The only thing that comes to mind is something similar to the source below, which illustrates place values in system with base 10, or (10).
If we define a number as X.Y, (or X,Y in some other system)
Place value would determine whether the particular number stands for
- units, tens, hundreds, thousands, etc. (if on the left of dot/point/comma - whatever you denote as a sign for whole vs. decimal numbers)
- tenths, etc. (if on the right of the dot/point/comma).
Another words, thats why you know that 23 should be read as twenty-three, and not thirty-two, or whatever.
Another important note is that 1 = 10^0, 10 = 10^1, 100 = 10^2, 1000 = 10^3, etc., while 1/10 = 10^(-1), 1/100 = 10^(-2), 1/1000 = 10^(-3) etc., which brings us to the real thing, ie. to the fact that place value actually represents powers of the base of a particular system. So, from the dot to the left, we have base^(powers from 0, 1,... to infinity), and from the dot to the right, we have base^(powers from -1, -2, -3, to negative infinity). (Don't confuse these positive and negative powers with good and evil, ha, ha!)
Now when you know this, you also know that 23 is actually twenty-three, because it is in decade system, ie. a system with a base (10) => 23 = 2*10^1 + 3*10^0 = 2*10 + 3*1 = 20 + 3 = 23
Therefore, if you are working with binary (base: 2) or any other system, the trick is the same as with (10) - and you'll instantly know which number represents what, and how to turn it into system (10), which makes things so much easier. This is related with turning (10) into any other system, but I'll rather leave it for another question ;D
Warning:
Please note that I referred to "dot", but you should use whatever you use in your country, in order to distinguish whole part from the decimal one, as well as with thousands (e.g. 1.234,56 vs 1,234.56)
2006-12-07 21:47:58
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answer #2
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answered by Mirta G 2
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the amount that the place is worth
ex. 1000
furthest left is the thousandth place, then the hundreth, then the tenth, then the ones
2006-12-07 21:04:43
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answer #3
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answered by Anonymous
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