What is the square root of #100000#? I got as far as #100sqrt(10)#, but what is the final answer?
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#sqrt(10)# does not simplify further.
#10 = 2*5# has no more square factors, so #sqrt(10)# cannot be simplified further.
So the "final answer" may be simply #sqrt(100000) = 100sqrt(10)# unless you want a decimal approximation or an expression using a continued fraction...
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#sqrt(10)# is an irrational number. It cannot be expressed as a fraction. Its decimal representation does not terminate or recur.
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If you use a calculator, it will give you an approximation like:
#sqrt(10) ~~ 3.16227766#
Hence #sqrt(100000) = 100sqrt(10) ~~ 316.227766#
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Note that #10 = 3^2+1# is of the form #n^2+1#. As a result, it has a very regular continued fraction expansion:
#sqrt(10) = [3;bar(6)] = 3+1/(6+1/(6+1/(6+1/(6+1/(6+...)))))#
We can get rational approximations to #sqrt(10)# by truncating this continued fraction early.
For example:
#sqrt(10) ~~ [3;6] = 3+1/6 = 19/6 = 3.1bar(6)#
#sqrt(10) ~~ [3;6,6] = 3+1/(6+1/6) = 3+6/37 = 117/37 = 3.bar(162)#
Use binomial expansion
#sqrt 10= (9+1)^(1/2)#
#=3(1+1/9)^(1/2)#
#=3(1+(1/2)(1/9)+((1/2)(1/2-1))/(2!)(1/9)^2+...)#, and upon simplification,
#=3+1/6-1/216+1/3888- ..#
The sum to four terms is #3.16229.. #.
The magnitudes of the ratio of consecutive terms is more than 10.
So, easily the sum here might be correct to 5-sd, rounded. And So #
5-sd #sqrt 10 = 3.1623#.
