The square root of a number #n#, written #sqrt(n)#, is the number #x# such that #x^2=n#.
Actually, any positive number #n# has two square roots, namely #sqrt(n)# and #-sqrt(n)#. The notation #sqrt(n)# denotes the principal square root, which is the positive one.
Notice that:
#3^2 = 9 < 13 < 16 = 4^2#
Hence:
#3 < sqrt(13) < 4#
So #sqrt(13)# lies strictly between #3# and #4#, so is not an integer, whole number or natural number.
Is it a rational number?
Suppose #sqrt(13) = p/q# for some positive integers #p, q#.
Note that #p/q > 1#, so #p > q > 0#
Without loss of generality, we may suppose that #p, q# are the smallest pair of positive integers with #p/q = sqrt(13)#.
Then we have:
#13 = p^2/q^2#
Multiplying both sides by #q^2# we get:
#13q^2 = p^2#
So #p^2# is a multiple of #13#.
Since #13# is a prime number, that must mean that #p# is a multiple of #13# too. So there is some positive integer #m# with:
#p = 13m#
So we have:
#13q^2 = p^2 = (13m)^2 = 13^2m^2#
Dividing both ends by #13m^2# we get:
#q^2/m^2 = 13#
So:
#q/m = sqrt(13)#
Now #p > q > m > 0#, so #q, m# are positive integers smaller than #p, q# whose quotient is also #sqrt(13)#.
This contradicts our supposition that #p, q# was the smallest such pair and hence our supposition that #sqrt(13) = p/q# for some pair of positive integers.
So #sqrt(13)# is not rational. In other words it is irrational.
