How do you find the exact square root of 41?

1 Answer
George C.
Oct 24, 2015

#sqrt(41)# is an irrational number. It cannot be expressed as one integer divided by another or by a terminating or repeating decimal.

#sqrt(41) ~~ 6.40312423743284868648#

Explanation:

#sqrt(41)# can be represented as a repeating continued fraction:

#sqrt(41) = [6;bar(2,2,12)] = 6+1/(2+1/(2+1/(12+1/(2+1/(2+...)))))#

We can use this expression to get rational approximations for #sqrt(41)#

For example,

#sqrt(41) ~~ [6;2,2] = 6+1/(2+1/2) = 6+2/5 = 6.4#

Alternatively, use a Newton Raphson type approach:

To find approximations for #sqrt(n)#, start with a reasonable approximation #a_0# and iterate using the formula:

#a_(i+1) = (a_i^2+n)/(2a_i)#

So if we started with #n=41#, #a_0=6#

#a_1 = (a_0^2+n)/(2a_0) = (6^2+41)/12 = 77/12 = 6.41dot(6)#

#a_2 = (a_1^2+n)/(2a_1) = ((77/12)^2+41)/(2*(77/12))#

#= (77^2 + 41*12^2)/(2*77*12) = (5929+41*144)/1848#

#= (5929+5904)/1848 = 11833/1848 ~~ 6.40314#

Related questions
See all questions in Simplification of Radical Expressions
Impact of this question
8619 views around the world
You can reuse this answer
Creative Commons License