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mathematically can someone please explain why squaring the circle was proved impossble by the way that they were attempting to solve it?

Matthew Conroy
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Curios and Knowledgeable
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    See Squaring the circle : "In 1882, the task was proven to be impossible, as a consequence of the Lindemann–Weierstrass theorem which proves that pi (π) is a transcendental, rather than an algebraic irrational number; that is, it is not the root of any polynomial with rational coefficients" – Mauro ALLEGRANZA Dec 04 '16 at 19:23
  • Hi I've lent it to someone but there's a great little book which starts with explaining the classical construction problems from scratch (and then gets to galois theory, a real gem). HIghly recommended: https://www.amazon.co.uk/Classical-Problems-Mathematical-Association-Textbooks/dp/088385032X – Mehness Dec 04 '16 at 19:25
  • Possible duplicate of this – Vidyanshu Mishra Dec 04 '16 at 19:49

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The result relies fundamentally on the theory of field extensions of $\mathbb{Q}$. Basically, what you need to observe is that a point in the plane is constructible in one step if and only if it lies in $K_1^2\subset\mathbb{R}^2$ where $K$ is a field extension of $K_0=\mathbb{Q}$ obtained by adjoining roots of a quadratic polynomial with coefficients in $K_0=\mathbb{Q}$. Then you show that a point is constructible in $n$-steps if and only if it lies in $K_n^2\subset\mathbb{R}^2$ where $K_n$ is a field obtained by adjoining a root of a quadratic polynomial with coefficients in some $K_{n-1}$. That is, we just iterate the process of adjoining roots of a quadratic polynomial to our current field $n$ times.

The important point is that whenever you adjoin a root of a degree two polynomial in this way, you create a field extension whose degree is at most $2$ over the base field. So any field obtained via this procedure is a finite field extension of $\mathbb{Q}$.

Then once you know that $\pi$ is trancendental over $\mathbb{Q}$, it follows that $\pi$ doesn't lie in any field extension of $\mathbb{Q}$ with finite degree. Therefore, the point $(0,\pi)$ cannot be constructed in finitely many steps with a ruler and compass.

D Wiggles
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