Random math proof marathon.

Tell me if it's in the wrong forum, because I don't notice any others.

I'll start.

prove that where represents the th Fibonacci number.

Please do not do calculus in this forum.

Limits are too advanced for here. I don't know them.

Limits
I barely know them, and may be using them in the incorrect context (but probably not), but here's a link. They seem pretty simple.
Anyway, I'm trying to say as approaches infinity, whether it comes out that way or not.

I agree, limits seem too complex for this forum.

Posting this in the high school threads is appropriate I think, but not here.

Yeah.. this definitely needs moving..
Solution 1
Problem 2

We can list all the modulos of 4 and square them.

1^2=1 mod 4
2^2=0 mod 4
3^2=1 mod 4

As you can see, only 0 and 1 mod 4 are possible outcomes. I don't have a new problem .

Problem 3Click to reveal hidden text

Solution

New Problem

Solution 3
Edit: snipeed...

Ugh, you guys should give problems that people like me can understand since this is in the middle school forum.

The greatest area of a rectangle comes when it is a square. I will try to prove that. The sum of the length and width is 2s. Let L be s+x and W be s-x. Then we get Area = L*W = s^2-x^2. x^2 is a positive number for any x not equal to zero, so the maximum area would be s^2.

Well.. problem 4 is prove the binomial theorem without induction

Unfortunately Theorem 1 takes some Galois Theory which might be slightly out of the range of MC National Sprint tests. (maybe even too advanced for target )

Indeed, the second more general generalization is also on math154's blog.

It's #2 on his most recent entry at http://www.artofproblemsolving.com/blog/80001.

Also,

Nice "inequality."

Hey you made fun of me for writing "field" instead of "ring," and you cannot expect me to show any mercy for you writing "inequality" instead of "equality."

Hi, If you are really interested and determined, then I suggest this book http://www.amazon.com/Abels-Theorem-Problems-Solutions-International/dp/1402021860 It is very understandable, doesn't assume prior knowledge and it is basically focused on developing the theory that is then used to prove Abel's theorem.

As promised, I'll define algebraic closures and algebraically closed fields and explain what they mean in the context of the algebraic numbers. I present the concepts in a different order than dinoboy does, but we cover essentially the same ground. By continuing to use Theorem 3 for symmetric polynomials, we will avoid the use of the dimension of a field extension altogether.

Suppose we are given a field containing a subfield . For example, our larger field may be and our contained field . Another way of saying " is a subfield of " is " is a field extension of ." We use the term field extension when we emphasize what must be additionally included (in fancier terms, "adjoined") in , in order to obtain . We've already seen one example: an algebraic field extension is when algebraic elements are adjoined.

We define the algebraic closure of in to be the set of all numbers in that are algebraic over (which means that satisfies a polynomial equation with coefficients in , and that (which lives inside since and are in ) is an algebraic extension of ). We will write the algebraic closure of in as . (This is different than the that dinoboy defines!)

We have already considered a special case of this definition. When we defined the algebraic numbers to be those numbers in which satisfy polynomials with coefficients in , we were really considering the algebraic closure of in . Again, the algebraic numbers are precisely the algebraic closure of in . Furthermore, are the algebraic numbers which happen to be real, which is a subfield of the algebraic numbers.

Our proof that the algebraic numbers form a field (remember that we checked that algebraic implies are algebraic) works just as well for any field extension of . If the following arguments are confusing, think of as and as .

Theorem 6 If is a field extension of , then .

This theorem says that, after taking the algebraic closure of in once, taking the algebraic closure again does not introduce any additional elements to . After all, it is clear that contains because taking an algebraic closure can't take away any elements already in our field. Any new elements in that are not in must satisfy polynomial equations with coefficients in , because of the definition of algebraic closures. Our goal is to show that, if this happens for some element , then must satisfy some more complicated polynomial equation whose coefficients are in . This will imply that was already in to begin with.

Suppose that satisfies a polynomial equation , such that each coefficient lies in , which means that satisfies its own polynomial equation with coefficients in . We can assume the leading coefficient is 1 because we may divide by . Denote all the roots of by , where is the degree of .

We have introduced a lot of subscripts for the purpose of talking about those numbers that satisfy the same polynomial equations as do the coefficients of . All these polynomial equations have coefficients in . The number means the -th root of the polynomial in that is assumed to satisfy.

Now define a (very large) set of polynomials where , where the are obtained by replacing coefficients in with for some . Since there are choices for this, for each , we obtain our value for . I claim that is a polynomial with coefficients in which has as a root.

Since , we only need to show that , whose coefficients we know to be in , are in fact in . View as a polynomial in the and , with coefficients in . By the construction of , is unchanged when we switch and , for any fixed . Using this for , Theorem 3 lets us write as a polynomial in , the symmetric polynomials , and the remaining , now with . Repeating this procedure for , we write as a polynomial in , the symmetric polynomials , the symmetric polynomials , and the remaining , now with . Continuing in this way, for each from to , we write as a polynomial in and the symmetric polynomials for each fixed . Because the 's have coefficients in , the 's are all in , so is a polynomial in with coefficients in , as desired.

The credit for this argument, which is much simpler than what I had initially planned, goes to geoishard. My original argument involved even more symmetric polynomial bashing, working with symmetric polynomials for the roots of directly, instead of our way of writing in terms of the (and ).

Now we say a field is algebraically closed if any polynomial with coefficients in actually has a root in . Theorem 2 says that is algebraically closed, and the paragraph following that theorem implies that all polynomials with in , where is algebraically closed, can actually be factored into linear factors within that field. In a somewhat confusing turn of terminology, we define the algebraic closure of a field , without reference to a larger field , to be an algebraically closed field containing such that the algebraic closure in is itself. In other words, the algebraic closure of is the smallest algebraic field extension of that is also algebraically closed. This is the same as dinoboy's .

Theorem 7 Every field has an algebraic closure , unique up to -isomorphism.

This means that given two algebraic closures and (both of which must contain ), there is a one-to-one correspondence between the elements of outside and the elements of outside , which matches up with addition and multiplication. In other words, and are really the same field extension of , just with the newly-adjoined elements given different names.

The general method of attack for Theorem 7 is refreshingly direct, and relies on Zorn's lemma, as dinoboy mentions. Specifically, given an irreducible polynomial with coefficients in a field (irreducible means can't be factored into smaller-degree polynomials with coefficients in ), it's possible to "adjoin a root of ," in the sense that we construct a new field where the only relationships satisfies are those required by . In this way, if has degree , we may express each element of in the form

where the 's lie in , using the remarks following Theorem 2 (expressing high powers of as sums of lower powers, because of ). In fact, since is not technically in any field containing yet, we define to be all expressions of this form. Multiplication and addition work fine, because of the distributive property, followed by reducing the high powers of using .

Division also works fine, because given an element of , we can view it as a polynomial in , with degree less than . The Euclidean algorithm works as well for polynomial division as it does for integer division, and running it in reverse for the relatively prime polynomials and (no common polynomial factors because is irreducible) gives polynomials and such that . Substituting gives , so is an inverse for , provided we use to reduce high powers of . Furthermore, happens to be unique if . Hence as defined is in fact a field extension of , obtained by adjoining , which satisfies an arbitrary (irreducible) polynomial equation .

I think dinoboy's exposition assumes this notion of field extension. Note that our irreducible is called the minimal polynomial of .

To informally construct , just "keep adjoining roots of polynomials which don't yet have solutions." That is, at each step, consider a new (irreducible) polynomial with coefficients in our current field , which is an algebraic extension of . Adjoin a root of using the procedure described above, to obtain a field , and repeat. Zorn's lemma is the axiom that allows us to consider the field we get after doing this "infinitely often." More precisely, each of these operations defines the next member of a chain of increasingly large field extensions, each containing the last. Now every chain of field extensions so defined has an upper bound, which is the union of all the fields in that chain of field extensions. Under these conditions, Zorn's lemma allows us to find a maximal element, where there are necessarily no more 's to adjoin, which means no more irreducible polynomials to solve. This is our , because this means that all polynomials have roots in that field.

We can use Zorn's lemma again to construct an -isomorphism between any two algebraic closures of , by taking the identity map of to itself and expanding this element-by-element to include the whole algebraic closures. Zorn's lemma gives us the result where there are no more elements to be included in the isomorphism, so we've covered everything. (This is very informal, but I hope it gives a hint of the flavor of "Axiom of Choice" arguments.)

To tie all these definitions and results together, let's return to our examples of , , and . We can consider the algebraic closure of in , and the algebraic closure of in . (Caution: I don't think the notation is standard.) Note that is the intersection of with ; it's what we get when we adjoin every real algebraic number. Of course, is the ring of algebraic numbers.

Since is algebraically closed, is actually a genuine algebraic closure of . This is because any polynomial in factors into linear factors in , and contains those because they are algebraic numbers. Since is clearly an algebraic field extension of by construction of the algebraic closure of in , we see that is an algebraic closure of , and by uniqueness we say it is the algebraic closure of , up to -isomorphism. In symbols, .

But an algebraic closure of in is not always algebraically closed, that is, it's not always an algebraic closure of , without reference to a field extension. For instance, is not algebraically closed because has no solutions. Simply put, since we only adjoined the elements of that were algebraic over , we never bothered to adjoin a root of , which doesn't have real roots! So is not algebraically closed.

There is, of course, more than one way to arrive at these results, and dinoboy's exposition is probably more standard, because the dimension of an algebraic field extension is a very important concept. The dimension of a field extension over is the dimension of as a vector space over , in the same sense that (ordered triples, with addition and multiplication by a constant operating component-wise) is an -dimensional space. The liberal use of symmetric polynomials is one way to circumvent the use of these ideas from linear algebra.

Those interested in these beginnings of field theory or algebraic number theory might read this article, which considers a greater variety of field extensions of , and their applications to solving number theory problems in the integers.

darn I thought this was the Superior Algebra forum for a second

all right back to MathCounts-level problems

Problem 13

Ugh what is Theorem 7 .-.

also dinoboy, you typoed the Zorn's lemma URL in your post.

SolUTIon
ProBELLm 14

I don't think you can just cite .

partisal salution

In my calculus class, about a third of the year was spent proving statements from the axioms, so how you can you just cite these?

You could also use Peano axioms for Problem 13
Click to reveal hidden text

wow all the abstract algebra ruined this thread

seriously we're trying to prove that on a MathCounts-level forum

what is this

I think the more accurate statement is ruined the thread, as the abstract algebra improved it.