(Unknown Author)

Cho a,b,c là 3 nghiệm của phương trình thỏa mãn (Let a,b,c be 3 roots of the equation, such that)

Chứng minh rằng (Prove that)

Lời giải (Proof)

(Vietnamese)

Từ định lý Viète ta có:

Từ đó, ta có thể tính được

Vậy  cũng là 3 nghiệm của phương trình 

Vậy

Ta có:

Vậy

(English)

Using Viète theorem

We can compute that

So are also 3 roots of the equation

So

We have,

So that,





The proof is completed.

I just talked about characteristic functions as masks on other functions. Given a function and a subset , we can mask the function to the subset by multiplying it by the characteristic function . I want to talk a little more about these functions and how they relate to set theory.

First of all, it’s easy to recognize a characteristic function when we see one: they’re exactly the idempotent functions. That is, , and if then must be for some set . Indeed, given a real number , we can only have if or . That is, or for every . So we can define to be the set of for which , and then for every . Thus the idempotents in the algebra of real-valued functions on correspond exactly to the subsets of .

We can define two operations on such idempotent functions to make them into a lattice. The easier to define is the meet. Given idempotents and we define the meet to be their product:

This function will take the value at a point if and only if both and do, so this is the characteristic function of the intersection

We might hope that the join would be the sum of two idempotents, but in general this will not be another idempotent. Indeed, we can check:

We have a problem exactly when the corresponding sets have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it off to define

We can multiply this out to check its idempotence, or we could consider its values. If is not in , then , and we find — it takes the value if and otherwise. A similar calculation holds if , which leaves only the case when . But now and both take the value , and a quick calculation shows that does as well. This establishes that

We can push further and make this into an orthocomplemented lattice. We define the orthocomplement of an idempotent by

This function is wherever is , and vice-versa. That is, it’s the characteristic function of the complement

So we can take the lattice of subsets of and realize it in the nice, concrete algebra of real-valued functions on . The objects of the lattice are exactly the idempotents of this algebra, and we can build the meet and join from the algebraic operations of addition and multiplication. In fact, we could turn around and do this for any commutative algebra to create a lattice, which would mimic the “lattice of subsets” of some “set”, which emerges from the algebra. This sort of trick is a key insight to quite a lot of modern geometry.

Robert Marzano found that non-linguistic representations, especially ones that have students compare, contrast, or classify information significantly influence student achievement in a positive way. The Southwest Georgia RESA has put together an excellent compilation of math graphic organizers. Many are in Word so that you can tweak them to fit your students’ abilities. They are categorized very well so that you can find exactly what you are looking for.

Go to the bottom of the site and you will see links to other places to find math graphic organizers and examples made by other teachers. If you teach concepts on Algebra basics, decimals, perimeter, area, volume, rational or rational numbers, click here to look at their content maps/units/lessons. Don’t let the grade levels deceive you… some would work well in the middle school.

Introduction to Complex Numbers

Complex numbers are not really complex! In fact they are reasonably simple to understand and operate upon. The concept is definitely strange on a first look, but is damn powerful and has diverse ramifications in various branches of mathematics. Now, to illustrate the point that these numbers are really simple, we are gonna define them in terms of quantities already known.

In coordinate geometry a point is referred to using its coordinates in the form . Thus to every point in the plane there corresponds a unique pair of real numbers and vice versa. Complex numbers are these ordered pairs of real numbers with certain operations defined on them. The formal definitions for these operations are as follows:

A complex number is an ordered pair of real numbers denoted by with the operations  of additions and multiplication defined on them as follows:

Moreover is called the real part and is called the imaginary part of the complex number . When use a single letter to denote a complex number, its real part is denoted by and the imaginary part is denoted by .

With these definitions of addition and multiplication, it is easy to check that the complex numbers of the form behave exactly the same as the real number with regard to these operations. Thus we have

Therefore the numbers of the form are identified with the real numbers and thus the set of all complex numbers properly contains the set of real numbers.

It is also easy to see that for any non-zero complex number (meaning at least one of the and is non-zero) there is complex number

such that and this is called the reciprocal of and denoted by or .

It can be easily verified by the reader that the complex numbers form a field under the operations defined on them with as the multiplicative identity and as the additive identity.

The Imaginary Unit

Lets the consider the complex . Clearly we have

so that the complex number acts as the square root of !. This number is called the imaginary unit and using it any complex number can be written in the form

Henceforth the complex numbers will be written in this form (one writes if y is some numeric value like 2.3).

Argand Diagram

Since complex numbers are just ordered pair of real numbers they can be used to represent points in a plane as in coordinate geometry. It turns out that this is very fruitful and the source of all the usefulness of complex numbers in mathematics. The coordinate plane is then called the complex plane or the Argand Diagram, X-axis is called the real axis and Y-axis is called imaginary axis.

In this setting if is a point in the complex plane then the complex number representing acts as the vector where is the origin. The rule of addition of complex numbers matches exactly the rule of addition of complex numbers.

Fig 1. Geometrical Interpretation of Addition of Complex Numbers

The geometrical interpretation of multiplication of complex numbers requires us to cast the complex number in a different form. If is the point corresponding to and makes angle with the real axis then

where is the length of line segment . Consequently we have

and then the complex number can be written in terms of parameters (actually called polar coordinates and the form called polar form) and as

where is called the modulus of and denoted by and is called the amplitude or argument of and denoted as am(z) or .

Fig 2. Polar Form of Complex Numbers

Multiplication in this form is easily seen to be

so that the moduli get multiplied and the arguments get added during multiplication. The number is also written in the form

which is just a formal notation which helps us remember the above product rule. (One does not need to think that some kind of imaginary power to a base is involved here.)

Therefore in geometrical language, multiplying a given complex number represent point by another complex number amounts to scaling the vector by a factor of and then rotating it in counterclockwise direction by angle .

The Complex Number

The number represents a point on the unit circle . If takes all the values from to , traverses the full circle. Let us now locate a point on the unit circle such that the it represent the complex number . Let the point represented by be called .

Fig 3. Constructing a Polygon

Similarly we represent points corresponding to the complex number for each value of . From the construction it is easily seen that the radial vectors are equally spaced out with the angle between consecutive radial vectors being . It clearly follows that all the lengths are equal. If for some value of , say the radial vector turns out to be the same as the initial radial vector then we have a regular polygon of sides inscribed in the unit circle. Since we get

The simplest solution (apart from ) for the above equations is .

It now follows that construction of a regular polygon of sides is intimately connected with finding a complex number such that . Such numbers are called roots of unity and the next post will deal exclusively with them and the polynomial equations satisfied by them.

Introduction

What do we exactly mean by the term “Euclidean Constructions”? Informally the term refers to the geometrical constructions done using the ruler (also called straightedge) and compass. Such constructions are studied as part of high-school (7th to 10th grade) mathematics curriculum and I hope most readers are familiar with the construction of bisection of line segment, bisection of an angle and construction of equilateral triangles.

To formalize the term, we shall allow only the following steps in any Euclidean construction:

1) Given two points, the ruler can be used to join them to make a line segment. Note that the ruler cannot be used to draw a segment of pre-determined length as we have to treat the ruler as unmarked.

2) Given two line segments, the ruler can be used to extend them (if needed) to find their point of intersection (if it exists).

3) Given a point P and a line segment AB, the compass can be used to draw a circle (or an arc) with center P and radius equal to the length of line segment AB. Note that the compass can not be used to draw circles of pre-determined radius.

4) Given a circle and a line segment, the ruler can be used to extend the line segment (if needed) to find its points of intersection with the circle if any.

Fig 1. Geometrical Constructions with Euclidean Tools

Algebraic Meaning of Constructibility

We now switch to the language co-ordinate geometry (analytic geometry). Two points O and A are arbitrarily chosen on the plane. O is referred as origin and the distance OA is chosen as the unit of length. The segment OA is extended on both sides to form the X-axis. A line perpendicular to X-axis and passing through origin is drawn and is called Y-axis. The positive half of the X-axis is the one in direction from O to A and other half is negative. Similarly if turn anticlockwise from the positive half of X-axis we reach the positive half of the Y-axis (the other half being negative).

The co-ordinate plane is now set up with both the axes and the origin O and a unit length OA. Now to every point in the plane corresponds a pair of real numbers and and the point is denoted by where (called abcissa or x-coordinate of the point) is the signed distance (positive if it lies towards positive half of X-axis, negative otherwise) of the point from Y-axis and (called ordinate or y-coordinate of the point) is the signed distance (positive if it lies towards the positive half of Y-axis, negative otherwise) of the point from X-axis.

Fig 2. Coordinate Plane

Having established a coordinate plane we can now figure out the implications of the Euclidean constructions described above in the language of algebra. First of all we note that starting with unit length OA we can construct a line segment AB whose length is a rational number (Integers are easy, just add many segments equal to OA one after another. For rationals just note that a line segment can be divided into a number of equal parts by drawing some parallel lines and using Thales theorem). Therefore it is possible to locate a point , whose coordinates are rational numbers, by using Euclidean tools only.

As we can see any new point in the plane is introduced in one of the following ways:

1) Intersection of two lines: Assuming that both lines are obtained as extensions of line segments whose end points have rational coordinates, it follows the equations of the both lines are of the form where are rational numbers. Consequently the coordinates of the point of intersection of these lines are also rational.

2) Intersection of a line and a circle: We can assume that the center of circle is having rational coordinates and the radius as the length of segment where and have rational coordinates. In that case is rational and so the equation of the cirle

has rational coefficients. Similarly the equation of the given line has rational coefficients. Therefore the solutions of these equations will ultimately be either rational or be roots of quadratic equations with rational coefficients. Hence the coordinates of the points of intersection will either be rational or be roots of quadratic equations with rational coefficients.

3) Intersection of two circles: Here the equations of both the circles will be having rational coefficients and therefore as in the previous case the coordinates of the points of intersection will be either rational or be roots of quadratic equations with rational coefficients.

Using points so obtained one can calculate new lengths (which could be used as radius of a circle in further constructions). Since the distance formula

also involves the extraction of a square root, it is clear that if one starts with points having rational coordinates then after one step of Euclidean construction either one gets points with rational coordinates or the coordinates of the points obtained are roots of  quadratic equations with rational coefficients. Same holds true of the distances between any pair of old and new points so constructed.

If we apply one more step of Euclidean construction then we have the following possibilities for the coordinates of new points and distance between a pair of points so obtained:

1) Either these are rational.

2) OR, these are roots of quadratic equations with rational coefficients.

3) OR, these are roots of quadratic equations whose coefficients themselves are roots of quadratic equations with rational coefficients.

Generalizing the above reasoning for steps of Euclidean constructions we can say the following about the coordinates of points so obtained and the distances between them:

1) Either these are rational.

2) OR  there are roots of quadratic equations of the following types:

where the coefficients of equations of type are rational and for the coefficients of equations of type are the roots of an equation of type .

We say that a real number is constructible if one can construct a line segment of that length starting from a line segment of unit length using Euclidean constructions only. It is now clear that all constructible numbers must be roots of quadratic equations of the above types. Since solution of a quadratic equation involves the extraction of a square root, it is clear that constructible numbers can be expressed as a finite combination of rationals with the following operations : addition, subtraction, multiplication, division and square roots. Moreover, given two line segments of lengths and we can always get line segments of lengths , and using Euclidean constructions only. Thus any number which can be expressed finitely using rationals and operations of “+, -, *, /” and square roots is constructible.

So we have the following result:

A real number is constructible if and only if can be expressed finitely as a combination of rationals with the following operations: addition, subtraction, multiplication, division and square root extraction.

It is also clear from the above discussion that finding the expression for a constructible number will involve solving a series of quadratic equations where solutions of one equation will be used as coefficients in the next equation. Thus we if have only two equations to solve and we wish to remove the irrationalities involved in the coefficients of 2nd equation then we need to eliminate them from both the equations. This will always result in either a quadratic equation or a quartic equation (degree 4) with rational coefficients and we shall never have 3rd degree equation involved. (Readers are requested to check this using some real equations and elimination). The logic behind this conclusion looks obvious after we check a few examples, however a proper proof would require us to develop field theory and lot of “Abstract Algebra” stuff.

An informal reasoning would suffice here. Roughly equations of type will lead to numbers having only one square root in their expression and when such numbers are added to the set of rationals (in the sense we can combine them with rationals using usual operations of +, -, *, /) then it will form a field say F1.

If an equation of type splits as a factor of two linear polynomials having coefficients from the field F1, then the roots of this equation will also be in F1, and in this the elimination of coefficients from equations of type and type will result in a quadratic equation with rational coefficients. If this is not the case (i.e. an equation of type is irreducible over the field F1) then the elimination of irrational coefficients will lead to a quartic equation with rational coefficients.

Proceeding in this fashion one can conclude the following:

A constructible number is algebraic (i.e. it is the root of a polynomial equation with rational coefficients) and moreover, the minimal polynomial (the polynomial of least degree with rational coefficients satisfied by the number in question) for will have a degree which is power of 2.

This is a very powerful theorem and its exact proof is somewhat cumbersome and full of discussions of extensions fields and irreducibility and other stuff which we don’t need here. Needless to say that the formal proof does not contain any new idea apart from those used in our informal reasoning and therefore we will be content with the informal version only.

The power of this theorem is exhibited in a negative fashion i.e. it can be used to prove that certain numbers are not constructible. For example consider . Clearly we have and therefore is a root of the polynomial equation . It is easy to establish that is irreducible over the rationals and therefore it is the minimal polynomial for . Since the degree of this polynomial is 3 which is not a power of 2, therefore is not constructible.

Construction of angles

We next focus on the problem of construction of angles. Suppose we wish to construct an angle using Euclidean tools. We know this is possible if . To address the problem for general , let’s assume that angle can be constructed in this fashion. We can then have a line segment AB of unit length and we can construct a ray AX such that . From a perpendicular is drawn to ray which intersects it at point . Then it clear that

so that and are constructible numbers.

Fig 3. Construction of Angles

Conversely if we assume that is constructible, then is also constructible and consequently the point can be located on the coordinate plane using Euclidean tools and thus we can construct ( being any point on X-axis). We thus have the following result:

An angle is constructible using Euclidean tools if and only if (or ) is a constructible number.

Construction of Regular Polygons

Construction of a regular polygon is ultimately linked to the construction of its central angle (angle subtended by a side at the center of its circumcircle). If a regular polygon is constructible using Euclidean tools then we can find its center by drawing bisectors of two adjacent sides and finding their point of intersection. Joining this center with two adjacent vertices we can construct the central angle. For a regular polygon of sides the measure of the central angle is .

Again on the other hand if we assume that an angle of measure is constructible by Euclidean tools then we can draw a circle of unit radius and draw two rays from its center such that angle between them is . These rays will then intersect the circle in two points, say and . and will then act as adjacent vertices of the regular polygon. Using as the radius and center we can draw an arc to cut the original circle at which is the 3rd vertex of the polygon. Continuing in this fashion all the vertices of the regular polygon can be obtained and the vertices can be joined to form the desired regular polygon.

We can now conclude (using the result of the previous section):

A regular polygon of sides can be constructed by using Euclidean tools if and only if (or equivalently ) is a constructible number.

We shall investigate the constructibility of along the lines of Gauss in the forthcoming posts.

Introduction

After studying elliptic integrals and formulas for π, we shall now focus on one of the most beautiful gems discovered by Gauss at the age of 17. Gauss proved that the construction of a regular polygon of 17 sides is possible by using an unmarked ruler and a compass only (henceforth these will be known as Euclidean tools and such constructions will be called Euclidean constructions). This is quite remarkable because since 2000 years or so from the time of Euclid the only polygons which were constructible in such a fashion were having sides 3, 4, 5, 6, 8, 10, 12, 15. Gauss added a new number in this series namely 17 and generalized his results to add further numbers. Legend has it that Gauss was so excited by this discovery that he decided to make  a career as a mathematician.

The result Gauss established after deep research is the following:

A regular polygon of sides is constructible by Euclidean tools if and

where is non-negative integer and are distinct primes of the form

.

(such  primes are known as Fermat primes and as of now only 5 such primes namely 3, 5, 17, 257, 65537 are known)

First of all we make some observations about the above result:

1) The conditions is obvious as we cannot have a polygon with 2 sides or less.

2) The factor is obvious in the formula for because if we can construct a regular polygon of sides then we can obviously construct a regular polygon of sides. (Readers should supply a simple proof of this statement on their own.)

3) Each of the primes (other than 2, i.e. Fermat primes) in the prime factorization of (the formula for actually gives its prime factorization) occurs only once.

4) There may be no Fermat primes involved in the expression for and then with .

5) The condition as mentioned above is sufficient. In other words if is of the form as specified above then the construction of polygon is possible. It does not say anything about the case when is not in the specified form. However, it turns out that the condition is also necessary and Gauss did not have a proof of it (which is somewhat surprising because it is rather easy to establish the necessity of the condition). Thus when is not of this particular form then a regular polygon of sides can not be constructed by using Euclidean tools only.

6) Gauss only proves that the construction is possible using Euclidean tools, but does not provide any steps of the geometrical construction. The actual steps become quite cumbersome as increases and the final post in this series will provide a construction of regular polygon of 17 sides.

After this brief explanation of the Gauss’ result we can now ponder as to how Gauss might have proved it and follow his footsteps in this series of posts.

Today is the first snow day of the year! Very exciting, actually, since I really wasn’t expecting it. We got a lot of snow – a foot, maybe a little more? – but I remembered the more recent DC school closings, ever since my high school started following the DC public schools. We NEVER had a day off! But seriously, I still haven’t finished my final paper for grad school, so it’s really good.

at the beginning of the storm, out the window

It snowed all day Saturday so JJK and I basically did nothing. I think the highlight was pancakes and bacon, followed closely by meeting Julia Cad for hot chocolate between our apartments.

Anyways, last week was a big rollercoaster. I’m a little bit burnt out, to be honest. Depressingly unmotivated, exhausted, yadda yadda yadda. But I’ve been co-teaching with Andrew, the one who’s slowly taking over my corrective reading classes. Mostly he teaches by himself now, and I’m working with other students in the back, or giving “sit up straight and take your hoodie off!!” looks to the corrective reading kids – who I will continue to think of as “mine,” no matter what.

What has been stressing me out this week was the meeting I had with my PD (program director, aka TFA advisor). She’s really great, and helpful, but she also clued me in to some things that I should have been doing for my SPED kids that I haven’t been doing – like modifying their tests, progress monitoring in a more specific way, etc. She also talked with me about what it means that I’m not teaching corrective reading anymore – what I AM doing is teaching resource room material. Resource rooms in some cases replace the core subject classes (english, math) but in my case will supplement the core classes.

Which means I have to know what the gen-ed teachers are going to be teaching, which means that I have to lesson plan for all the core subjects. And given he spread of students joining my resource rooms, that means I have to prep for English 1, English 2, English 4, Geometry, Algebra 1, and Algebra 2.

I had a little bit of a stress-fest when I got home after that.

Anyone who knows me knows I fear math. My students could tell you that. They ask me questions and my response is usually “well, let’s look it up,” because seriously who remembers the rules for exponents and stuff like that? Factoring?? I’m lost. So that will be a challenge.

Most frustrating was Friday, when the algebra 2 class that I sometimes co-teach had a quiz. Sha, who I’ve written about before, is on my caseload and in one of my resource rooms. All week we worked on exponents. We learned the “by heart,” we said. We went one rule at a time, and he knew them, dammit, he really did.

So I popped into the class and checked his quiz after he was done. ALL of the problems were wrong. All of them. But I looked at the problems and saw that he’d been on the right track. For 3(x^2)(x^3), he’d written 3^5. Meaning, he did look and see that he should’ve added the exponents. He just forgot to copy down the x! This was good, and my PD had been telling me that I should look at my kids’ quizzes and tests to re-grade them for partial credit, and correct application or correct prior knowledge. Soo, I took Sha in the hallway to see what he could earn in partial credit, at least for me to know that he’d mastered this particular objective for his IEP.

He was (understandably) frustrated and annoyed with me. I literally begged him to try again, to look one more time.

“Whatchu want from me, Miss, it’s wrong. It’s over.”

“But you know this Sha, you do, you know it by heart, remember? You just didn’t take your time and you didn’t use your brain skills. Use your brain and remember what you and I talked about 3rd period and look at this problem one more time! Tell me what you should do with these exponents!”

I talked him through them, reminding him what we’d talked about without giving answers, etc. He had the knowledge, he just struggled with the application. Needless to say, last period on a Friday, it was just as frustrating for me than it was for him. He knew it hours ago!! I then got irrationally mad at the math teacher for making the quiz so hard. He doesn’t understand that if you give kids (x^3)(x^2), they know what to do with it, but if you then throw in another variable, or a number, they get lost. ESPECIALLY if that’s not how you taught it, and I’ve been in his class and that’s not how he taught it. He then countered with “Well, what, you want me to give them 3 + 3 their whole lives? This is what high school algebra looks like!” Which is a good point. And then I got mad at him for not recognizing the fact that Sha did know the rules he was supposed to know, he just had a hard time applying them. “This is a wrong answer, and in math their is no in between,” was his response, and in my head I’m like “I know, that’s why math sucks!” but I didn’t say that. And he’s not really wrong, but neither am I.

Oh well.

So, today’s snow day is quite welcome, although it will make Tuesday and Wednesday all the more frustrating in terms of classroom management.

I’ll be going home Wednesday, catching a ride with Julia Cad, and I couldn’t be more excited!

Happy Holidays Everyone

xoxo

SBT

p.s. The cats also enjoyed the snow. In fact, they enjoyed it so much that they pulled the curtains down to get a better look. Little brats. But they’re so cute!!

snuggly snow kitties

My girlfriend is taking an abstract algebra course this year, and they had the following “longterm project” to think about for the term: Show that every group of size less than 60 is either Abelian or not simple. If we assume Burnside’s theorem that a group of order (p and q are primes) is solvable, this gives every case except for 30 and 42. (A proof of Burnside’s theorem can be found here on p.19, Exercise 6 of one my old solution manuals, but you’ll need Serre’s book to see what the cited theorems are). But since I had never done this exercise, I wanted to write a post about how to do this using only techniques from a first course in algebra (i.e., Sylow theorems).

Let’s recall what the Sylow theorems say. First, let G be a finite group of order n. Let p be a prime and write where p does not divide m. A maximal subgroup of G whose order is a power of p is a Sylow p-subgroup. Let be the number of Sylow p-subgroups. Then we have the following theorem.

Theorem. 1. The size of a Sylow p-subgroup is .
2. All Sylow p-subgroups are conjugate to one another via inner automorphisms of G.
3. The index of the normalizer of any Sylow p-subgroup is . In particular, divides m and is congruent to 1 modulo p.

The third statement is usually enough. If we can show that for some p, then the corresponding Sylow p-subgroup (assuming p divides n) must be normal since any conjugate of it must have the same size. There were three general facts that have relatively short proofs that take care of most of the cases. Then the rest was a (quick) case-by-case analysis. If the reader knows of another quick general fact that makes the problem more efficient, leave a comment!

Fact 1. Any p-group of size bigger than p is not simple.

Proof. The center of a p-group is never the identity subgroup. This is usually a standard fact proved in a first course, so I’ll omit it. Using this, either the group is Abelian (and any nontrivial proper subgroup gives us the desired normal subgroup), or the center gives us the desired normal subgroup.

Fact 2. A group of order pq with p<q primes is not simple.

Proof. The number of Sylow q-subgroups divides p and is congruent to 1 modulo p. Hence there can only be 1 of them.

Fact 3. A group G of order 2k with k odd has a subgroup of index 2 (which is normal), and hence is not simple.

Proof. Let P(G) be the group of permutations of G thought of as a set. We have an injective homomorphism defined by sending g to the permutation obtained by left multiplication by g. By the Sylow theorems (or more simply, Cauchy’s theorem), we have a nonidentity element g of G of order 2. The cycles of F(g) are each of size 2, so in disjoint cycle notation, F(g) is a product of k 2-cycles, and hence is an odd permutation. So the kernel of the sign homomorphism has index 2 and is the desired subgroup.

This accounts for everything except the following numbers: 12, 20, 24, 28, 36, 40, 44, 45, 48, 56.

Using part 3 of Sylow’s theorems, we can immediately get rid of the following cases:
20: There can only be 1 Sylow 5-subgroup.
28: There can only be 1 Sylow 7-subgroup.
40: There can only be 1 Sylow 5-subgroup.
44: There can only be 1 Sylow 11-subgroup.
45: There can only be 1 Sylow 5-subgroup.

In the cases of 24, 36, and 48, there are either 1, 4, or 16 Sylow 3-subgroups (16 is only a possibility for 48). In the last case, the Sylow 3-subgroups have size 3 and hence pairwise intersect in the identity element. So their union consists of 33 elements. A Sylow 2-subgroup doesn’t intersect any of them except in the identity, so contributes 15 more elements. So we only have one Sylow 2-subgroup in this case. Otherwise, there’s either 1 or 4. We assume that it’s 4. In this case, let X be the set of Sylow 3-subgroups. We get a homomorphism (where again P(X) is the permutation group of X) by the conjugation action. By Sylow (2), the kernel is a proper subgroup of G. Since P(X) has size 24, in cases 36 and 48, the map is not injective and hence the kernel gives us our desired normal subgroup. Otherwise, if G has size 24 and the kernel is trivial, then G must be the symmetric group , which we know is not simple (take the alternating subgroup for a normal subgroup).

All that is left to do is 12 and 56. These can be handled by counting arguments. First we do 12. There is either 1 or 4 Sylow 3-subgroups. In the case of 4, their union consists of 9 elements. A Sylow 2-subgroup has size 4, so adds another 3 elements, giving 12 in total. Hence there would be 1 Sylow 2-subgroup in this case. In a similar way, we can do 56. There is either 1 or 8 Sylow 7-subgroups. In the case of 8, their union gives us 49 elements. A Sylow 2-subgroup has size 8, so it contributes an additional 7 elements, giving 56. So there can only be 1 Sylow 2-subgroup in this case.

So that’s the end of the proof. We can extend the exercise further, i.e., the next non-Abelian simple group has order 168. Courtesy of Wikipedia, I’ll list the first few non-Abelian simple groups and their orders. The notation denotes the finite field with q elements (q being a prime power), the notation denotes the determinant 1 matrices over this field, and the notation denotes the quotient of this group by the center where I is the identity matrix (if q is even, then ). Also denotes the alternating subgroup of the corresponding symmetric group.

Order 60: , which is isomorphic to and
Order 168:
Order 360: , which is isomorphic to
Order 504:
Order 660:
Order 1092:
Order 2448:
Order 2520:
Order 3420:
Order 4080:

Have fun!

-Steven

When people find out I am a math teacher, if they don’t tell me how much they hate math and how they were never “good at it” (this comment, BTW, usually culminates with them vomiting on my shoes), they invariably offer that they were good in Geometry and bad in Algebra, or vice versa.

I never ask  people which they preferred in school; for some reason they almost always volunteer their preference.  If you happen to be of this same mind – that you preferred Algebra to Geometry in school or vice versa, consider the following reason why this may be so.

Geometry is very spatial.  Nary a Geometry problem is not accompanied by a figure, usually involving angles, triangles, quadrilaterals, or circles with chords and tangent lines, all chock full of capital letters denoting vertices and such.  True, there may be some algebra involved in the solution of the geometry problem (which, incidentally, is why Algebra generally precedes Geometry in most high school coursework), but  diagrams and figures reign dominant in Geometry.

Algebra, on the the hand, is typically void of diagrams and figures and relies instead on a student’s ability to manipulate symbolic expressions in solving problems.

Therein lies the distinction.  The left hemisphere of the brain is dominant in the cognitive processes using LOGIC, SYMBOLS and LANGUAGE.  The right hemisphere of the brain deals with SPATIAL, CREATIVE, and MUSICAL processing.  That is to say, Algebra and Geometry are processed in different hemispheres of the brain:  Algebra is a left-brain pursuit while Geometry is right-brained.

Of course, some of Geometry requires left-brain logic and some of Algebra requires right-brain spatial/linguistic processing (as in the case of solving word problems).  But on whole, doing Algebra has more in common with writing an English paper than with doing an Geometry problem – at least cognitively.

I myself am a case in point.  I am much more an Algebraist than a Geometer by trade and I am also a writer (not only in the case of this blog, but also for a forthcoming book I am writing that will published by Corwin Press).

I am willing to bet that those folks who liked Geometry tended to pursue artistic careers (sculptures, painters, photographers, architects and musicians) and those who liked algebra tend to pursue left-brain careers (scientists, economists, writers and lawyers).

If you had a preference, which did you prefer in school:  Algebra or Geometry?    dven.

NNNNNNNNNNNNNNNNNNNNNNNNNNN

Wow, PDE really changed their site. If you have anything bookmarked even as of last month, it will not work now. It took me some time to figure out where everything is now located.

Here’s how you navigate to the PSSA resources.

1. Go to http://www.education.state.pa.us/

2. Click on “Programs” on the left side of your screen.

3. Click on “Programs O-R” on the left side of your screen. (PSSA starts with “P”)

4. Click on “Pennsylvania System of School Assessment (PSSA)” on the left.

5. Scroll down, and here are the PSSA resources.

PDE did chose to release new sample questions this year (unlike last year). The quick links to the released PDF’s are:

6th Grade Math

7th Grade Math

8th Grade Math

11th Grade Math

These are great resources to get a sense for how the questions are worded. Another great place to grab some questions are from Study Island. Pick the content area and instead of going to test mode, select “worksheet” and print questions from there. Our past data analysis has shown that the benchmarks are very closely correlated to the actual test grade, so the questions must be a fair represenation of those used on the test.

Ugh, today I spent my WHOLE math lesson converting fractions to percentages to decimals.  Why can’t they just be happy with who they are?

“Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives.”-author unknown.

So true.  So true.

part 1                                                            part 2

We will show that if is a normed domain, then is as well. Since is a domain if is, we need only construct a norm on to complete the proof. To this end, let be a norm on . We define by , where . We proceed to verify that has the three properties of a norm.

If , then for . This implies that for , so all the ’s are zero, making the zero polynomial. Conversely, if , then clearly .

If , then and for . This implies that , where is a unit in , so is a unit in . Conversely, if is a unit in , then , where is a unit in , so .

Now suppose and . By allowing some of the ’s and ’s to be zero, we can assume that . Now,

We conclude that is a norm on , so is a normed domain.

As good as start as any…

Adding, multiplying, dividing, subtracting to name a few are just mappings from a set into its self. We can start by formalizing this idea.

Definition. A binary operation on a set is a mapping .

Now an group theory object.

Definition. A groupoid is a pair where is a non-empty set and is a binary operation in .

Actually, the word “groupoid” has more than one meaning in mathematics, this definition should not be confused with a groupoid in category theory.

Some simple examples of the algebraic structure groupoid:

1. where and . is clear not empty and is easily shown a binary operation in .
2. is not a groupoid, since .
3. One to think about. Let be any non-empty set and be the set of all subsets of . Are and groupoids?

Denote the order of a groupoid by , this is as always the number of elements in the set.

Now, groupoids don’t seem very exciting or interesting yet. Well, maybe there is a little something in the third example that ‘blows my skirt up’ a little bit. What will develop groupoids in to something more interesting? Putting more restrictions on these groupoids will lead to semigroups and then to groups. Associativity will give semigroups and identities and inverses will take it farther to groups.

A start, but ugly. I don’t like the way the latex drops down below the bottom of the text. Suggestions?

Loyal readers will recall that I almost succeeded in making a Traditional American Thanksgiving dinner, lacking only a turkey for reasons beyond my control: the butcher ordered a frozen rather than a fresh turkey, so I had to substitute chickens.

In speaking with New Zealanders, I found out that the TAT dinner is extremely close to the traditional New Zealand Christmas dinner—cranberries and all—differing only in the dessert. So I figured I wouldn’t have any trouble getting a turkey for Christmas.

Both Rick and Emily will both be at the hospital all day, so we thought about having the big dinner the next day, on Boxing Day, which is on Saturday this year. We’re out of town until late Christmas Eve afternoon, the butcher is never open Saturdays, much less major holidays, so obviously a fresh turkey was out of the question. Considering the Thanksgiving result, I was happy to order a frozen bird, and I was willing to give the butcher a second chance, so a couple of weeks ago I ordered a 4.5-5 kilogram frozen turkey.

Today I went to pick up the turkey, and was waited on by the woman who had taken my order over the phone. It was only 3.5 kg, even though my order was attached. What had happened? “I guess the boss told them it would be okay.” Both my home and cell phone numbers were on the order form, so I could have been consulted, and no, it was not okay. Did they have a turkey of the size I wanted? Of course not—just a 6 kg one.

The butcher I’d dealt with at Thanksgiving got involved, and he gave me a serious price break, so I ended up taking the big one. I know that the third time is a charm, but it’s just as well that there’s no third American holiday involving turkey. I don’t think the butcher would survive. 

At least we have a turkey, and cranberries for the sauce. The only issue is dessert, and we’ve decided to skip the pies and do the New Zealand thing. Traditionalists serve fruitcake, and there are dozens of kinds to choose among, but my family won’t go for that. The other main choice is Pavlovas. They’re named after a petite Russian ballerina of the early 20th century, and they’re simply meringues topped with whipped cream and fruit. There’s some dispute whether they were invented in New Zealand or Australia (one of these days I’ll write an entry or six or eight on the rivalry between the two countries). We’re planning on the strawberry sort, and I hope by then we can find fruit that’s not crunchy.

In the meantime, we’re making a few traditional Christmas cookies. A family favorite is brownies topped with mint frosting and bitter chocolate. With a firm layer topped by two soft ones, it’s something like the slices so popular here. Today I went to the store to get ingredients. There was no spearmint extract, just peppermint, but I could live with that, since that was the original flavor before I amended it. What surprised me was the total lack of unsweetened baking chocolate—although that goes along with the huge New Zealand sweet tooth. I ended up getting some Lindt 85% dessert chocolate and adjusting the recipe. When I next get some algebra tutees, I’ll give them the problem I had to solve, of how much 85% to use and how much less sugar. They’ll be happy, I know.

Math gets a bad reputation for being cold and “just about the numbers” but think of what a world without math would be.  No birthdays, anniversaries, calendars, time, money, computers ect. now wouldn’t that be sad.  But we do have math and it helps us live our lives.  Math can also be pretty, different functions can draw out some pretty fantastic pictures.  Lets take the functions f(x)=|x|+√1-x^2 and g(x)=|x|-√1-x^2   graph them at the same time, what did you get?

A cardioid, the mathematical term for a heart-shaped graph.  Check out Romantic Mathematics for other more complex functions that draw a better heart shape and other things about how math is romantic.  This site also has a very good explanation of fractals.

To continue, I am now going to have to use the language of sheaves. For it, and for all details I will omit here, I refer the reader to Charles Siegel’s post at Rigorous Trivialties and Hartshorne’s Algebraic Geometry. When I talk about sheaf cohomology, it will always be the derived functor cohomology. I will briefly review some of these ideas.

Sheaf cohomology

The basic properties of this are as follows.

First, if is a topological space and , then is a covariant additive functor from sheaves on to the category of abelian groups. We have

that is to say, the global sections. Also, if

is a short exact sequence of sheaves, there is a long exact sequence

 Finally, sheaf cohomology (except at 0) vanishes on injectives in the category of sheaves.

In other words, sheaf cohomology consists of the derived functors of the (left-exact) global section functor.

When is flasque (i.e., for any open with , restriction is surjective), we have

(This is basically because of a well-known fact about sheaves: if is exact and is flasque, then the sequence of global sections is also exact.)

Moreover, this gives a way to compute the cohomology of a sheaf . If we have a flasque resolution , then consider the complex of the global sections of the sheaves (and ). The key formula is then:

There are many interesting results about the cohomology of a coherent sheaf over schemes, though they are not really relevant to us now. We are interested in the analytic applications.

Cech cohomology

Cech cohomology gives a reasonable way to actually compute these cohomology groups. For instance, Hartshorne uses it to calculate the cohomology of line bundles on projective space. In general, Cech cohomology does not equal derived functor cohomology, though it does in certain cases (in algebraic geometry, if the scheme is separated, then it is ok).

In the analytic case, we are working with paracompact spaces, so Cech cohomology will always be acceptable, as we will see.

Let be a covering of , and let be a sheaf on . Consider the chain complex

with boundary maps defined as follows. Let with corresponding

for every -tuple . Define

where the hat denotes omission, as usual. (Also, when .) This can be checked to be a complex. Then we define the Cech cohomology groups

as the cohomology of this (co)chain complex.

There is another way to interpret the Cech cohomology groups that better illustrates their connection with regular cohomology. Given , consider the sheaves defined by

The boundary map is defined the same as before, so we have a complex of sheaves

Proposition 1 The above complex is exact.

Exactness at the first step is basically the definition of a sheaf: cocyles in represents collections of elements of that agree on the common intersections, so piece together to a section of .

Now consider the other more general case. We need to check exactness on the stalks, say at . Choose with . Let be such that ; we can assume . We need to lift on some neighborhood containing to get some that goes to via .

Thus, for all -tuples , define

Then is a section of on . In this way, we get a map from , and it is easy to check that it is a chain homotopy, which implies exactness.

Cech cohomology versus derived functor cohomology

I will now discuss some examples of Cech cohomology. The notation remains the same.

Proposition 2 .

This is really just the sheaf axiom; see the beginning of the proof of exactness of the Cech complex of sheaves.

The next proposition gives us some reason to suspect a relation between tehse two types of cohomology.

Proposition 3

There is a natural transformation of -functors

if . 

Let

be an injective resolution of . Then by a basic result in homological algebra, there is a unique commutative diagram of resolutions

which induces a corresponding commutative diagram on the global sections. Taking the cohomology of the complex then gives the result.

Next up: the Leray theorem. 

Incidentally, I’m not sure how I should categorize this post.  I’m using “algebraic geometry” because that subject seems to depend the most heavily on sheaves.

No assignment over winter break – enjoy, enjoy!

Solving inequalities can sometimes be a difficult concept. Chris Boyer just finished and used a excellent resource to assess student understanding. Instead of simply assigning his class a few problems out of the book, he used an activity from a book called FACEing Algebra written by Kristen Dewit, a math teacher located in California. The above picture is the product of solving 27, yes 27 inequality problems that involved fractions and negatives. Instead of grumbling that they had to do 10 or 15 problems, his classes were excited to do all 27. Why? Because of the product. The activity outweighed the work – it was fun for them. Each solution tells the student a part to draw or color to shade it. A second benefit for Mr. Boyer is that he can quickly see where mistakes are based on how the face looks.

Click on the link to see a sample from the book and actual student examples from Chris’s class. You can order Kristen’s book at www.facingmath.com. She has Algebra, Geometry (H.S.), and a cool Unit Circle Kit for some Trig. applications. She’s working on an Alg II and another Algebra I. Let me know if you want to see more, we have the Algebra and Geometry (H.S.) Thank you Mr. Boyer for sharing!

I almost forgot, after seeing this done in one class, a few more teachers liked the idea. Some are beginning to adapt the project for collaborative classes. They are drawing the face for the students, and they simply have to solve the coloring part.