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September 12th, 2011, 05:04 AM   #1
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prove the floor of two numbers is the same as....

let [] be a floor function

For all real number x and all integers m, show that [x + m] = [x]+ m
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September 12th, 2011, 05:48 AM   #2
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Re: prove the floor of two numbers is the same as....

[x + m] = [[x] + {x} + m] = [x] + m, where {x} denotes the fractional part of x, i.e. {x} = x - [x].

edit: I think that's right; the fractional part is defined to be positive or zero.
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September 12th, 2011, 03:53 PM   #3
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Re: prove the floor of two numbers is the same as....

?? so it's proven just like that? that simple?
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September 13th, 2011, 01:31 PM   #4
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Re: prove the floor of two numbers is the same as....

Not all real numbers have fractional parts right?
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September 13th, 2011, 01:53 PM   #5
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Re: prove the floor of two numbers is the same as....

Integers don't, unless you count 0/?.

x is real.
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September 14th, 2011, 02:17 AM   #6
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Re: prove the floor of two numbers is the same as....

so i think you guys need to help explain on how the answer of derived?
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September 14th, 2011, 02:42 AM   #7
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Re: prove the floor of two numbers is the same as....

I think a little more argument is needed...

First, if A is an integer and 0 ? B < 1, then [A+B] = [A] = A.
Also, x = [x] + {x} where [x] is an integer and 0 ? {x} < 1.

[x+m]
= [[x]+{x} + m]
= [[x]+m + {x}],
and [x]+m is an integer, because an integer plus an integer is an integer. And 0 ? {x} < 1.
Let A = [x]+m and B = {x} and compare with the first statement above.
Therefore [x+m] = [[x]+m] = [x]+m.
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September 14th, 2011, 07:28 AM   #8
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Re: prove the floor of two numbers is the same as....

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