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Old 05-05-2004, 07:35 PM   #1 (permalink)
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Default Am I right on this graph?

On this graph,

0 = 0
1 = 1
2 = 10
3 = 11
4 = 100
5 = 101
6 = 110
7 = 111
8 = 1000
9 = 1001
10 = 1010
11 = 1011
12 = 1100
13 = 1101
14 = 1110
15 = 1111
16 = 10000
17 = 10001
18 = 10010
19 = 10011
20 = 10100

Does this mean that the number on the right is what the number on the left look like in binary language? For example the number 20 is 10100 in binary language. If im wrong than tell me what this graph really represents.

thank you
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Old 05-05-2004, 07:47 PM   #2 (permalink)
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Yes the numbers on the right are binary for the numbers on the left.
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Old 05-06-2004, 08:59 AM   #3 (permalink)
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yeah they're right

I know < is left and > is right
I was a bit confused at which way he meant

I think even numbers are 8-bit
like
0 = {0000000[0]}
1 = {0000000[1]}
2 = {000000[10]}
and so on
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Old 05-06-2004, 10:38 AM   #4 (permalink)
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Default Re: Am I right on this graph?

Quote:
Originally posted by NanoWarrior

0 = 0
1 = 1
2 = 10
3 = 11
4 = 100
5 = 101
6 = 110
7 = 111

Does this mean that the number on the right is what the number on the left look like in binary language? For example the number 20 is 10100 in binary language. If im wrong than tell me what this graph really represents.

thank you
Technically speaking, that is incorrect. The author or where ever you got that information from took a very lazy & abbreviated route of explaing that.

It's more commonly shown (when using binary numbers) in groups of 4 digits. So just as they have shown 8 & 9 below, they should have shown it like thus:


0 = 0000
1 = 0001
2 = 0010
3 = 0011
..
8 = 1000
9 = 1001
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Old 05-06-2004, 06:12 PM   #5 (permalink)
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just think of the octet 0000 0000 which is 1 byte. now u just put in the numerators for it like so:

0 0 0 0 0 0 0 0 = 0
128 64 32 16 8 4 2 1 = 255

soooo.... if you want, say the number 5 look at the chart and what bits flipped would give you said number? Well, 1 + 2 is 3... nope...

so 1 + 4 = 5.... so now flip bit place 1 and 4 for the following

0000 0101 = 5

another example would be the number 182... by using the above we can find out which numbers to use to come up with that... its best actually to subtract... so starting at 128, look and see if that will go into 182? Yes, it will. so flip the bit representing 128.
Now 182 - 128 = 54.... next in line is the it for 64. Well, 64 is bigger than 54 and we can't have negative numbers, so NOPE! leave that bit alone. so next is 32. will 32 go into 54? yes, it will. 54-32 = 22. So flip the bit representing 32. how about 16? will 16 subtract from 22? It will, also so flip that one and now we have 5 left. So far we have

1011 xxxx

so now finish with the last 4. will 8 go into 5? nope. so leave that bit. will 4? yes, flip the 4 spot. you now have 1 left. so we know 2 wont go into 1 but 1 will leaving with 0! so the out come is this:

1011 0110
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Old 05-06-2004, 06:13 PM   #6 (permalink)
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this is also how to configure IP addresses and how to subnet using the 2n method to find addresses from a subnet and so on. the math gets a little more complicated.
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Old 05-06-2004, 08:19 PM   #7 (permalink)
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how does this graph,

0 0 (OFF OFF) = Decimal 0
0 1 (OFF ON) = Decimal 1
1 0 (ON OFF) = Decimal 2
1 1 (ON ON) = Decimal 3
-----------------------------------------------------
relate to this graph?

Decimal Binary Representation
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
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Old 05-06-2004, 09:13 PM   #8 (permalink)
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Quote:
Originally posted by killians45
how about 16? will 16 subtract from 22? It will, also so flip that one and now we have 5 left.
Does'ent 22 - 16 = 6 instead of 5? When you said 182 on top, does this mean you are caculating what the number 182 is represented in the binary language?
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Old 05-06-2004, 09:50 PM   #9 (permalink)
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oh, oops! heh, well my math isn't that great!


just remember that as far as a byte you have 8 bits
8 bits = 1byte
so, assuming that we have
xxxx xxxx
8765 4321

bit /value1

1 = 1
2 = 2
3 = 4
4 = 8
5 = 16
6 = 32
7 = 64
8 = 128

so if you have the binary of 0001 0110
you have flipped the bit in the 2, 3 and 5th position... or as you also
asked, these are turned 'on'.
looking at the values, the 2nd position has a value of 2, the 3rd
position has a value of 4, and the 5th position has a value of 16. So the decimal equivilant of 0001 0110 = 22

You can also work backwards from this. If you have number 25, then simple math will get the value. bit places 8, 7, and 6 are to large to fit into that value. So those will all be zeros. However, the 5th bit has a value of 16. That is less than 25, so we flip it, or turn it 'on'. That leaves 9. Next in line is the number 4th bit value of 8. That is also less than nine. So the 4th bit is on/flipped. Now we have 1 left. 3rd and 2nd bit are to large and we CANT have a negative number here. So that leaves the 1st bit value of 1. This zeros it out. So flip it/turn it on. So now we have the following bits flipped/on.... bit 5, 4 and 1... or values 16, 8, and 1 (16+8+1=25). So lets goto the xxxx xxxx and fill in the x's. anything not flipped/turned on gets a zero and anything flipped/on gets a one, like so:

0001 1001 bytes = 25 dec
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Old 05-06-2004, 10:52 PM   #10 (permalink)
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Thanks for that great info, your info is Greatly presented by you.

However I have three more questions

1. When you say 25....are you calculating what the actual number 25 is represented in binary language? Or the decimal?

2. How would I calculate what the letter t or a is in binary language? In other words how would I find out what the letter t or number 3 is represented in binary language? Every character on the keyboard is a byte in size I know.

3. Since the anwser was actually 6 instead of five....the 5 answer would look more like this 1011 0100, but if it was a six like it should of been it would of looked like this 1011 0110, right?

thanks
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