Mental arithmetic is, with the apparent exception of a few shockingly talented people (whom I’ve read were in high demand among astronomers and physicists in the days before mechanical calculators), a matter of approximation and simplification: how can you get something tolerably close to the answer you want while doing problems that are easy enough to do in your head?

Ralph Drabble starts off well here by estimating the traffic signal to last about 90 seconds. Obviously it’s not 90 seconds every time, but it’s reasonable to suppose that the occasions he has very short waits are balanced by times he has longer waits than usual. In any case replacing the whole spectrum of possible waits with one time is a good move, particularly since he chose an easy number. 90 isn’t very easy in general, but it is one and a half minutes, and since he figures he passes the light twice a day, that’s 3 minutes total per day at the light. 3 is a very easy number to work with, or at least much easier than 90.

He also supposes he passes the light twice every single day, 365 days per year. This is another good simplification. Surely there are days he doesn’t leave the house, or is on vacation, or so; but there are surely days he has to make several trips away from home. It would be a bit surprising if they balanced to exactly one pair of visits per day for an entire year, but it’s much easier to work with 1 pair of crossings than, say, a more precise 1.224 or whatever the actual average might be. Also his approximating a year as 365 rather than 365-and-a-bit-under-a-quarter days is wise. Given that we’ve already approximated time spent at the traffic light per day at 3 minutes, rather than, say, 2.85 minutes, it’s pointless to estimate the year more exactly than that.

He goes wrong in the next step, in figuring that as it’s 3 minutes a day, times 365 days per year, times 20 years, that he then needs to multiply 3 by 365 by 20. That’s not truly wrong, no, since that is the calculation to do. But it’s a bad approach to use for mental arithmetic since 365 is an awful number to multiply by. I’d refuse to multiply 365 by 3 in my head. If I had to, I would instead note that 365 is 400 minus 35. Then, by the distributive property, 365 times 3 must be 400 times 3 minus 35 times 3, or 1200 minus 105, or 1095. I hope. Distribution is one of mental arithmetic’s greatest powers. It effectively lets you turn multiplication by a multiple-digit number into a set of multiplications of single-digit numbers, at least if you’re willing to believe that 400 times 3 is no harder than 4 times 3, with some addition or subtraction on the side. It’s also good for noticing that some number is close to, say, one-third of a hundred, or two-fifth of ten thousand, or so.

(The other way I might multiply 365 by something, if I had to: it strikes me that 365 is pretty near to 333 plus 33. That looks worse, except, 333 times 3 is just about 1000, and 33 times 33 is just about 100. If I know I’ll be multiplying 365 by some multiple of three, I can leap directly to 1100 and not be too far off. I suppose this spoils my claim that he should have multiplied 3 by 20 first, but it’s pretty clear he didn’t see what 3 times 365 might be.)

But here’s the right way to figure it: remember that multiplication commutes. We can change the order of the things we’re multiplying in any way we like. So 3 minutes times 365 days-per-year times 20 years is the same thing as 3 times 20 times 365. And, ah, 3 times 20 is 60. 60 is nice to have here because 60 minutes is one hour, so we can change that 60 to 1, and get a result not in total minutes spent at the light but instead total hours spent at the light. And 1 times 365 is easy enough you don’t even need to do it in your head. He’s spent something like 365 hours at the light. Commutation is another great tool for mental arithmetic, and it works in addition as well as multiplication, so if you do much of it you’ll pause before calculating to figure whether swapping some multiples around might give you an easier problem. Often it does.

Granted, nobody has much idea how long 365 hours is. But there are 24 hours in a day, and 24 goes into 365 … well, that’s a pain. But 365 is pretty near 360 and everything divides into 360. I exaggerate but not by much; it’s likely that the great number of things that divide evenly into 360 is part of why we have 360 degrees in a circle, and so many early calendars set the year at 360 days plus a handful of intercalary days.

24 going into 360 is a nice, easy proposition: that’s 15. So there were 15 days spent at the traffic light, which sounds like a lot until you remember that’s spread out over 20 years, which over 365 days per year over 20 years comes out to about three minutes a day, not very long at all.

PS: I got the calculations right. Of course, I would say that. Or I used the calculator wrong.

Some photos from “Excelsior”, our homeschool cooperative

AABEILN

Near the top of the board was TAVERNA, and it was possible to hook above the first six letters or below the first two letters. There were other spaces on the board to place words, but this was clearly the most fertile. The full board looked like this:

On my rack, the letters weren’t in alphabetical order (as above), so I missed a seven-letter word that would have garnered 78 points. Instead, I played ABLE for a paltry 13 points.

After my turn, the Teacher feature showed me the word I should have played:

ABELIAN

Kickin’ myself. I’ll get over not seeing BANAL, LANAI, or even LEV. But how does a math guy miss ABELIAN? I would not put up a fight if someone wanted to rescind my Math Dorkdom membership card.

What loves letters and commutes?
An abelian Scrabble player.

(That’s a joke. Please don’t play Scrabble while driving.)

Lemma: Let be a quotient map. Then

saturated and open open.

Proof: If is saturated, then , so is open by definition of a quotient map. Same for closed.

Now I want to discuss the motivation behind the definition of quotient topology, and why we want the topology to arise from a surjective map. It is possible to define a topology on in the same way as the quotient topology, given any function which may not be surjective, but this doesn’t mean much to us. And here is why.

Lemma: Let be a function. Then induces an equivalence relation on , given by

.

No proof required! Now let be the set of equivalence classes under this relation. If is the map that takes elements to their equivalence class, then we put a topology on known as the strongest topology on such that is continuous. In other words, a topology that has the most sets possible in it so that is continuous. Of course a simple way to achieve this is to declare that every set is defined to be open if is open. It is maximal because if we had another open set such that was continuous, then would have to be open, so was in our topology anyway.

Now, if is surjective, there is a bijection given by ; a statement which is easy enough to check. A good way to define a topology on would be to ‘pretend’ that it is just and use its topology. That is what we do, by saying is open is open. But open is open. Now, since the diagram

commutes, we have   open. This means that the topology we put on is precisely the quotient topology.

To recap, we associate the space with the set of equivalence classes induced by a surjective map . Every element in corresponds to a unique element in , and there is a natural topology defined on this set. If is not surjective, then cannot be bijective, and the ideas above fall apart.

We explained above that a function induces an equivalence relation on its domain. In fact, every equivalence relation induces a surjective function. If is an equivalence relation on , then given by is a surjection. So roughly speaking, surjections and equivalence relations on are equivalent.

Hopefully the following driving-related math jokes will make you feel a little less depressed.

Which American highway is related to the length of the vector (1, 8, 1)?
Route (root) 66.

 Perhaps the next one hits a little too close to home.

A large group of math majors were headed to a party, but they only had one car. Consequently, one person served as the driver and shuttled folks to the party in groups. When the last of them arrived, they realized they didn’t really feel like interacting with any of the other party-goers. So, they repeated the process of returning home one group at a time, in their only car.

In short, they commute but don’t associate.

How is the group described above like the averaging function x AVG y = ½(x + y)? The averaging function is commutative:

x AVG y = y AVG x

However, it is not associative:

(x AVG y) AVG z ≠ x AVG (y AVG z)

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When talking about abelian groups specifically, we usually write them additively: the group operation applied to and is , and then we can build expressions like .  The proof I give below is due to J. S. Milne, who in turn says it’s similar to Kronecker’s original proof.  Of course, I’ve added more detail in places where I thought it was necessary, and taken it out where I thought it wasn’t.  There are other, more common proofs, typically using matrices, but I find them unwieldy and inelegant.

For arbitrary groups, we had constructed this nice presentation by generators and relations that expressed the group as a quotient of a free group by another free group.  A group is finitely generated if the first free group can be chosen to have a finite basis, and finitely presented if both free groups can.  What does the generators-and-relations presentation of an abelian group look like?  We have to require that any two of the generators commute (check for yourself that this makes the whole group commutative), so the relations must generate every element of the form , for generators.

(Aside: this is called the commutator of and , written , and the subgroup of a group generated by its commutators is called the commutator subgroup, .  So for any group, we can get an abelian quotient group .  This is called the abelianization of , written .)

If we’re given that a group is abelian, this implies a certain redundancy in its generators-and-relations presentation.  So maybe there’s a better way of forming such a presentation.  A good solution is to consider free abelian groups — the abelianizations of free groups.  Elements of such a group are of the form , where are integers and are generators.  As you might expect, for any set , there is a free abelian group with it as basis, and bijections of the bases give isomorphisms of the groups.  In fact, the free abelian group with basis is just (recall that this is the “direct sum” over of , meaning the group of -tuples of integers only finitely many of which is nonzero).  The only finitely generated free abelian groups, then, up to isomorphism, are .

For vector spaces, we had these nice things called “bases.”  We required a basis to span the vector space, so every vector was a linear combination , where were real numbers.  This generalizes directly to abelian groups — just require to be integers instead.  We also required “linear independence” — if , then all .  Here we have to be more careful, since in , for example, we have for any .  So most abelian groups won’t have linearly independent sets.  What we can require is that if , then all .  We define a basis for an abelian group to be a spanning set that satisfies this requirement.

Unlike vector spaces, it isn’t immediately clear that abelian groups have bases.  If an abelian group does have a basis, then we can write it as a direct sum of the groups generated by each element of the basis, , and conversely.  This is a pretty nice way to express a group.  To prove that bases exist, we argue by induction on the number of generators of .  Since is finitely generated, there must be some minimum number of generators (and this is necessarily the cardinality of a basis, if it exists).  If this is , then we’re done: .  If it’s , then we will prove that .  Then the second summand is generated by generators, so by induction, we can find a basis for it with generators and write it as a direct sum, and the proof will be done.

Consider all the -element generating sets for and choose the one where has the smallest possible order.  If is not the direct sum , then we have, for some , , where .  Let be the gcd of the , and let .  Then has order .  If we can show that is the first element in some -element generating set, we have a contradiction.

So suppose that are a generating set for and are relatively prime integers (their gcd is 1).  We will prove the existence of a that generate with .  First observe that by replacing with if necessary, we can assume all the are nonnegative.  Now we use induction on the sum of the .  If this is , then for some , so can be the relabelled.  If it is more, then we must have at least two nonzero — reorder so that they are and , with .  But now is still a generating set, are still relatively prime, and their sum is strictly less than the original sum of the .  So by induction, there is a generating set with .  Clearly this is the generating set we want.

Now go back to our original situation and find the generating set .  We now have , so is the first member of a -element generating set and has order , which is smaller than the order of , which is a contradiction since is minimal.  So, in fact, .  By induction, .

But each either has infinite order or order .  So is respectively , or .  And so, reordering, we have:

• Classification Theorem for Finitely Generated Abelian Groups.Every finitely generated abelian group is a direct sum of cyclic groups, that is, of the form .  The first summands are the torsion subgroup, and the last one is the free subgroup.

Ordinarily I’d be glad to stop here, but the theorem is usually stated with a little more detail, so I’ll go on.  If you’re interested in finite group theory, this says that every finite abelian group is a direct sum of finite cyclic groups , and so the question becomes how to canonically write the .  There are actually different ways to do this: in , for relatively prime, the element has order , and the group has elements, so generates the whole group, and we have .

So a cyclic group of order with prime factorization is isomorphic to .  In particular, a finitely generated abelian group can be written as a direct sum of a free subgroup and a finite set of cyclic groups of prime power order.  There is only one way to do this, as can be seen by counting elements: if there are elements of order , then occurs in of the factors, and so on.  We call the the elementary divisors of .

Alternatively, starting with an elementary divisor decomposition, we could let where there is one copy of each distinct prime from among the invariant factors, with its highest exponent.  Then we could let be the product of all the distinct prime powers remaining, again with their highest exponents, and so on.  If the elementary divisors are then we have .  We have , because at each stage we’re collecting cyclic groups of relatively prime order.  The also have the property that divides for each .  We call them the invariant factors — since the are unique, so are the .  Both of these decompositions are occasionally useful, especially for finite abelian groups.

We haven’t forgotten the on the end!   is the rank of the group, something like its dimension, and it is also invariant of the group.  If you know some more advanced commutative algebra, it’s easy to see this by tensoring with (or any field of characteristic not occurring in the ), but then this page probably isn’t for you.  It’s a nice exercise to come up with a simpler, group-theoretic proof that is unique.

I’ll end with a word on non-finitely-generated abelian groups.  There are very many of these, and at a certain stage it becomes hard to tell the difference.  For example, and are both non-finitely-generated, and includes as a subgroup, but what exactly is the difference?  This is the stage when we have to appeal to analysis, with notions like limits: we can get, say, by defining it as a solution of the polynomial with rational coefficients, but we can’t do this for or .  If we want to keep talking on algebraic lines, we have to consider different ideas of “finite generation”: for example, finite-dimensional vector spaces are infinitely-generated abelian groups, so we introduce the vector space “picture” to find a way to make them finite.  Now that we have a good picture of abelian groups, I’ll start generalizing substantially and show a couple more pictures of structures, leading the way to more advanced algebra.

Then we learned about the properties closure, identity, commutative, associative, and distributive.  Closure is any whole number times another whole number will equal a whole number, or any even number times another even number will equal an even number.  Identity it is 1 being the unique number, anything times one is itself. Commutative is when you can flip the numbers and it still means the same thing like, 2×4=4×2=8.  Associative if any three numbers are in the same order it doesn’t matter where the parenthesis go, 2x(3×10)=(2×3)x 10.  Then the last one is distributive it would look like this…ax(b+c)=(axb)=(axc).  We also learned how to do mental calculations, which I think are really nice.  I found this website that tells you what the multiplication properties are and you can play games and take quizzes and stuff about it.  the link is http://www.aaamath.com/pro74bx2.htm

http://terrytao.wordpress.com/2009/12/26/a-demonstration-of-the-non-commutativity-of-the-english-language/

1. The operation on defined by .
2. The operation on defined by .
3. The operation on defined by .
4. The operation on defined by
   .
5. The operation on defined by .

1. Not commutative since but .
2. Commutative since .
3. Commutative since .
4. Commutative since
   

   	=
   	=
   	=	.

5. Not commutative since but .

Definition. A Semigroup is a associative groupoid.

Now we can move on identity elements of semigroups.

Definition. An element in a semigroup is a identity element, or a unit element, if for all in the semigroup.

These identity element are denoted with a 1 when the binary operation is multiplication or a 0 for addition. Now suppose  and are an identity elements in a semigroup . Then and thus an identity element is unique!

Definition. A monoid is semigroup with an identity element.

Example. Which of the following groupoids are monoids?

So we have a groupoid with identity element, a monoid, but we all know that we need a little more to go on to groups. However, I am not going to head there just yet.

Let be a homomorphism of Abelian groups.  Show that for each .

Proof:

Since f is a homomorphism, and since , an Abelian group, then:

(by associativity)

(f is a homomorphism)

The last line is true because (but that’s another proof).

If you can’t remember the midpoint, distance, and slope formulas for Algebra, maybe these notes i’ve made will help. Also, if you don’t understand the 4 major properties: Commutative, identity, associative, and distributive, that’s included too.  Hopefully the sheet explains it to you well.

To get the notes, you may click the picture above. This will give you the JPEG (Image) version of the notes.

To get the PDF version of the notes, click here.

*Please send me a comment if the notes are helpful or not.

The applet/game is called Factorize. Check it out.

You’re given a number and asked to illustrate the factors by selecting the appropriate dimensions on a grid. After you select a dimension (area of squares), you press the “Enter” button.

I have to admit that when you’re working with larger numbers the grid squares become a little difficult to highlight with the mouse, but I think this is a great visual activity for students.  It also helps demonstrate the Commutative property, you simply have to de-select, the “Do Not Show Commutative Property” button.

Let be a ring, a multiplicative set, and an ideal of disjoint from ; then there exists a prime ideal of containing and disjoint from .

. 

DEF. let       

.

DEF. idempotent where