane a little more

[79wjd]

So each digit gets it own 01011 ... from the binary code 

 

If I want to create say 64 digit how bit will that be !? and how to calculate it !? ( starting from 0  )

Well, 

 

0000000000 = 0 

0000000001 = 1

0000000010 = 2

0000000011 = 3 

0000000100 = 4

0000000101 = 5

0000000110 = 6

0000000111 = 7

0000001000 = 8

 

and so on

[ichihgo]

Well, 

 

0000000000 = 0 

0000000001 = 1

0000000010 = 2

0000000011 = 3 

0000000100 = 4

0000000101 = 5

0000000110 = 6

0000000111 = 7

0000001000 = 8

 

and so on

 

are they 10 bits for a reason !? 

[79wjd]

are they 10 bits for a reason !? 

no 

[othertomperson]

Counting in 64 bit sequences starting from 0? Do you literally mean

 

00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000

00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000001

00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000010

00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000011

 

etc?

[HeathCliff]

are they 10 bits for a reason !? 

 

Usually,  in binary, the most common amount of 'bits' are shown in groups of 8. We can refer to them as an 'octet'. With these 8 bits, we can make a maximum number of 255 (presuming all the bits in the octet are '1'). 

 

Search for Professor Messer's binary maths video. It should cover all the basics. As for your question, 64 can be represented as '01000000'.

[Speedbird]

Usually,  in binary, the most common amount of 'bits' are shown in groups of 8. We can refer to them as an 'octet'.

Yeah, they're also known as bytes.

[ichihgo]

Usually,  in binary, the most common amount of 'bits' are shown in groups of 8. We can refer to them as an 'octet'. With these 8 bits, we can make a maximum number of 255 (presuming all the bits in the octet are '1'). 

 

Search for Professor Messer's binary maths video. It should cover all the basics. As for your question, 64 can be represented as '01000000'.

 

Okay gonna see it

[vm'N]

There are 10 types of people

Those who understand binary, and those who dont

[Ciccioo]

If I want to create say 64 digit how bit will that be !? and how to calculate it !? ( starting from 0  )

if you mean "how many bits do you need to write 64?" then the answer is 6

you can calculate it as the base 2 logarithm of the number you want to represent, rounded up

[Midnight]

if you mean "how many bits do you need to write 64?" then the answer is 6

you can calculate it as the base 2 logarithm of the number you want to represent, rounded up

 No, you need 7 bits for 64.

[wolfsinner]

With N bits you can represent up to 2^N-1 (no sign, no negative values).

 

So, 2^N-1 = v <=> N = log₂(v+1) rounded up is the value you're seeking.

[madknight3]

When working with numbers that are 2^x (two to the power of a number) you just shift the bit to the left.

 1 = 00000001 = 2^0 2 = 00000010 = 2^1 4 = 00000100 = 2^2 8 = 00001000 = 2^316 = 00010000 = 2^432 = 00100000 = 2^564 = 01000000 = 2^6

And so on. You can write out each number in binary like djdwosk97 posted to see that this is true.

 

Understanding this allows you to figure out numbers a little quicker by breaking a number into these pieces. Here's an example.

54 = 32 + 16 + 4 + 2So take the binary of each of those numbers54 = 00100000 +     00010000 +     00000100 +     00000010And added together the result is54 = 00110110

In the case of odd numbers, you simply will have the extra 1 = 00000001 on the end (which is still just a part of 2^x as I listed above)

23 = 16 + 4 + 2 + 1So add together the binary23 = 00010000 + 00000100 + 00000010 + 00000001So the result is23 = 00010111

You can double check with a binary calculator to make sure your right. For example if you type in "23 to binary" in Google it will show you "0b10111" which is just another way of writing binary. But if you look after the 0b you'll see that the answer is indeed 10111