SOLUTION: Prove that there are infinitely many natural numbers n such that √(19n+9) is irrational. Thank you in advance!

Question 1182915: Prove that there are infinitely many natural numbers n such that √(19n+9) is irrational.
Thank you in advance!
Answer by ikleyn(52866)   (Show Source): You can put this solution on YOUR website!
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Prove that there are infinitely many natural numbers n such that is irrational.
~~~~~~~~~~~~~~~

                The proof is very simple.

                I'd say, it is UNEXPETEDLY simple.

Consider any integer positive number  " n ",  which has the last digit of 2.


(In other words, consider any integer positive number n such that  n = 2 (mod 10) ).


For example, such numbers "n" are 12, 22, 32, 42, 52 . . . and so on.


Then the number  19n  has the last digit of  8,  and the number  (19n+9)  has the last digit of 7,  because  8 + 9 = 17 = 7 (mod 10).


    +-------------------------------------------------------------------------+
    |   But  NO  ONE  such integer positive number is a perfect square  (!)   |
    +-------------------------------------------------------------------------+



(Notice that a square of an integer number may have the last digit only 0, 1, 4, 5, 6, 9, but may not have the last digit of 2, 3, 7, 8).


Therefore, all the numbers  (19n +9)  with n = 2 (mod 10) are not perfect squares and, THEREFORE, create/produce  IRRATIONAL numbers .

PROVED.

and Solved.

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