Exercise 2.6

For a rational number r let ⌊ r ⌋ be the largest integer less than or equal to r , e.g., ⌊ 1 2 ⌋ = 0 , ⌊ 2 ⌋ = 2 , and ⌊ 3 + 1 3 ⌋ = 3 . Prove ord p n ! = ⌊ n p ⌋ + ⌊ n p 2 ⌋ + ⌊ n p 3 ⌋ + ⋯ .

Answers

Proof. The number N k of multiples m of p k which are not multiple of p k + 1 , where 1 ≤ m ≤ n , is

N k = ⌊ n p k ⌋ − ⌊ n p k + 1 ⌋ .

Each of these numbers brings the contribution k to the sum ord p n ! = ∑ ⁡ i = 1 n ord p i . Thus

ord p n ! = ∑ k ≥ 1 k ( ⌊ n p k ⌋ − ⌊ n p k + 1 ⌋ ) = ∑ k ≥ 1 k ⌊ n p k ⌋ − ∑ k ≥ 1 k ⌊ n p k + 1 ⌋ = ∑ k ≥ 1 k ⌊ n p k ⌋ − ∑ k ≥ 2 ( k − 1 ) ⌊ n p k ⌋ = ⌊ n p ⌋ + ∑ k ≥ 2 ⌊ n p k ⌋ = ∑ k ≥ 1 ⌊ n p k ⌋

Note that ⌊ n p k ⌋ = 0 if p k > n , so this sum is finite. □

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2022-07-19 00:00
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