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Do all non-perfect squares have an irrational root?

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In the integers, a perfect square is one that has an integral square root, like 0 , 1 , 4 , 9 , 16 , … The square root of all other positive integers is irrational. In the rational numbers, a perfect square is one of the form a b in lowest terms where a and b are both perfect squares in the integers. So 0. read more

The roots of [math]x^2=n[/math] for integer [math]n[/math] are roots of the monic integer polynomial [math]x^2-n[/math]. Thus, they are either integer or irrational. In other words, [math]n[/math] must be the perfect square of an integer, or the square of an irrational number. read more

In the integers, a perfect square is one that has an integral square root, like $0,1,4,9,16,\dots$ The square root of all other positive integers is irrational. In the rational numbers, a perfect square is one of the form $\frac ab$ in lowest terms where $a$ and $b$ are both perfect squares in the integers. read more

Well, suppose there is an integer N with a rational, non-integer square root. This square root could be written as p/q, where p and q have no common factors and q is not equal to 1. This means than N = (p/q)^2 = p^2/q^2. read more

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