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I am trying to prove that the set of all ruler-and-compass constructible points is a subfield of $R$. I have one difficulty though:

  • Given two numbers $\sqrt{a}$ and $\sqrt{b}$ , how can I construct their product and their inverse with rule and compass?
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  • $\begingroup$ How far have you gotten? Where are you stuck? $\endgroup$ Commented Apr 21, 2017 at 19:23
  • $\begingroup$ Actually I have no idea where to start in the product case. In the inverse case I found out a good idea. $\endgroup$
    – João
    Commented Apr 21, 2017 at 19:43
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    $\begingroup$ For product and inverse you also need to know a segment of unit length or how to construct one from the two numbers you have. $\endgroup$
    – g.kov
    Commented Apr 21, 2017 at 19:47
  • $\begingroup$ @JoãoGabriel It would be helpful if you include all ideas you have thought of in the question. $\endgroup$
    – shardulc
    Commented Apr 21, 2017 at 20:04
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    $\begingroup$ For the inverse: math.stackexchange.com/questions/1090733/… And the product: math.stackexchange.com/questions/139340/… $\endgroup$ Commented Apr 21, 2017 at 20:09

3 Answers 3

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Think similar triangles. Make a $1, \sqrt a$ right triangle and then a similar one that is $\sqrt b, \sqrt{ab}$

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One way to construct a product, an inverse and also a square root is to use Intersecting chords theorem:

enter image description here

But we also need to know a segment of unit length.

For example, if $|AB|=a$, $|BC|=b$, and a unit length $|BD|=1$, then $a\,b=|BE|$. But if a unit length is instead $|BE|=1$, then $a\,b=|BD|$.

The same construction can be used to get an inverse of a number as well as a square root.

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In the next picture it is $AM×BM=1$ enter image description here Proof: The figure formed by the tangent to the unitary circle and the two parallel lines tangent to the unitary circle itself, can be completed to a rhombus whose sides touch the unitary circle, we know that the center of the circle tangent to the sides of a rhombus is the point of intersection of its diagonals and we know that the two diagonals of the rhombus are perpendicular

So $AO⊥BO$.

We know that the tangent to a circle at a point is perpendicular to the radius of the circle that includes the point of tangency

So $OM⊥AB$.

$OM$ is the height of the hypotenuse in a right triangle $∆AOB$, so: $OM^2=AM⋅BM$ $⇒AM×BM=1$

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