#sqrt(17)# is irrational essentially as a consequence of #17# being prime - that is having no positive factors apart from #1# and itself.
Here's a sketch of a proof:
Suppose #sqrt(17) = p/q# for some integers #p#, #q#, with #q != 0#.
Without loss of generality, #p, q > 0# and #p# and #q# have no common factor greater than #1#.
[[ If they did have a common factor, then you could divide both by that common factor to get a smaller #p_1# and #q_1# with #sqrt(17) = p_1/q_1# ]]
Then #p^2 = 17q^2# and since #p^2# is a multiple of #17# and #17# is prime, #p# must be a multiple of #17#.
Let #k = p/17#
Then #17q^2 = p^2 = (17k)^2 = 17*17k^2#
Divide both ends by #17# to find:
#q^2 = 17k^2# hence #q# is a multiple of #17#.
So both #p# and #q# are divisible by #17#, contradicting our assumption that #p# and #q# have no common factor greater than #1#.
So there is no such pair of integers #p# and #q#.
