@mattbrowne If it’s not transcendental then it’s algebraic, which means it’s the root of an equation of the form:
ax^n + bx^(n-1) + cx^(n-2) +...+ ux + v = 0
where all the coefficients a, b, etc are integers and n may be large but must be finite.
I didn’t mean to imply that all algebraic numbers are the square root of a rational. The cube root of 2, for example, is not the square root of any rational number even though it’s algebraic because it satisfies x^3 – 2 = 0.
Cantor showed that the algebraics are countable just like the integers and rational numbers. Transcendental (non-algebraic) numbers, on the other hand, are uncountable like the set of real numbers itself. So in a sense, there are “more” transcendental numbers than there are algebraic numbers, even though they’re both infinite sets.
@cazzie ”...pie is round.” When it comes to figuring the area of a circle, “pie are square”!