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I've heard that when $\pi$ was proved irrational, that squaring the circle was not proved impossible. This lead me to believe that you could construct a square with an irrational area. Is this possible?

It is possible to create a polynomial with integer coefficients that satisfies an irrational area ($x^4=2$), so that leads me to believe it is doable, although my knowledge of the precise relationship between polynomial and areas is sketchy at best.

Also, it is easy to construct a square of irrational side lengths (consider 4 1x1 squares in a grid, you can make a square side length $\sqrt{2}$ using the diagonals) but a square with an irrational area eludes me yet it seems a possible extension.

What about any shape with an irrational area, is that possible to construct? Does that shape have to be regular to be constructible?

Gridley Quayle
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  • Your question specifies constructing a square, but you later ask "what about any shape with an irrational area". The latter is even easier: Euclid, Book 1, Prop.1 shows how to construct an equilateral triangle of side 1. Its area is $\sqrt3/4$. – Rosie F Feb 09 '20 at 21:31

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It's easy to construct a square with a lenght of a side of $1+\sqrt{2}$, i.e. with an irrationnal area of $3+2\sqrt{2}$

Tryss
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  • Don't know how I missed that, thanks :). Is it possible to construct a non-regular shape of irrational area? – Gridley Quayle Feb 19 '15 at 17:27
  • @GridleyQuayle Cut the square in half along a diagonal, or take a quarter of the area by joining a vertex to the middle of an appropriate side to give three distinct side lengths. – Mark Bennet Feb 19 '15 at 17:49
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Once you have decided on a unit length, you can construct square roots of arbitrary line segments. In particular you can create a line segment of length $\sqrt{\sqrt 2}$ and then erect its square -- which of course will have area $\sqrt 2$.

Merely showing that $\pi$ is irrational does not prove that the circle can't be squared. This was only proved as a result of $\pi$ being transcendental, since all constructible numbers are algebraic.

hmakholm left over Monica
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    Do you know the proof for all constructional numbers being algebraic? – Gridley Quayle Feb 19 '15 at 17:21
  • @GridleyQuayle: By induction on the length of the construction. At each step your figure will consist of some lines that have equations with algebraic coefficients, and some circles with algebraic centers and radii. Every basic geometric operation (draw a new thing, or find the intersection between two things you already know) preserves that property. – hmakholm left over Monica Feb 19 '15 at 17:26
  • It is in Pierre Wantzel's 1837 proof. As all constructible ratios are solutions to certain polynomial equations with certain properties, all constructible ratios are algebraic numbers. – Michael Ejercito Feb 29 '24 at 22:58
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Construct a square.

Label the points $A,B,C$, and $D$ clockwise around the square

Then, place the needle of the compass on corner $A$, and construct an arc length connecting the corner $C$ to a line passing through points $A$ and $B$ outside the square. Call this intersection between the arc and this line point $E$

Then use the compass to construct a square from the line segment between points $E$ and $B$

This new square has a side length $1+\sqrt2$ times the length of the side of the first square.

So its area is $(1+\sqrt2)^2=3+2\sqrt2$, which is irrational.

Michael Ejercito
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