 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/3/e/ii/solution

Put

k = ⌊ y^{1/ n} ⌋

. Because

y > 0

k

is a nonnegative integer, and

k \leq y^{1/ n} < k + 1.

t \mapsto t^{n}

t \geq 0

, so

k^{n} \leq y < (k + 1)^{n}

. Both powers are integers. Taking the floor function therefore gives

k^{n} \leq ⌊ y ⌋ < (k + 1)^{n} .

Taking nonnegative

n

th roots now gives

k \leq ⌊ y ⌋^{1/ n} < k + 1

. Consequently

⌊ ⌊ y ⌋^{1/ n} ⌋ = k = ⌊ y^{1/ n} ⌋ .

The argument also covers

0 < y < 1

, when both outer floors are zero.

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