What is the square root of 67.98?

To calculate an approximation for #sqrt(67.98)#, find an approximation for #sqrt(6798)# and divide by #10#

#80^2 = 6400 < 6798 < 8100 = 90^2#

To find the square root of a number #n#, you can choose a first approximation #a_0# and iterate using the formula:

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

I prefer to work with integers, so I express #a_0 = p_0/q_0# and iterate using the formulae:

#p_(i+1) = p_i^2 + n q_i^2#

#q_(i+1) = 2 p_i q_i#

Then #p_i/q_i# is the same as #a_i#.

If the resulting #p_(i+1)# and #q_(i+1)# have a common factor larger than #1# then divide both by that before the next iteration.

Let #n=6798#, #p_0 = 80#, #q_0 = 1#

Then:

#p_1 = p_0^2 + n q_0^2 = 80^2 + 6798 * 1^2 = 6400+6798 = 13198#

#q_1 = 2 p_0 q_0 = 2 * 80 * 1 = 160#

These are both divisible by #2#, so do that to get:

#p_(1a) = p_1/2 = 6599#

#q_(1a) = q_1/2 = 80#

Next iteration:

#p_2 = p_(1a)^2 + n q_(1a)^2 = 6599^2 + 6798*80^2 = 43546801 + 43507200 = 87054001#

#q_2 = 2 p_(1a) q_(1a) = 2 * 6599 * 80 = 1055840#

Stopping here, we get:

#sqrt(6798) ~~ 87054001 / 1055840 ~~ 82.44999 ~~ 82.45#

So

#sqrt(67.98) ~~ 8.245#

For an alternative method of finding that #sqrt(6798) ~~ 1645/20 = 82.45# see:
http://socratic.org/questions/given-an-integer-n-is-there-an-efficient-way-to-find-integers-p-q-such-that-abs-#176764