Rational Roots

or the puzzle, we are given an existence theorem: in the form of a question:  Let there be two irrational numbers A and B and the formula C = B A. Can C ever be a Rational Root?

Solvers have the freedom to decide on the values for A and B so long as neither can be expressed as a ratio of p / q where p and q are integers.  Using spreadsheet typographic conventions...

Let A = 2^(1/2) and B = X^(2^(1/2)), where X is an integer.
We write C = (X^(2^(1/2)))^(2^(1/2)) = X^2
Since X is an integer, X^2 is an integer, for example (get out your calculator)...
Let X = an integer, say 5, then X^2 = 25, noting that 2^(1/2) = 1.41421356...
We write (5^1.41421356...)^
1.41421356... = 25 = 5^2...voilą!
...and the existence appears to be established.  We seem to be ready to say Q.E.D. despite a possible issue...
Whereas A = 2^(1/2), the square root of two, is definitely an irrational number, B may not be an irrational number as specified in the puzzle.  Best not choose some 'wrong' value for X, such that B = X^(2^(1/2)) happens to be rational.
Go ahead with your Q.E.D., though.  "Existence" does not require all values of X to be accommodated.  Our solution, then, is...
 Yes

ntegers can be produced in many ways, not just consecutively starting with one.  Some solvers will be interested in exploring the statistical attributes of Rational Roots by modifying the original question in the puzzle ("What percentage of integers have Rational Roots?") to address various other sources of integers...

• What percentage of Fibonacci numbers have Rational Roots
• Seeded with 0 and 1, only one, the 13th (144) qualifies.
• What percentage of Factorials have Rational Roots
• None, since all are products of u times v, where u is never equal to v.
• What percentage of Totorials have Rational Roots
• Only the 8th (36) and no others.
• What percentage of prime numbers have Rational Roots
• Well, none, of course.

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