Rational Roots hat percentage of integers have Rational Roots?  Solvers will recall that a number r is said to be rational if and only if r can be expressed as p / q where p and q are integers.  An irrational number cannot be expressed as a ratio of integers.  Surely, the most famous is , the ratio of circumference of a circle to its diameter.  Of course, all integers are rational numbers but not vice versa.

Meanwhile, the general word "root" has been appropriated for the question above to represent taking the Nth root of an integer I to produce a root R, such that R N = I, and R can be either rational or irrational...

Square Root: For N = 2, R represents the square root of I, such that for I = 9, R = 3, an integer.   For I = 10, R = 3.1622776602..., which solvers can readily prove is irrational "Square Root," selected from a gallery of magnificent images
by Paul DeCelle. Published here by permision of the artist.
Cube Root: With N = 3, R represents the cube root of I, such that for I = 27, R = 3, and for I = 9, R = 2.080083823..., which is an irrational number.

Nth Root: Let N be an integer and R represent the Nth root of I, such that R N = I.  Clearly, if R is an integer, I will be an integer.  We have seen in the examples above, for some irrational values of R, I will nevertheless be an integer.  What about a rational value of R that is a non-integer?  Sophisticated solvers will enjoy proving that for integer values of both N and I, the Nth root if I must be either an irrational number or an integer -- but not something that looks like a fraction.

The opening question,"What percentage of integers have Rational Roots?" is tantamount to asking, "What percentage of integers are perfect integer powers?"  Taking perfect squares for starters, we find only three (1, 4, 9) in the first ten integers and three (16, 25, 36) in the next thirty integers, so perfect squares are not especially abundant.  Indeed, the 'numerical distance' between successive perfect squares can be derived as follows:
I 2 - (I - 1) 2 = I 2 - (I 2 - 2 I + 1) = 2 I -1,
...which can be converted into a percentage of integers = 100 / (2 I -1) and seen to decrease as I increases toward infinity and eventually 100 / (2 I -1) vanishing to zero.  Thus, 0% is the answer to the question, at least for N = 2.   Here is a graphical representation, showing that the first 10 perfect squares are subsumed by 10% of the corresponding consecutive integers and the first 100 perfect squares are subsumed by 1% of the corresponding consecutive integers (a mouthful not likely to be welcome at a cocktail party). For N = x, we merely take notice that I= I X - 2 I 2, so by a generalized version of the same formula, the percentage of integers that have Rational Roots tends to zero for all values of x.

 Oh, but wait.  For every integer there is a perfect square.  That means there must be an infinite number of perfect squares, thus an infinite number of Rational Roots.  Exclamation point optional.
Something of a paradox, it seems.  Not unlike Paradox Lost (and found). e have seen that in the Land of Numbers, it is possible to go from an integer to an irrational number by simply extracting an integer root of the integer.  Here is an amazement: It is also possible to go the other way -- from an irrational number to an integer -- by extracting an irrational root of an irrational number…

Let there be two irrational numbers A and B.  Given the formula C = B A...
 Can C ever be a Rational Root?