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Nov 5 2015 10:22am
how can I prove this? I don't know where to start when it comes to proving. Thank you

[]=floor

Prove that [3x] = [x] + [x + 1/3] + [x + 2/3]
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Nov 5 2015 11:50am
For example you can write x = [x] + y, with 0 ≤ y < 1.

Then you must discuss 3 different cases :
0 ≤ y < 1/3 :
then [3x] = [3[x] + 3y] = 3[x], since 3y < 1 ;
and [x] + [x+1/3] + [x+2/3] = [x] + [x] + [x] = 3[x], since both y+1/3 and y+2/3 < 1.

1/3 ≤ y < 2/3 :
then 1 ≤ 3y < 2, and 2/3 ≤ y+1/3 < 1, and 1 ≤ y + 2/3 < 4/3, so :
[3x] = [3[x] + 3y] = 3[x] + 1
and [x] + [x+1/3] + [x+2/3] = [x] + [x] + [x] + 1 = 3[x] + 1.

and at last : 2/3 ≤ y < 1 :
then 2 ≤ 3y < 3, and 1 ≤ y+1/3 < 4/3, and 4/3 ≤ y +2/3 < 5/3, so :
[3x] = [3[x] + 3y] = 3[x] + 2
and [x] + [x+1/3] + [x+2/3] = [x] + [x] + 1 + [x] + 1 = 3[x] + 2.




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