Insanely simple proof of irrationality of sqrt(N) when N is not a perfect square

Proving that people still discover things about the oldests of proofs, this is from Geoffrey C. Berresford in 2008. [Claims abound of it being much older, so bonus kudos to whoever finds the oldest reference.]

It is almost stupidly easy (and hence is very smart).

Prove: If \sqrt N is a rational number then N is a perfect square.

Suppose: \sqrt N is a rational number \frac{B}{A} in lowest terms.

Doing a little algebra…

… we can say \frac{B}{A} = \frac{NA}{B}

Since \frac{B}{A} is in lowest terms, there’s an integer c such that Bc = NA and Ac = B.

Since Ac = B we also know c = \frac{B}{A}. That is, \frac{B}{A} is an integer, so \sqrt N is an integer and N is a perfect square.

Taking the contrapositive, if N is a not a perfect square then \sqrt N is not a rational number.


19 Responses

  1. Very nice!

  2. A colleague (Jan Wehr) pointed out to me this is related to a more compact proof…

    Let B / A be lowest rational form of sqrt(N).
    B / A = sqrt(N) / 1
    B^2 / A^2 = N / 1

    Since B and A share no factors B^2 / A^2 is also in lowest terms. N/1 is clearly in lowest terms. So B^2 = N, that is, N is a perfect square.

    • I think there is a problem with this proof. Squaring B / A = sqrt(N) / 1 and concluding that B^2 / A^2 = N / 1 and then equating the denominator or one fraction as the multiple of another (A^2 * c = 1, so that both A and c = 1) asuumes N is an integer. It should instead be written as B^2 / A^2 = N_B / N_A, where N_B and N_A are integers. Continuing as before yields B^2 * c = N_B and A^2 * c = N_A, which doesn’t help.

      Consider similar reasoning, replacing sqrt(N) with x:
      x / 1 = 3 / 2 (lowest form expression of the fraction)
      x^2 / 1 = 9 / 4 (this is still in lowest form, as you note)
      But it’s not valid to conclude from this that there is some integer c such that
      9 * c = x^2
      4 * c = 1
      (There is no such integer c. Indeed, if we attempt to conclude that c = 1 from this and use that result, we would also conclude that
      9 * 1 = x^2
      x = 3,
      which is clearly not true, as we stated x = 3 / 2 at the outset).

  3. This is a proof by Geoffrey C. Berresford (Am Math Monthly, Vol. 115, No. 6 (June-July 2008), p. 524). For more, see

  4. when I was a TA at UC Riverside about a decade ago the professor who taught pre-calc always did this for sqrt(2) in one of his early lectures.

  5. I learned that proof in the late 1960s, and it was considered an old technique even then. Why someone gets author credit for failing to do library research and reinventing the wheel is beyond me.

  6. The subtle differences seem to be differences in writing style, which has changed a bit over the past 160 years, rather than differences in the math, so I’d argue that it is indeed the same proof, and pretty much the same proof that the Greeks used (though I don’t have any classic Greek to read the original).

    • The second proof relies on “when squaring the numerator and denominator which previously did not have common factors still do not have common factors and is therefore still irreducible”. The first proof instead relies on “if the fractions are in the same equivalence class there is a multiple that makes the numerators and denominators equal”. This is nontrivial.

      Did you check out the rest of that book you linked to? It’s actually pretty neat.

    • I second Jason. These are not the same proof. The proof quoted in the book relies on the fundamental theorem of arithmetic, but the proof given by Jason only relies on the claim “X/Y in lowest terms and X/Y = Z/W implies there is integer c with Z=cX and W=cY.” They probably generalize a bit differently. If not, the equivalence doesn’t seem trivial to me.

  7. “Since B/A is in lowest terms, there’s an integer c such that Bc=Na and ac=b.”

    Sorry I don’t understand your proof (I’m a student learning about proofs), but why is there an integer c? Aren’t you assuming what you’re trying to prove?

    • Just the raw statement
      B/A = NA/B
      implies that the fractions are equivalent, but not that the numerators and denominators are equal, because the fractions might be, say 1/2 and 2/4. However, we do know when fractions are equivalent and one of them is in lowest terms we can multiply the lowest-terms fraction by some integer to get the other fraction. So 1/2 = 2/4 but we can do 1 * 2 / 2 * 2 to get 2/4 = 2/4. Or 2/5 = 6/15 but we can multiple 2 * 3 / 5 * 3 to get 6/15. This only works when one of the fractions is in lowest terms, though; think of 2/4 and 3/6.

      So where did c come from? Well, since we know the fraction property is true for some integer, we can peg that integer at c. We haven’t assumed anything about c other than it is “some integer” so we aren’t assuming anything beyond what we can.

      Now if you’re wanting a proof of that property, it’s fair to ask for one, but it’s considered to be one of those “already proved” things in most proofs of the irrationality of square roots, just like we can assume the fundamental theorem of arithmetic (which I proved a few posts ago) is also true.

  8. This is genius. I love the mathematical thinking behind this proof. I am starting to see the beauty of proofs, where at first I just hated them because I did not know where to start chipping away the problem.


  9. I can provide a counter example sqrt(59)

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