What is Mathematics: Solution Chapter 3

Before the solutions :)

The solution presented on the blog is my personal solutions for the exercises in the book ‘What is Mathematics: An Elementary Approach To Ideas And Methods’ by Herbert Robbins and Richard Courant, please leave a comment if you spot any mistakes or you have questions on the solution. Thanks in advance!

Fundamental Geometrical and Algebra

Since $2^n \rightarrow \infty$, prove as a consequence that $\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}$ (with $n$ square roots) converges to 2 as $n \rightarrow \infty$.
\[\begin{align*} x &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + x \end{align*}\]
And we can get x = 2.
Show that every field contains all the rational numbers at least. (Hint: If $a \neq 0$ is a number in the field $F$ , then $a/a = 1$ belongs to F , and from 1 we can obtain any rational number by rational operations.)
Ex: From $p=1+\sqrt{2}, q=2-\sqrt{2}, r=-3+\sqrt{2}$ obtain the numbers $\frac{p}{q}, \quad p^2, \quad (p-p^2)\frac{q}{r}, \quad \frac{p+qr}{p-r}, \quad \frac{p+r}{q+pr}$ in the form $a+b\sqrt{2}$.
\[\begin{aligned} \frac{p}{q} &= \frac{1+\sqrt{2}}{2-\sqrt{2}} \\ &= \frac{(1+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} \\ &= \frac{2+2\sqrt{2}+\sqrt{2}+2}{4-2} \\ &= \frac{4+3\sqrt{2}}{2} \\ &= 2 + \frac{3\sqrt{2}}{2} \end{aligned}\] \[\begin{aligned} p + p^2 &= (1+\sqrt{2}) + (1+\sqrt{2})^2 \\ &= (1+\sqrt{2} + 1 + 2\sqrt{2} + 2) \\ &= 4 + 3\sqrt{2} \end{aligned}\] \[\begin{aligned} (p-p^2)\frac{q}{r} &= \left[(1+\sqrt{2}) - (1+\sqrt{2})^2\right]\frac{2-\sqrt{2}}{-3+\sqrt{2}} \\ &= \left[1+\sqrt{2} - (1+2\sqrt{2} + 2)\right]\frac{(2-\sqrt{2})(\sqrt{2}+3)}{(-3+\sqrt{2})(\sqrt{2}+3)} \\ &= \left[-2-\sqrt{2}\right]\frac{2\sqrt{2}+6-2+3\sqrt{2}}{2-9} \\ &= \left[-2-\sqrt{2}\right]\frac{4-\sqrt{2}}{-7} \\ &= \frac{6}{7} + \frac{2\sqrt{2}}{7} \end{aligned}\] \[\begin{aligned} \frac{pqr}{1+r^2} &= \frac{(1+\sqrt{2})(2-\sqrt{2})(-3+\sqrt{2})}{1+(-3+\sqrt{2})^2} \\ &= \frac{\sqrt{2}(-3+\sqrt{2})}{1+9-6\sqrt{2}+2} \\ &= \frac{2-3\sqrt{2}}{12-6\sqrt{2}} \\ &= \frac{(2-3\sqrt{2})(12+6\sqrt{2})}{(12-6\sqrt{2})(12+6\sqrt{2})} \\ &= \frac{24+12\sqrt{2}-36\sqrt{2}-36}{144-72} \\ &= \frac{-12-24\sqrt{2}}{72} \\ &= -\frac{1}{6}-\frac{1}{3}\sqrt{2} \\ \end{aligned}\] \[\begin{aligned} \frac{p+qp}{q+pr^2} &= \frac{(1+\sqrt{2})+(2-\sqrt{2})(-3+\sqrt{2})}{(2-\sqrt{2})+(1+\sqrt{2})(-3+\sqrt{2})^2} \\ &= \frac{(1+\sqrt{2})+(-6+2\sqrt{2}+3\sqrt{2}-2)}{(2-\sqrt{2})+(1+\sqrt{2})(11-6\sqrt{2})} \\ &= \frac{-7+6\sqrt{2}}{(2-\sqrt{2})+(5\sqrt{2}-1)}\\ &= \frac{-7+6\sqrt{2}}{1+4\sqrt{2}} \\ &= \frac{(-7+6\sqrt{2})(1-4\sqrt{2})}{(1+4\sqrt{2})(1-4\sqrt{2})} \\ &= \frac{-7+28\sqrt{2}+6\sqrt{2}-48}{1-32} \\ &= \frac{-55+34\sqrt{2}}{-31} \\ &= \frac{55}{31}-\frac{34}{31}\sqrt{2}\\ \end{aligned}\]
Represent $(\sqrt{k})^3 , \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} , \frac{\sqrt{2}\sqrt{k}+\frac{1}{\sqrt{2}}}{(\sqrt{k})^3-3}, \frac{(1+\sqrt{k})(2-\sqrt{k})(\sqrt{2}+\frac{1}{\sqrt{k}})}{1+\sqrt{2}k}$ in the form $p+q\sqrt{k}$.
\[(\sqrt{k})^3 = k\sqrt{k}\] \[\begin{aligned} \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} &= \frac{(1+k)(1-\sqrt{k})}{(1+\sqrt{k})(1-\sqrt{k})} \\ &= \frac{1+\sqrt{k}+k-k\sqrt{k}}{1-k} \\ &= \frac{1+k}{1-k} - \frac{1+k}{1-k}\sqrt{k} \end{aligned}\] \[\begin{aligned} \frac{\sqrt{2}\sqrt{k}+\frac{1}{\sqrt{2}}}{(\sqrt{k})^3-3} &= \frac{(\sqrt{2}\sqrt{k} + \frac{\sqrt{2}}{2})(k \sqrt{k} + 3)}{(k \sqrt{k} - 3)(k \sqrt{k} + 3)} \\ &= \frac{k^2\sqrt{2} + 3\sqrt{2} \sqrt{k} + \frac{\sqrt{2}}{2} k \sqrt{k} + 2 \frac{\sqrt{2}}{2}}{k^3 - 9} \\ &= \frac{k^2\sqrt{2}}{k^3 - 9} + \frac{3\sqrt{2} + \frac{\sqrt{2}}{2} k}{k^3 - 9}\sqrt{k} \end{aligned}\] \[\begin{aligned} \frac{(1+\sqrt{k})(2-\sqrt{k})(\sqrt{2}+\frac{1}{\sqrt{k}})}{1+\sqrt{2}k} &= \frac{(2 - \sqrt{k}+ \sqrt{2k} - \sqrt{k})(\sqrt{2} + \frac{1}{\sqrt{k}})}{1 + \sqrt{2k}} \\ &= \frac{2 \sqrt{2} + \frac{2 \sqrt{k}}{k} + 2 \sqrt{k} - 1 + 2\sqrt{k} + \sqrt{2} - \sqrt{2}k - \sqrt{k}}{1 + \sqrt{2k}} \\ &= \frac{2 \sqrt{2} + \sqrt{k} - \sqrt{2}k - 1}{1 + \sqrt{2}k} + \frac{\frac{2}{k}-\sqrt{2}+1}{1+\sqrt{2}k}\sqrt{k} \end{aligned}\]
If two segments of lengths $1$ and $a$ are given, give actual constructions for $1 + \alpha + \alpha^2$, $(1+\alpha)/(1-\alpha)$, $\alpha^3$.

$5.1$: To construct $1 + a + a^2$, first let’s assume $\alpha>1$ first construct 1 as OA, then extend from A to B with length $\alpha$, therefore we have $1+\alpha$.

the next is to construct $\alpha^2$. We first construct a right triangle OCD, where angle OCD is right angle, OC is 1, and OD is $\alpha$, and then extend OC to OA, the length of OA is $\alpha$, then construct a line from A that is perpendicular to OA, exten OD and the line will intersect with previouly contsructed line on point B. then obviously we have $\triangle OCD \sim \triangle OAB$.

Therefore we can eaisly have the ratio equation that is $\frac{OC}{OA}=\frac{OD}{OB}$, where we can see OB is $\alpha^2$, using compass we then can construct another extension on the basis of $1+\alpha$ and get $1 + a + a^2$.

$5.2$: To construct $(1+\alpha)/(1-\alpha)$, first let’s assume $0<\alpha<1$construct $1+\alpha$, using the same way as described in 5.1, and then construct $1-\alpha$ by constructing unit length 1 OA and OB with length $\alpha$, then we can get $1-\alpha$

Then we construct right triangle, first one side is $OB=1+\alpha$, and another side OA with length $1-\alpha$ which is perpendicular to AB, then we construct OC which is $1$, from point C construct a line CD that is parallel to CD. then we can get $\triangle OAB \sim \triangle OCD$.

Therefore we can eaisly have the ratio equation that is $\frac{OD}{OB}=\frac{OC}{OA}$, which we can gete $\frac{OD}{1+\alpha}=\frac{1}{1-\alpha}$, then we can get OD is $(1+\alpha)/(1-\alpha)$.

$5.3$: To construct $\alpha^3$, construct $\alpha^2$ by using the step in 5.1 and then construct $\alpha^3$ by using the same way using triangle similarity $\triangle OCD \sim \triangle OAB$.
The lines $x + \sqrt{2}y - 1 = 0$, $2x - y + \sqrt{2} = 0$, have coefficients in the field $a + b\sqrt{2}$. Calculate the coordinates of their point of intersection, and verify that these have the form $a + b\sqrt{2}$. Join points $(1, \sqrt{2})$ and $(\sqrt{2}, 1 - \sqrt{2})$ by a line $ax + by + c = 0$, and verify that the coefficients are of the form $a + b\sqrt{2}$. Same with respect to field $p + q\sqrt{k}$ for the lines $\sqrt{(1+\sqrt{2})x} + \sqrt{2} y = 1$, $(1+\sqrt{2})x - y = 1 - \sqrt{1+\sqrt{2}}$, and the points $(\sqrt{2}, -1)$, $(1+\sqrt{2}, \sqrt{1+\sqrt{2}})$ respectively.

To find intersection:
\[\begin{cases} x + \sqrt{2}y - 1 &= 0 \\ 2x - y + \sqrt{2} &= 0 \end{cases}\]
Easily get:
\[\begin{cases} x &= \frac{7}{7} - \frac{3}{7} \sqrt{2} \\ y &= \frac{2}{7} + \frac{3}{7} \sqrt{2} \end{cases}\]
And they are both in the filed of $a + b\sqrt{2}$.

Join 2 points $(1, \sqrt{2})$ and $(\sqrt{2}, 1-\sqrt{2})$, The line is $ax + b = y$. Then we can have:
\[\begin{cases} a + b &= \sqrt{2} \\ \sqrt{2} a + b &= 1 - \sqrt{2} \end{cases}\]
Easily get:
\[\begin{align*} a &= -3 + \sqrt{2} \\ b &= 3 + \sqrt{2} \end{align*}\]
And they are both in the form of $a + b\sqrt{2}$.

To find intersection of:
\[\begin{cases} (1 + \sqrt{2})x + \sqrt{2} y &= 1 \\ (1 + \sqrt{2})x - y &= 1 - (1 + \sqrt{2}) \end{cases}\]
Say $k=1+\sqrt{2}$
\[\begin{cases} \sqrt{(k-1)x+1} + \sqrt{2}y &= 1 \\ kx-y&=1-\sqrt{k} \end{cases}\]
Then we can get
\[\begin{cases} x &= \\ y &= \end{cases}\]
Which are in the form of $p+q\sqrt{k}$

Join $(\sqrt{2}, -1)$, $(1+\sqrt{2}, \sqrt{1+\sqrt{2}})$, with $k = 1+\sqrt{2}$:
\[\begin{cases} \sqrt{2} a + b &= -1 \\ k a + b &= \sqrt{2} k \end{cases}\]
Say $k=1+\sqrt{2}$ Easily get:
\[\begin{cases} a &= \frac{-1 - \sqrt{k}}{\sqrt{2} - k} \\ b &= \frac{-1 \sqrt{k}}{\sqrt{2} - k}\sqrt{2}+1 \end{cases}\]
which is of the form $p + q\sqrt{k}$.
Let $F$ be the field $p + q\sqrt{2 + \sqrt{2}}$, where p and q are of the form $a + b\sqrt{2}$, a, b rational. Represent $\frac{1 + \sqrt{2 + \sqrt{2}}}{2 - 3\sqrt{2 + \sqrt{2}}}$ in this form.

Say $k = 2 + \sqrt{2}$, we can get:
\[\frac{1 + \sqrt{k}}{2 - 3k}\]
Rewriting:
\[\frac{(1 + \sqrt{k})}{(2 - 3\sqrt{k})}\]
Multiplying numerator and denominator by $(2 + 3\sqrt{k})$:
\[\frac{(1 + \sqrt{k})(2 + 3\sqrt{k})}{(2 - 3\sqrt{k})(2 + 3\sqrt{k})}\]
Expanding:
\[\frac{2 + 5\sqrt{k} + 3k}{4 - 9k}\] \[= \frac{2 + 3k}{4 - 9k} + \frac{5}{4 - 9k} \cdot \sqrt{k}\]
Therefore it is the form of field $p+q\sqrt{2+\sqrt{2}}$
Consider the circle with radius $2\sqrt{2}$ about the origin, and the line joining the points $(\frac{1}{2}, 0)$, $(4\sqrt{2}, \frac{\sqrt{2}}{2})$. Find the field $F'$ determined by the coordinates of intersection of the circle and the line. Do the same with respect to the intersection of the given circle with the circle with radius $\frac{\sqrt{2}}{2}$ and center $(0, 2\sqrt{2})$.

$8.1$: Say the line is $ax + b = y$,
\[\begin{cases} \frac{1}{2} a + b = 0, \\ 4\sqrt{2} a + b = \frac{\sqrt{2}}{2}. \end{cases}\]
Easily get the equation of the line,
\[\left( \frac{32}{127} + \frac{2\sqrt{2}}{127} \right) x - \frac{16}{127} - \frac{\sqrt{2}}{127} = y. \quad\]
The equation of the circle is
\[x^2 + y^2 = 8.\]
With (1) and (2), we get
\[x^2 + \left[ \left( \frac{32}{127} + \frac{2\sqrt{2}}{127} \right) x - \frac{16}{127} - \frac{\sqrt{2}}{127} \right]^2 = 8.\]
By calculating it we can get:
\[\begin{cases} x_1 &= \frac{548 + 64 \sqrt{2} - \sqrt{2501773094 - 19209248 \sqrt{2}}}{18257}\\ y_1 &= (-16/2318639 - \sqrt{2}/2318639) (17161 - 128 \sqrt{2} + 2 \sqrt{2501773094 - 19209248 \sqrt{2}}) \\ x_2 &= \frac{548 + 64 \sqrt{2} - \sqrt{2501773094 - 19209248 \sqrt{2}}}{18257} \\ y_2 &= (16/2318639 + \sqrt{2}/2318639) (-17161 + 128 \sqrt{2} + 2 \sqrt{2501773094 - 19209248 \sqrt{2}}) \\ \end{cases}\]
Therefore the field extension F is needed to accommodate these two values is
\[F = \mathbb{Q}(\sqrt{2}, \sqrt{2501773094 - 19209248 \sqrt{2}})\]
$y$ can be gained by the coefficient with $\sqrt{2}$ therefore $y$ is also in the same filed $F$.

$8.2$: For the circle with radius $\frac{\sqrt{2}}{2}$ and center $(0, 2\sqrt{2})$.
\[\begin{cases} C_1 &= x^2 + y^2 = 8 \\ C_2 &= x^2 + (y-2\sqrt{2})^2 = \frac{1}{2} \end{cases}\]
Therefore we can get:
\[\begin{cases} x_1 &= \frac{-3 \sqrt{\frac{7}{2}}}{8} \\ y_1 &= \frac{31}{8 \sqrt{2}}\\ x_2 &= \frac{3 \sqrt{\frac{7}{2}}}{8} \\ y_1 &= \frac{31}{8 \sqrt{2}} \end{cases}\]
The field is $F = \mathbb{Q}(\sqrt{2}, \sqrt{7})$
Verify that, starting with the rational field, the side of the regular $2^m$-gon is a constructible number, with $n = m - 1$. Determine the sequence of extension fields. Do the same for the numbers: $\begin{align*} &\sqrt{1+ \sqrt{2}+ \sqrt{3} + \sqrt{5}}, \\ &\frac{(\sqrt{5} + 11)}{(1+\sqrt{7-\sqrt{3}})} \\ &(\sqrt{2} + 3)(\sqrt[3]{2} + \sqrt{1+\sqrt{2+\sqrt{5}}} + \sqrt{3-\sqrt{7}}) \end{align*}$

The side of a regular $2^m$-gon is given as: $\begin{equation*} 2^n \cdot \sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt{2}}}} \end{equation*}$

where there are $n$ square roots.

Starting with the rational field $F_0$, let $k_0 = 2$, obtaining: $\begin{equation*} F_1 = 2 + \sqrt{2}. \end{equation*}$ Similarly, setting $k_1 = \sqrt{2 + \sqrt{2}}$, we obtain $F_2$, and continuing this process iteratively $n$ times, we reach the desired expression.

Since $2^n$ is inside the field $F_n$ by a product of $2$, the number is constructible.

Let $F_0$ denote the rational field. $\begin{align*} &F_0: \text{rational numbers}, \\ &k_0 = 2 \Rightarrow F_1 = 1 + \sqrt{2}, \\ &k_1 = \sqrt{3} \Rightarrow F_2 = 1 + \sqrt{2} + \sqrt{3}, \\ &k_2 = 3 \Rightarrow F_3 = 1 + \sqrt{2} + \sqrt{3} + \sqrt{5}. \end{align*}$

Thus, continuing this extension process, we obtain a constructible number.
Find the equations with rational coefficients for:$(a) x = 2 + \sqrt{3}$, $(b) x = \sqrt{2} + \sqrt{3}$, $(c) x = \frac{1}{\sqrt{5} + \sqrt{3}}$

(a) $x = 2 + \sqrt{3}$
\[x - \sqrt{2} = \sqrt{3}\]
Squaring both sides:
\[(x - 2)^2 = (\sqrt{3})^2\] \[x^2 - 4x + 4 = 3\] \[x^2 - 4x + 1 = 0\]
(b) $x = \sqrt{2} + \sqrt{3}$
\[x^2 = 2 + 2\sqrt{6} + 3\]
Squaring again:
\[(x^2 - 5)^2 = 24\] \[x^4 - 10x^2 + 25 = 24\] \[x^4 - 10x^2 + 1 = 0\]
(c) $x = \frac{1}{\sqrt{5} + \sqrt{3}}$
\[\frac{1}{x^2} = 5 + \sqrt{3}\] \[\frac{1}{x^2} - 5= \sqrt{3}\] \[\frac{1}{x^4} - 10\frac{1}{x^2} + 25=3\]
Rearrange:
\[22x^4 - 10x^2 + 1=0\]
Find by a similar method equations of the eighth degree for:$(a) x = \sqrt{2 + \sqrt{2 + \sqrt{2}}}, (b) x = \sqrt{2} + \sqrt{1 + \sqrt{3}}, (c) x = 1 + \sqrt{5 + \sqrt{3+\sqrt{2}}}$

(a) $x = \sqrt{2 + \sqrt{2 + \sqrt{2}}}$
\[x^2 = 2 + \sqrt{2 + \sqrt{2}}\] \[(x^2 - 2)^2 = 2 + \sqrt{2}\] \[x^4-4x^2+4-2=\sqrt{2}\] \[x^4 - 4x^2 + 4 = 2 + 2\sqrt{2}\]
Squaring again:
\[(x^4 - 4x^2 + 2)^2 = 2\]
Rearrange:
\[x^8 - 8x^6 + 20x^4 - 16x^2 + 2 = 0\]
(b) $x = \sqrt{2} + \sqrt{1 + \sqrt{3}}$
\[(x-\sqrt{2})^2 = 1+\sqrt{3}\] \[x^2-2\sqrt{2}x+1=\sqrt{3}\] \[x^2 +1 - \sqrt{3} = 2\sqrt{2}x\] \[(x^2+1-\sqrt{3})^2 = 8x^2\]
Rearrange:
\[(x^2+1)^2-2\sqrt{3}(x+1)+3=8x^2\] \[(x^4-6x^2+4)^2 = 3[2(x+1)]^2\]
Rearrange:
\[x^8-12x^6+32^4-72x x^2+4=0\]
(c) $x = 1 + \sqrt{5 + \sqrt{3 + \sqrt{2}}}$
\[(x-1)^2= 5 + \sqrt{3 + \sqrt{2}}\] \[(x^2-2x-4)^2 = 3 + \sqrt{2}\] \[((x^2-2x-4)^2-3)^2 = 2\]
Rearrange:
\[x^8-8x^7+8x^6+64x^5-86x^4-232x^3+152x^2+416x+167=0\]
To prove the theorem for $x$ in a field $F_k$ with arbitrary $k$, we use induction. The goal is to show $x$ satisfies an equation of degree $2^k$ with coefficients in $F_k$. The statement for $k=l$ completes the proof.

Base Case $l = 1$:

For $l = 1$, we need to show that $x$ satisfies a quadratic equation over $F_{k-1}$. Since $F_k$ is a quadratic extension of $F_{k-1}$, any element $x \in F_k$ can be written as $a + b\sqrt{w}$, where $a, b \in F_{k-1}$. This element $x$ satisfies the quadratic equation:
\[(x - a)^2 = b^2 w\]
which is a degree $2^1$ equation over $F_{k-1}$. Thus, the base case holds.

Inductive Step: $l=k$ Assume that for some $l \geq 1$, every element $x \in F_k$ satisfies an equation of degree $2^l$ over $F_{k-l}$. We need to show that $x$ satisfies an equation of degree $2^{l+1}$ over $F_{k-(l+1)}$.

Since $F_{k-l}$ is a quadratic extension of $F_{k-l-1}$, we can write $F_{k-l} = F_{k-l-1}(\sqrt{w})$ for some $w \in F_{k-l-1}$. By the induction hypothesis, $x$ satisfies a polynomial $P(x) = 0$ of degree $2^l$ with coefficients in $F_{k-l}$. Each coefficient of $P(x)$ can be written as $c_i + d_i \sqrt{w}$, where $c_i, d_i \in F_{k-l-1}$. Separating the polynomial into terms with and without $\sqrt{w}$, we get:
\[Q(x) + \sqrt{w} R(x) = 0\]
where $Q(x)$ and $R(x)$ are polynomials with coefficients in $F_{k-l-1}$. Applying the automorphism $\sigma$ (which sends $\sqrt{w}$ to $-\sqrt{w}$) gives:
\[Q(x) - \sqrt{w} R(x) = 0\]
Multiplying these two equations eliminates $\sqrt{w}$:
\[(Q(x))^2 - w (R(x))^2 = 0\]
This results in a polynomial equation of degree $2 \cdot 2^l = 2^{l+1}$ over $F_{k-l-1}$. Thus, the inductive step holds.

By induction, for each $l$ from $1$ to $k$, $x$ satisfies an equation of degree $2^l$ over $F_{k-l}$. When $l = k$, this gives an equation of degree $2^k$ over $F_0$, which is the desired theorem.
Exercise: Show that this construction actually yields $y =x/3$.

Based on the graph in page 138:
\[\angle BOA + \angle BOC + \angle COD = 180^\circ\] \[\angle OBC = \angle OCB = 2y\]
So we can have:
\[\begin{aligned} y + (180^\circ-4y)+x&=180^\circ\\ y&=x/3 \end{aligned}\]
Exercises: The following ia a description of Mohr’s constructions. Check their vadility, Wht do they solve the Mascheroni problem?

1). On a segment AB of length p erect a perpendicular segment BC. (Hint: Extend AB by a point D such that AB = BD. Draw arbitrary circles around A and D and thus determine C .)

2). Two segments of length $p$ and $q$ with $p > q$ are given somewhere in the plane. Find a segment of the length $x=\sqrt{p^2-q^2}$ by making use of 1).

3). From a given segment a construct the segment $a/ \sqrt{2}$. (Hint: Observe that $(a\sqrt{2})^2 = (a\sqrt{3})^2 -a^2$.)

4). With given segments p and q find a segment $x=\sqrt{p^2-q^2}$. (Hint: Use the relation $x^2 =2p^2 — (p^2 — (p^2-q^2)$.) Find other similar constructions.

5) Using the previous results, find segments of length $p + q$ and $p — q$ if segments of length $p$ and $q$ are given somewhere in the plane.

6) Check and prove the following construction for the midpoint $M$ of a given segment $AB$ of length $a$. On the extension of $AB$ find $C$ and $D$ such that $CA = AB = BD$. Construct the isosceles triangle $ECD$ with $EC = ED = 2a$, and find $M$ as the intersection of the circles with diameters $EC$ and $ED$.

7) Find the orthogonal projection of a point $A$ on a line $BC$.

8) Find $x$ such that $x:a=p:q$. if $a$, $p$ and $q$ are given segments.

9) Find $x = ab$, if $a$ and $b$ are given segments.

What is Mathematics: Solution Chapter 3

Before the solutions :)

Fundamental Geometrical and Algebra

Constructible Number and Number Fields

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