Exercise 10.1.3

Proof.

(a)

Let $D$ be the line passing through $0,\alpha ,\beta$, and $C$ the circle of radius $|\beta |$ with center $\alpha$. Then $D$ and $C$ intersect in two points, one of them being $\alpha +\beta$, which is so constructible.

(b)

As $a$ is constructible, so is $\mathit{ia}$. We can construct the parallel to the line $1,\mathit{ia}$ passing through $i$. The intersection point of this parallel with the $x$-axis gives a point $d>0$. As the triangles $(0,d,i)$ and $(0,1,\mathit{ia})$ are similar, $d∕1=i∕\mathit{ia}$, so $1∕a=d$ is constructible.

(c)

We have proved in the text that $𝒞$ is a subgroup of $(ℂ,+)$, so $𝒞\cap ℝ$ is a subgroup of $(ℝ,+)$.

Let $a,b\in 𝒞\cap ℝ$. If $a=0$ or $b=0$ then $a+b\in 𝒞\cap ℝ$. If $a\ne 0,b\ne 0$ then $|a|=\pm a\in 𝒞\cap {ℝ}_{\text{+}}^{\text{∗}}$, so $|\mathit{ab}|=|a||b|\in 𝒞\cap {ℝ}_{\text{+}}^{\text{∗}}$, therefore $\mathit{ab}=\pm |\mathit{ab}|\in 𝒞\cap ℝ$.

Let $a\in 𝒞\cap ℝ,a\ne 0$. then $|a|\in 𝒞\cap {ℝ}_{\text{+}}^{\text{∗}}$, so $1∕|a|\in 𝒞\cap {ℝ}_{\text{+}}^{\text{∗}}$. Therefore $1∕a=\pm 1∕|a|\in 𝒞\cap ℝ$.

Conclusion: $𝒞\cap ℝ$ is a subfield of $ℝ$.

(d)

We can construct the orthogonal line $D$ to the $x$-axis passing through $1$, the perpendicular bisector of $(0,2)$. As $r$ is constructible, so is $1+r$ (Exercise 10.1.3(a)). The intersection of the perpendicular bisector of $(0,1+r)$ with the $x$-axis gives the number $(1+r)∕2$, which is so constructible. The intersection of the circle with radius $(1+r)∕2$ centered in $(1+r)∕2$ with $D$ gives $\beta$ with positive $y$-coordinate, and so $\beta$ is constructible.

The end of the proof of Theorem 10.1.4 shows that $\sqrt{r}$ is constructible.