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Old 07-21-2007, 01:22 PM   #11 (permalink)
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Default Re: Understanding Binary Code

And you failed to mention, that, its powers of 16 in Hex...

....16^5...16^4...16^3...16^2...16^1... (Am i forgeting a 1?)
.... A.........4.........7........C........3


A47C3 == 16^5 X 10 + 16^4 X 4 + 16^3 X 7 + 16^2 X 12 + 16 X 3
So whatever the number comes out to be.

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Old 07-30-2007, 01:42 PM   #12 (permalink)
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Default Re: Understanding Binary Code

Ugh, college computer science nightmares. Thanks for reminding me why i switched from CS to MIS.

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Old 07-30-2007, 07:28 PM   #13 (permalink)
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Default Re: Understanding Binary Code

I learned this stuff in high school computer science. I personally believe hex is easier to under than binary to a person who knows basic base ten math.

The difference base sixteen (hex) and base ten is that you don't move to the next "place" until you reach 16 instead of 10. In base ten if you add 1 to 9 you will move to the next place and put a 1. In hex you simply just change the 9 into A, once you reach 16 you move to the next place and put a 1, and so on.

Also, the places work as exponents in all base systems. In base ten every place is 10 to the (what ever place that number is in)th power. So the ones place would be 10^0, because anything to the 0th power equals 1 (other than zero as that would equal zero). So if it's 8, that would mean:

8 x 10^0. then you simplify to 8 x 1 because 10^0 equals 1.

The tens place would be 10^1, so if you had a 8 it would mean:

8 x 10^1, simplify that to 8 x 10.

The hundreds place would be 10^2, thousands would be 10^3, etc...

So 583 would equal (5 x 10^2) + (8 x 10^1) + (3 x 10^0). Simplify that to

(5 x 100) + (8 x 10) + ( 3 x 1) = 500 + 80 + 3 = 583

In base two instead of the "places" being x^10, it would be x^2, and hex would be x^16. Sounds more complicated than it actually is.
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Old 08-01-2007, 12:16 PM   #14 (permalink)
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Default Re: Understanding Binary Code

i was taught it in maths

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