All unnatural square roots are irrational

Theorem: If n is a natural numberMathworldPlanetmath and n2 is not whole, then n2 must be irrational.

Proof Ad absurdum: Assume there exists a natural number n that n2 is not whole, but is rational.

Therefore n2 can be notated as an irreducible fraction: md

Now break the numerator and denominator into their prime factorsMathworldPlanetmath:


Because the fraction is irreducible, none of the factors can cancel each other out.

For any i and j, midj.

Now look at n:


Because n is a natural number, all the denominator factors are supposed to cancel out,

but this is impossible because for any i and j, midj.

Therefore n2 must be irrational.

Unfortunately this means that a (non-integer) fraction can never become whole by simply squaring, cubing, etc.

I call this unsatisfying fact my ”Greenfield Lemma”.

Title All unnatural square roots are irrational
Canonical name AllUnnaturalSquareRootsAreIrrational
Date of creation 2013-03-22 17:37:03
Last modified on 2013-03-22 17:37:03
Owner ubershmekel (18723)
Last modified by ubershmekel (18723)
Numerical id 12
Author ubershmekel (18723)
Entry type Theorem
Classification msc 11J82
Classification msc 11J72
Synonym The square of a fraction is always a fraction
Related topic RationalBriggsianLogarithmsOfIntegers