How do you simplify #sqrt(442)#?

1 Answer
Nov 7, 2015

Answer:

#sqrt(442)# cannot be simplified, but it has a simple continued fraction expansion:

#sqrt(442) = [21;bar(42)] = 21 + 1/(42+1/(42+1/(42+...)))#

Explanation:

#442 = 2*13*17# has no square factors, so we cannot simplify the square root.

It is an irrational number, so it cannot be expressed in the form #p/q# for any integers #p# and #q#. Neither will its decimal expansion terminate or repeat.

However, it is of the form #n^2+1#, since #442 = 441+1 = 21^2+1#.

As a result, the continued fraction for its square root takes a particularly simple form:

#sqrt(442) = [21;bar(42)] = 21 + 1/(42+1/(42+1/(42+...)))#

In general, any positive integer of the form #n^2+1# has square root:

#sqrt(n^2+1) = [n;bar(2n)]#