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Is #sqrt(17)# rational or irrational ?

Answer 1

#sqrt(17)# is an irrational number.

That is, it is not expressible in the form #p/q# for some integers #p# and #q# with #q != 0#.

#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#.

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Answer 2

Genesis Allen

(sqrt{17}) is an irrational number.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.