Shoutbox

quote:

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Originally posted by Chrono
bah sorry, i thought those where regular parenthesis My mistake

Edit:
Ok it's been a few years already since i solved these kind of problems for the last time, so i dunno if it's right

It works for 0:

floor(0) + floor(1/2) = 0

Now let's say it works for n

floor(n/2) + floor((n+1)/2) = n

then we can add 1

floor(n/2) + floor((n+1)/2) +1 = n+1

floor((n+2)/2) + floor((n+1)/2) = n+1

floor(((n+1)+1)/2) + floor((n+1)/2) = n+1

tadda

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You suck at induction

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then we can add 1

floor(n+1/2) + floor(((n+1)+1)/2) = n+1
floor((n+1)/2) + floor((n+2)/2) = n+1
floor((n+1)/2) + floor(n/2 + 1) = n+1

floor(a + 1) = floor + 1 as 1 is an integer

floor((n+1)/2) + floor(n/2) + 1 = n+1
floor(n/2) + floor((n+1)/2) + 1 = n+1

substitue n from (1)
n+1 = n+1
therefor true for n+1 for all n
take n=0 (as proven), then n=1 is also true, then n=2 is also true and so on and so forth

similarly the same can be proven for -1 using the same methods and thus applies from n=0, hence this is proven where n is any integer

EDIT: Volv gets full marks...