perplexus.info :: Just Math : Rounding the Root

Rounding the Root (Posted on 2008-03-03)

Some notation:

[x] = truncated x

<x> = rounded x = [x + 0.5]

E.g., <1.7> = 2, [1.7] = 1

Show:

<sqrt(n)> = [sqrt(n + [sqrt(n)])]

for natural number n.

Submitted by FrankM

Rating: 4.0000 (3 votes)

Solution: (Hide)

Take m^2 = largest square not exceeding n. Then

n + m = (m + 1)^2 + (n - m^2 - m - 1)
so that the right hand side of the equation is equal to m or m+1 according as to whether

n < m^2 + m + 1 But the left hand side of the equation is equal to m or m+1 according as to whether

n < (m + 1/2)^2 = m^2 + m + 1/4
As n, m are integers, these conditions are equivalent.

Comments: ( You must be logged in to post comments.)

	Subject	Author	Date
	Kudos	Steve Herman	2008-03-04 15:16:43

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