Geometric Constructions

Geometric Constructions
In ancient Greece, there were many mathematicians who were concerned with the possibility of creating figures using only a straightedge and a compass. Now, after I read the chapter and worked through some of the exercises, the question that I felt had not been answered was this: What is a straightedge and compass, and how are they used? So, for the first part of my talk, I will deviate from the book and give you some context to work with. Let’s start with some definitions.

Vocabulary: Geometric Constructions
Definition: A Geometric Construction is a two-dimensional pattern that can be drawn using a straightedge and compass in a finite number of steps. Geometric Constructions depend on being accurate and precise. You are not allowed to measure angles with a protractor or measure lengths with a ruler.

Vocabulary: Geometric Constructions
In geometry, constructions utilize only two tools - the straightedge (an unmarked ruler) and the compass. Never draw freehand when doing a construction!

Vocabulary: Compass Definition: A compass is a tool which allows the user to draw a circle of any radius at any given point.

What can I do with a compass?
You can do two things with a compass:

You can do two things with a compass: Given one point in the plane, you may draw a circle, but you do not know the size.

You can do two things with a compass: Given one point in the plane, you may draw a circle, but you do not know the size. Given two points in the plane, you may draw a circle which has one point as the center and the other point on the boundary.

Use of the compass does not, in itself, imply that you know the radius of the circle.

Use of the compass does not, in itself, imply that you know the radius of the circle. The compass cannot be used to find angles.

Use of the compass does not, in itself, imply that you know the radius of the circle. The compass cannot be used to find angles. The compass collapses whenever it isn’t drawing a circle, so it cannot be used to copy distances. Of course, it has been proven that even if you allow the compass to copy distances, no new constructions are made possible. Still, we will work with the original rules.

Vocabulary: Straightedge
Definition: A straightedge is a tool which allows the user to draw a straight line of any length.

What can I do with a straightedge?
Given one point in the plane, you may draw a line through it, though you may not know its slope or its relationship to other points.

Given one point in the plane, you may draw a line through it, though you may not know its slope or its relationship to other points. Given two points in the plane, you may draw a line which connects them and extends as far as you like.

A straightedge is not a ruler.

A straightedge is not a ruler. A straightedge has no markings and cannot be marked.

A straightedge is not a ruler. A straightedge has no markings and cannot be marked. Also, a straightedge is only a single edge, and has no end which can be used to draw right angles.

A straightedge is not a ruler. A straightedge has no markings and cannot be marked. Also, a straightedge is only a single edge, and has no end which can be used to draw right angles. .

Vocabulary: Points How can we create a point in the plane?

Vocabulary: Points How can we create a point in the plane?
If we like, we can specify an arbitrary point, though we will not know its relationship to other points or lines.

Vocabulary: Points How can we create a point in the plane?
If we like, we can specify an arbitrary point, though we will not know its relationship to other points or lines. We usually create a point by one of three methods: intersecting two lines, intersecting a circle and a line, or intersecting two circles. The new point is the point of intersection. Now that we have the terminology, let’s go through some examples.

Vocabulary: Inscribe Used to draw one figure inside another so that it lies entirely within the boundary of the outer. Now that we have the terminology, let’s go through some examples.

Vocabulary: Arc Is a curve between two points on a circle.
Now that we have the terminology, let’s go through some examples.

A part cut off from a figure.
Vocabulary: Segment A part cut off from a figure. Now that we have the terminology, let’s go through some examples.

Two line segments or two rays that have a common endpoint.
Vocabulary: Angle Two line segments or two rays that have a common endpoint. Now that we have the terminology, let’s go through some examples.

Cut’s a figure in half equally
Vocabulary: Bisect Cut’s a figure in half equally Now that we have the terminology, let’s go through some examples.

Vocabulary: Perpendicular
Lines that intersect to form a right angle Now that we have the terminology, let’s go through some examples.

All sides are equal. Vocabulary: Regular
Now that we have the terminology, let’s go through some examples.

Vocabulary: Equilateral
A figure having all sides equal. Now that we have the terminology, let’s go through some examples.

Lines that never intersect
Vocabulary: Parallel Lines that never intersect Now that we have the terminology, let’s go through some examples.

Vocabulary: Circle A closed plane curve consisting of all points at a given distance from a point within it, called the center. Now that we have the terminology, let’s go through some examples.

Vocabulary: Protractor
Used to measure angles. Now that we have the terminology, let’s go through some examples.

Suggestions for working with a compass:
Place several pieces of paper under your worksheet to allow the compass point to remain stable. Hold the compass lightly and allow the wrist to remain flexible. If you cannot manage to move your wrist when drawing circles, try rotating the paper under the compass.

Geometric Constructions: Example 1
Suppose that I am given a line and I wish to find a perpendicular bisector at a given point on the line.

Step 1: Draw a circle of any radius around the point. This creates two more points, equidistant from the first.

Step 2: Draw circles around the two new points, with radius equal to the distance between them.

Step 2: Draw circles around the two new points, with radius equal to the distance between them. This creates two new points where the circles meet.

Step 3: Connect the two newest points with a line. Done!

The same method can be used to bisect a length; that is, find the midpoint between two points.

Draw two circles. With one point as the center and the other on the boundary.

Draw two circles.

This creates two new points.

Now connect the two new points with a line. Done!

Now connect the two new points with a line. Done! We will do one more basic example, and then a more complicated one.

For something different, suppose now that I am given a circle and would like to find its center.

First, draw any line which each intersects the circle twice.

Now draw a perpendicular bisector for the segment of the line inside the circle.

The midpoint of this line segment is the center of the circle.

Angle Bisection Another very important thing we can do with a straightedge and compass is bisect an angle.

With these abilities, we can create a pentagon.

Start with a circle and any diameter.

Find the perpendicular bisector. Note the corners.

Find the midpoint of the top half of the new line.

Connect the new point with the point on the edge.

Bisect the angle.

Find the perpendicular at the new point.

This gives us the first two corners of the pentagon.

Rinse and repeat with the new diameter.

Connect the dots.

Done! Of course, there are obvious ways to do this in fewer moves, but this way looks cooler.

Of course, it remains to prove that this is in fact a regular pentagon. Students are not responsible for this proof. I show it merely to demonstrate what sort of thing is involved.

It suffices to show that the bolded angle is 2π/5. Is this clear? (If not, I’ll try to explain.) Since we are only concerned with angles, the radius of the circle doesn’t matter. Let’s assume radius 1.

Notice that this angle is arctan 2. The vertical segment has length ½ and the third side has length 1. The tangent of the angle, therefore, is 2.

Then this angle is (arctan 2)/2. Clearly.

So this length is We have the angle for the small triangle. The vertical side has length ½. The tangent of the angle is the length of this side divided by ½ (multiplied by 2), so twice the length is equal to half the tangent of the angle.

Therefore this angle is The hypotenuse of this triangle has length 1, being the radius of the circle, and we have the length of the bottom side. So the cosine of the angle is simply the length of the bottom side.

Now, to simplify: Now we just simplify.

Now, to simplify: Using the half-angle formula for the tangent, this becomes Since the ½ is in there, the first step is to look up the half-angle formula for the tangent. This changes the problem into this, which is good!

Now, to simplify: Using the half-angle formula for the tangent, this becomes Drawing the triangle reduces this to: Finding the cosine of the arctangent is easy enough. If the tangent of an angle is 2, then the triangle (WLOG) has opposite length 2 and adjacent length 1. This means the hypotenuse has length root 5, and the cosine of the angle must be one over root 5. We’ve done this in calculus.

Now, to simplify: Using the half-angle formula for the tangent, this becomes Drawing the triangle reduces this to: Which simplifies to: Multiply the top and bottom of the fraction by root 5, and bring the 2 into the root.

Now, to simplify: And further to: Multiply the top and the bottom of the fraction by root 5 minus 1.

Now, to simplify: And further to: Which is, of course, (This took me forever to figure out.)

Now, to simplify: And further to: Which is, of course, Which is good.

Constructible Numbers
Since all of these things happen in the two-dimensional plane, the clear next step is to define exactly what we’re doing in terms of classical algebra. To do so, we restrict our attention to finding points and numbers in the plane. Those that we can find will be called constructible numbers. Now we leave behind the methods of the Greeks.

Vocabulary: Constructible Numbers
Definition: We call a number x Constructible if, given a compass, straightedge, and line segment of length 1, we can construct a line segment of length x in a finite number of steps.

It is fairly straightforward to confirm that, if x and y are constructible, then so are x + y, x – y, xy and x/y. The next example shows a construction for the square root of a number. These are exercises in the book.

Given x > 0, I will show a way to construct the square root of x. This is also an exercise from the book. x

Lengthen the line segment by 1. 1 x

Find the midpoint of the extended segment and draw a circle. 1 x

Draw a perpendicular and connect the dots. 1 x

Now a2 + b2 = (x + 1)2; b a c I won’t show it, but this is actually a right triangle. 1 x

Now a2 + b2 = (x + 1)2; 1 + c2 = a2; b a c 1 x

Now a2 + b2 = (x + 1)2; 1 + c2 = a2; and x2 + c2 = b2. b a c 1 x

Now a2 + b2 = (x + 1)2; 1 + c2 = a2; and x2 + c2 = b2. This becomes: 1 + 2c2 + x2 = (x + 1)2 Substituting in a squared and b squared in the first equation.

Simplifying: 1 + 2c2 + x2 = (x + 1)2 Foil out the right side.

Simplifying: 1 + 2c2 + x2 = (x + 1)2 1 + 2c2 + x2 = x2 + 2x + 1 Cancel x squared and 1 on both sides.

Simplifying: 1 + 2c2 + x2 = (x + 1)2 1 + 2c2 + x2 = x2 + 2x + 1 2c2 = 2x Divide both sides by 2.

Simplifying: 1 + 2c2 + x2 = (x + 1)2 1 + 2c2 + x2 = x2 + 2x + 1 2c2 = 2x c2 = x. There.

Therefore c is the square root of x. b a c Of course, you might be asking yourself, what good does it do to put a square root there? What if we want it somewhere else? 1 x

We might wonder how we can set a line segment where we want it. Suppose we want to duplicate the given length at the point. C A This is taken from Euclid’s Elements. B

Connect the point to one end of the segment. Draw circles around the ends of this new line. C A B

Find the point created by the intersection and connect it with lines. C D A Incidentally, this creates an equilateral triangle. B

Draw a circle and lengthen the side connected to the original length. C D A B E

Draw a circle around the top of the little triangle intersecting that last point. C D A B E

Extend the other side of the triangle to meet the outer circle. C D A B F E

That final extension is the required duplicated length. C D A B F E

That final extension is the required duplicated length. C A B F

Of course we can take it in any direction by drawing a circle. This is why some choose to allow the compass to copy lengths.

So now, given a unit length, we are able to construct all rational numbers, square roots of rationals, and combinations of such objects.

So now, given a unit length, we are able to construct all rational numbers, square roots of rationals, and combinations of such objects. Can we construct every number?

Review: [E:F] Does everyone remember what this notation means? (For fields?)

Review: [E:F] Definition: Let E be an extension of a field F. We say that E has degree n over F and write [E:F] = n if E has dimension n as a vector space over F.

Review: [E:F] Definition: Let E be an extension of a field F. We say that E has degree n over F and write [E:F] = n if E has dimension n as a vector space over F. So, for example, [C:R] = 2,

Review: [E:F] Definition: Let E be an extension of a field F. We say that E has degree n over F and write [E:F] = n if E has dimension n as a vector space over F. So, for example, [C:R] = 2, [Q( ):Q] = 5,

Review: [E:F] Definition: Let E be an extension of a field F. We say that E has degree n over F and write [E:F] = n if E has dimension n as a vector space over F. So, for example, [C:R] = 2, [Q( ):Q] = 5, and [R:Q] is infinite.

Recall that the way to construct a point in the plane is to intersect two lines, two circles, or a circle and a line. This are the ONLY ways. This is important.

Recall that the way to construct a point in the plane is to intersect two lines, two circles, or a circle and a line. Suppose that we are working over a field F. Then a line in F is of the form ax + by + c = 0; where a, b, c є F;

Recall that the way to construct a point in the plane is to intersect two lines, two circles, or a circle and a line. Suppose that we are working over a field F. Then a line in F is of the form ax + by + c = 0; where a, b, c є F; And a circle in F is of the form x2 + y2 + ax + by + c = 0; where a, b, c є F.

Using the formula for lines, it is easy to check that the point of intersection of any two lines is in F. ax + by + c = 0 a’x + b’y + c’ = 0

Using the formula for lines, it is easy to check that the point of intersection of any two lines is in F. ax + by + c = 0 a’x + b’y + c’ = 0 The second equation becomes ax + (a/a’)b’y + (a/a’)c’ = 0 Multiply by a/a’.

Using the formula for lines, it is easy to check that the point of intersection of any two lines is in F. ax + by + c = 0 a’x + b’y + c’ = 0 The second equation becomes ax + (a/a’)b’y + (a/a’)c’ = 0 Subtracting gives (b – ab’/a’)y + c – ac’/a’ = 0. This equation can be solved to give y in terms of a, a’, b, b’, c, c’; and then using that we can find x. The intersection of two lines in F is a point in F.

Taking two circles in F, x2 + y2 + ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0

Taking two circles in F, x2 + y2 + ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0 Solving for y in the first equation gives the quadratic formula, and substituting into the second equation allows us to find specific values for x.

Taking two circles in F, x2 + y2 + ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0 Solving for y in the first equation gives the quadratic formula, and substituting into the second equation allows us to find specific values for x. Because of the square root in the quadratic formula, the points of intersection may not be in F.

Taking a line and a circle in F, ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0

Taking a line and a circle in F, ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0 Solving for y in the first equation and substituting into the second allows us to find specific values for x.

Taking a line and a circle in F, ax + by + c = 0 x2 + y2 + a’x + b’y + c’ = 0 Solving for y in the first equation and substituting into the second allows us to find specific values for x. Again this requires the quadratic formula and points of intersection may not be in F.

Solving these systems requires only addition, subtraction, multiplication, division, and taking square roots. (Pause for effect.)

Solving these systems requires only addition, subtraction, multiplication, division, and taking square roots. Since these are the only ways to find points, we are not able to construct numbers which require other operations, such as taking cube roots. (Pause for effect.)

A real number c is constructible if and only if there is a series of subfields of the reals such that Fi+1= Fi( ) for some xi in Fi and c in Fn. This is the main and most important result of the chapter.

A real number c is constructible if and only if there is a series of subfields of the reals such that Fi+1= Fi( ) for some xi in Fi and c in Fn. Therefore [Fi+1:Fi] is 1 or 2 for all i; This is the main and most important result of the chapter.

A real number c is constructible if and only if there is a series of subfields of the reals such that Fi+1= Fi( ) for some xi in Fi and c in Fn. Therefore [Fi+1:Fi] is 1 or 2 for all i; so, if c is constructible, then [Q(c):Q] = 2k for some nonnegative integer k. This is the main and most important result of the chapter. Note that we can certainly come arbitrarily close to any number, but that this is not enough.

Ancient Problems Geometric constructions began with the Greeks. They thought that everything was constructible, and mainly spent their time finding constructions. Some of the important constructions that they couldn’t find included the following:

Ancient Problems Doubling the Cube: Given a cube, construct a cube with double the volume of the original. Trisecting an Angle: Given an angle, construct an angle which is 1/3 the original. Squaring a Circle: Given a circle, construct a square which has the same area.

Ancient Problems Later on, it was postulated that these three constructions were impossible. Now we are able to prove this.

Doubling the Cube Consider doubling a cube of volume 1. The new cube would have volume 2, and side length This implies that the cube root of 2 is constructible.

Doubling the Cube Consider doubling a cube of volume 1. The new cube would have volume 2, and side length This implies that the cube root of 2 is constructible. However, [Q( ):Q] = 3.

Doubling the Cube Consider doubling a cube of volume 1. The new cube would have volume 2, and side length This implies that the cube root of 2 is constructible. However, [Q( ):Q] = 3. Since 3 ≠ 2k for any k, is not constructible.

Trisecting an Angle Consider trisecting a 60º angle. It’s possible to create a 60º angle from scratch, so the problem is equivalent to creating a 20º angle from scratch. This implies that we can construct cos (20º). I can go into the reason for the last assertion if necessary. I like it, so maybe I will on the board.

Trisecting an Angle Consider trisecting a 60º angle. It’s possible to create a 60º angle from scratch, so the problem is equivalent to creating a 20º angle from scratch. This implies that we can construct cos (20º). In other words, [Q(cos 20º):Q] = 2k for some k.

Trisecting an Angle Recall the trigonometric identity
(Not that any of us ever knew it in the first place.)

Using 20º for the angle, this becomes Since cos (60) is ½.

Using 20º for the angle, this becomes Then cos(20º) is a zero of the polynomial

Trisecting an Angle However, is irreducible over Q. Therefore [Q(cos 20º):Q] = 3. By the mod-something irreducibility test.

Trisecting an Angle However, is irreducible over Q. Therefore [Q(cos 20º):Q] = 3. Thus cos 20º is not constructible, and so the 60º angle cannot be trisected.

Squaring a Circle Consider constructing a square of the same area as a circle of radius 1. The circle has area π, so the square would have sides of length Ask if they know the degree of the extension of Q by root pi in Q.

Squaring a Circle Consider constructing a square of the same area as a circle of radius 1. The circle has area π, so the square would have sides of length However, [Q( ):Q] is infinite. Therefore the construction is impossible. Is this right? It ignores the hint in the exercises.

Example 6 We show that a regular 7-sided polygon (a heptagon) is not constructible. If a heptagon is constructible, then so is the angle 2π/7, and so is cos(2π/7). Is this right? It ignores the hint in the exercises.

Example 6 We show that a regular 7-sided polygon (a heptagon) is not constructible. If a heptagon is constructible, then so is the angle 2π/7, and so is cos(2π/7). Consider the fact that Is this right? It ignores the hint in the exercises.

Example 6 This shows that cos(2π/7) is a zero of the polynomial
Is this right? It ignores the hint in the exercises.

However, once again, the function is irreducible over Q by the mod 3 irreducibility test. Is this right? It ignores the hint in the exercises.

However, once again, the function is irreducible over Q by the mod 3 irreducibility test. This means that [Q(cos(2π/7)):Q] = 3. Then 2π/7 is not constructible, and so neither is a regular heptagon. Is this right? It ignores the hint in the exercises.

Example 6 It is a natural extension of this to wonder exactly which regular polygons are constructible and which are not.

Construction of regular polygons
The following theorem was proved by Gauss. Theorem: Construction of the regular n-gon with a straightedge and compass is possible if and only if n = 2kp1p2…pt, for some nonnegative integer k and distinct primes pi where each pi is of the form 2m + 1. This theorem is proved in chapter 33.

Proof, “”

Proof, “” If we can construct a regular polygon of n sides, then as before, we can construct cos(2π/n).

Proof, “” If we can construct a regular polygon of n sides, then as before, we can construct cos(2π/n). Let ω = cos(2π/n) + isin(2π/n), a primitive nth root of unity.

Proof, “” If we can construct a regular polygon of n sides, then as before, we can construct cos(2π/n). Let ω = cos(2π/n) + isin(2π/n), a primitive nth root of unity. Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). The theorem is in the book, chapter 33.

Proof, “” If we can construct a regular polygon of n sides, then as before, we can construct cos(2π/n). Let ω = cos(2π/n) + isin(2π/n), a primitive nth root of unity. Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω). The lemma is in the book, chapter 33.

Proof, “” as before, we can construct cos(2π/n).
Let ω = cos(2π/n) + isin(2π/n), a primitive nth root of unity. Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω).

Proof, “” Let ω = cos(2π/n) + isin(2π/n), a primitive nth root of unity. Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω).

Proof, “” of unity. Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω).

Proof, “” Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω).

Proof, “” Theorem: Let ω be a primitive nth root of unity. Then GalQQ(ω)  U(n). Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω). From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n).

Proof, “” GalQQ(ω)  U(n).
Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω). From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n).

Proof, “” Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω). From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n).

Proof, “” Lemma: Let ω = cos(2π/n) + isin(2π/n). Then Q(cos(2π/n))  Q(ω). From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n). Invoking the Lemma, we have [Q(cos(2π/n)):Q] = Also from the Fundamental Theorem: the degree of K over F is equal to the quotient of the Galois groups of E over F and E over K.

Proof, “” Q(cos(2π/n))  Q(ω).
From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n). Invoking the Lemma, we have [Q(cos(2π/n)):Q] =

Proof, “” From the Fundamental Theorem of Galois Theory and the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n). Invoking the Lemma, we have [Q(cos(2π/n)):Q] =

Proof, “” the Theorem above, we have |GalQQ(ω)| = [Q(ω):Q] = (n).
Invoking the Lemma, we have [Q(cos(2π/n)):Q] =

Proof, “” |GalQQ(ω)| = [Q(ω):Q] = (n).
Invoking the Lemma, we have [Q(cos(2π/n)):Q] =

Proof, “” Invoking the Lemma, we have [Q(cos(2π/n)):Q] =

Proof, “” Invoking the Lemma, we have [Q(cos(2π/n)):Q] =
Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k;

Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n).

Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n).

Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1.

Proof, “” Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1.

Proof, “” Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2,

Proof, “” Recall that, if σ is an element of GalQQ(ω), then σ(ω) = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” = ωk for some k; so that σ(cos(2π/n) + isin(2π/n)) = cos(2πk/n) + i sin(2πk/n). If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” = cos(2πk/n) + i sin(2πk/n).
If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” If such σ is in GalQ(cos(2π/n))Q(ω), then we must have that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” that cos(2πk/n) = cos(2π/n). This means that either k = 1 or k = n – 1. This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” either k = 1 or k = n – 1.
This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2.

Proof, “” This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2. Thus, if an n-gon is constructible, then (n)/2 is a power of 2,

Proof, “” This means that |GalQ(cos(2π/n))Q(ω)| = 2, and so [Q(cos(2π/n)):Q] = (n)/2. Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2.

Proof, “” [Q(cos(2π/n)):Q] = (n)/2.
Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2.

Proof, “” Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2.

Proof, “” Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2. Suppose that , where each pi is a distinct odd prime.

Proof, “” Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2. Suppose that , where each pi is a distinct odd prime. Then

Proof, “” Thus, if an n-gon is constructible, then (n)/2 is a power of 2, which means that (n) is a power of 2. Suppose that , where each pi is a distinct odd prime. Then Since (n) is a power of 2, this implies that each ni is 1 and each pi – 1 is a power of 2. This is the end of the forward part.

Proof, “”

Proof, “” Suppose that n is of the form 2kp1p2…pt where each pi is a distinct odd prime of the form 2m + 1, and again let ω = cos(2π/n) + isin(2π/n).

Proof, “” Suppose that n is of the form 2kp1p2…pt where each pi is a distinct odd prime of the form 2m + 1, and again let ω = cos(2π/n) + isin(2π/n). We have that Q(ω) is the splitting field of some irreducible polynomial in Q. This is by Theorem 33.3 in the chapter.

Proof, “” Suppose that n is of the form 2kp1p2…pt where each pi is a distinct odd prime of the form 2m + 1, and again let ω = cos(2π/n) + isin(2π/n). We have that Q(ω) is the splitting field of some irreducible polynomial in Q. Hence by the Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” is a distinct odd prime of the form 2m + 1, and again let ω = cos(2π/n) + isin(2π/n). We have that Q(ω) is the splitting field of some irreducible polynomial in Q. Hence by the Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” let ω = cos(2π/n) + isin(2π/n).
We have that Q(ω) is the splitting field of some irreducible polynomial in Q. Hence by the Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” We have that Q(ω) is the splitting field of some irreducible polynomial in Q. Hence by the Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” irreducible polynomial in Q. Hence by the Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” Fundamental Theorem of Galois Theory, (n) = [Q(ω):Q] = |GalQQ(ω)|.

Proof, “” [Q(ω):Q] = |GalQQ(ω)|.

Proof, “”

Proof, “” Since (n) is a power of 2, now we can create a chain of subgroups starting with the identity, H0, and the subset of GalQQ(ω) which fixes cos(2/n), such that each Hi+1 has index 2 over Hi: H0  H1  …  Ht = GalQQ(ω).

Proof, “” of subgroups starting with the identity, H0, and the subset of GalQQ(ω) which fixes cos(2/n), such that each Hi+1 has index 2 over Hi: H0  H1  …  Ht = GalQQ(ω).

Proof, “” subset of GalQQ(ω) which fixes cos(2/n), such that each Hi+1 has index 2 over Hi: H0  H1  …  Ht = GalQQ(ω).

Proof, “” that each Hi+1 has index 2 over Hi:
H0  H1  …  Ht = GalQQ(ω).

Proof, “” H0  H1  …  Ht = GalQQ(ω).

Proof, “” H0  H1  …  Ht = GalQQ(ω).
By the Galois correspondence, there is a corresponding chain of fixed fields such that the degree of each field over its direct subfield is 2.

Proof, “” H0  H1  …  Ht = GalQQ(ω).
By the Galois correspondence, there is a corresponding chain of fixed fields such that the degree of each field over its direct subfield is 2. But recall from our discussion of constructible numbers that this is equivalent to being constructible.

n = 2kp1p2…pt

n = 2kp1p2…pt This means that it is possible to construct regular polygons with 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, or 20 sides. It is not possible to construct regular polygons with 7, 9, or 11 sides.

Final notes

Final notes It is interesting to note that any construction that can be created with a straightedge and compass can be constructed with a compass alone.

Final notes It is interesting to note that any construction that can be created with a straightedge and compass can be constructed with a compass alone. These are called Mascheroni constructions. For example, to find the midpoint between two points, see the figure. It gets rather complicated.

Final notes It is interesting to note that any construction that can be created with a straightedge and compass can be constructed with a compass alone. Furthermore, if we are allowed to start with one circle and its center, we can create any of these constructions with only a straightedge.

Final notes There are many other simple ways to modify the rules. If we allow a straightedge with two marks, for example, then all of the three classical constructions can be created.

The End

Homework Problem 1: a) Describe and draw a process for bisecting an arbitrary angle using a straightedge and compass. b) Describe and draw a process for creating a regular 16-sided polygon. Explain how the method can be extended to create any regular polygon with 2n sides. Clearly (a) can be used for (b). These last two slides will be deleted and instead given as a handout.

Homework Problem 2: a) (Problem 10 from the book.) Prove that a 40 degree angle is not constructible. b) (Problem 15 from the book.) Show that a regular 9-gon cannot be constructed with an unmarked straightedge and compass. Again (a) is used for (b). These last two slides will be deleted and instead given as a handout.

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