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Prove that [3x] = [x] + [x + 1/3] + [x + 2/3]

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Prove that [3x] = [x] + [x + 1/3] + [x + 2/3]

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Case-I: when x is an integer. 

 i.e x = n 

L.H.S = [3n] = 3n

R.H.S = [n] + [n + 1/3] + [n + 2/3]

= n + n + n = 3n

Case-II: when x is not an integer

i.e. x = n + f

L.H.S = [3[n + f] = [3 + 3f] = 3n + [3f]

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