Does someone know a proof (books, articles) that $\pi$ is not a quadratic irrational?

The proof should not use that $\pi$ is transcendental.

Any hints would be appreciated.

Martin Sleziak
  • 50,316
  • 18
  • 169
  • 342
  • 4,928
  • 2
  • 18
  • 42
  • 8
    Niven's proof, which is widely available, shows that $\pi^2$ is irrational (which isn't quite what you want, but it's a start). – Gerry Myerson Aug 18 '13 at 08:11
  • 3
    Shameless self-reference : http://math.stackexchange.com/questions/713467/proving-that-e-is-irrational/879692#879692. As a simple continued fraction is periodic if and only if it is a CF of some irrational quadratic, this proof essentially proved that $\pi$ is not a quadratic irrational. – Balarka Sen Aug 09 '14 at 08:15
  • @Balarksa Sen ,Sorry, but all I see in your link is a proof that the CF for $\tanh (1)$ is not periodic. which only implies that $\tanh (1)$ and $e$ are not quadratic over $Q$. Did I miss something? – DanielWainfleet Mar 09 '16 at 06:54
  • 1
    Lambert proved $\pi ^2$ is irrational , by use of $\tan (x) = \frac{x}{1-\frac {x^2}{3-\frac{x^2}{5-\frac{x^2}{...}}}}$ –  Mar 11 '16 at 16:05
  • Would the proof that $\pi$ is trancendental be sufficient? It doesn't *use* the fact, it proves it – Yuriy S Mar 15 '16 at 10:03
  • http://math.stackexchange.com/questions/103786/direct-proof-that-pi-is-not-constructible – Yuriy S Mar 15 '16 at 18:05
  • This is exactly the same question (also marked as duplicate) http://math.stackexchange.com/questions/1008993/is-there-a-direct-proof-that-pi-is-not-the-root-of-an-algebraic-equation-whose-d?lq=1 – Yuriy S Mar 15 '16 at 18:11

1 Answers1


As mentioned in the comments Lambert proved that $\pi$ is irrational (http://www.pi314.net/eng/lambert.php). Lambert showed that it's irrational by first demonstrating that the continued fraction expansion holds: $$\tan(x) = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - {}\ddots}}}}.$$ He then proved that if $x$ is non-zero and a rational number then the expression must be irrational. Consequently because $\tan(\frac {\pi}{4}) = 1$, it follows that $\frac {\pi}{4}$ is irrational and as a result that $\pi$ is irrational. (Refer to the Wikipedia article on the proof of $\pi$ being irrational by Lambert - Lambert's Proof)

Laczkovich's Proof is a simplification of Lambert's proof (refer here for more information).

I found a few resources detailing numerous proofs of $\pi$ being irrational, none of them utilising $\pi$ as transcendental. The Wikipedia article lists 6 different proofs (including Lambert's and Laczkovich's). There was also a question asked on this math forum about understanding the following proof of $\pi$ being irrational making use of polynomials and calculus (A simple proof that $\pi$ is irrational).

I hope this helps.

  • 2,267
  • 1
  • 22
  • 34
  • 3
    This doesn't answer the question... OP wants to know if there's a simple proof that $\pi$ isn't a quadratic irrational, i.e., an irrational root of a quadratic with rational coefficients. – mjqxxxx Mar 14 '16 at 16:59