What is the square root of #0.4# ?

Question

Abigail Allen · Accepted Answer

# = pm sqrt10 / 5 #

#approx pm 0.633 # We can first write #0.4# as # 4/10 # So #sqrt0.4 = sqrt(4/10) # #=> sqrt(4) / sqrt(10) # #=> 2 / sqrt 10 # At this point, we can justify the denominator: #=> 2/sqrt10 * sqrt(10)/sqrt(10) # #=> (2*sqrt10 )/ 10 # The #2/10 = 1/5 # #therefore pm sqrt(10)/5 # It must be #pm# as two negative multiply to form a poisitve #approx pm 0.633 #

Ethan Adkins · Answer

#sqrt(0.4) = sqrt(10)/5 ~~ 4443/7025 ~~ 0.6324555#

Like any non-zero number #0.4# has two square roots. Since it is positive, both of those square roots are real and by convention the principal square root is the positive one, which we denote by #sqrt(0.4)#. The other square root is the negative one #-sqrt(0.4)#.

The principal square root is usually meant to be referred to when the phrase "the square root of..." is used.

We can take the following steps to determine the most basic algebraic form:

#sqrt(0.4) = sqrt(2/5) = sqrt(10/5^2) = sqrt(10)/5#

This is an irrational number a little larger than #sqrt(9)/5 = 3/5#

Reasonable approximations

We can find rational approximations to #sqrt(10)# and hence #sqrt(10)/5# in a number of ways (at least 25 that I know of). Here's one using:

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#

So putting #a=3# and #b=1# we have:

#sqrt(10) = 3+1/(6+1/(6+1/(6+1/(6+...))))#

Then, we can cut this short early to obtain a reasonable approximation, for example:

#sqrt(10) ~~ 3+1/(6+1/(6+1/(6+1/6)))#

Instead of laboriously breaking down such a fraction, we can use an integer sequence defined recursively by:

#{(a_0 = 0), (a_1 = 1), (a_(n+2) = 6a_(n+1)+a_n) :}#

To begin with:

#0, 1, 6, 37, 228, 1405, 8658,...#

The ratio between successive pairs of terms tends to #3+sqrt(10)#

So:

#sqrt(10) ~~ 8658/1405-3 = 4443/1405#

So:

#sqrt(10)/5 ~~ 4443/(5 * 1405) = 4443/7025 ~~ 0.6324555#