AoPS Academy
and then factor the terms with x and then take log base 2 for both sides? I am not sure how to do it, so could you please help?
is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture and replaced with same amount of cocoa powder to form a new mixture 
If the ratio of coffee and cocoa in the mixture 
is 
then the ratio of cocoa in mixture to that in mixture is
)
in 
. If there's popular interest toward this direction, I'll type up another (hopefully shorter) post tomorrow.
be a field. We will denote 
to be the algebraic closure of . So what does this mean? Basically, is an arbitrary field such that it contains and all roots of polynomials with coefficients in lie in . By Zorn's Lemma one can in fact show the algebraic closure is unique in some sense ("up to isomorphisms which fix the basis field elements", if that means anything to you).
we fields and 
is a subset of 
. Then we define the dimension of over to be the maximal integer 
such that there exists elements 
such that for all 
, not all 
, we have:![\[a_1v_1 + a_2v_2 + ... + a_nv_n \neq 0 \]](http://latex.artofproblemsolving.com/6/9/5/695962adee7218cb07aab8328ede8561b69a8793.png)
. This definition looks like a funny way to define dimension (it is tempting to define it as the degree of the minimum polynomial of over , i.e. for ![$\mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/6/7/6/676ab530e9b483427222f4315d1ec46da88cb0ab.png)
over it is 
, but this definition is annoying to work with and also does not actually have to be a field in general for which this definition completely dies, but let's ignore that case for now) but it turns out to be incredibly powerful. Something important to note that is that for every element 
, there exists 
such that: ![\[b_1v_1 + b_2v_2 + ... + b_nv_n = x\]](http://latex.artofproblemsolving.com/d/0/f/d0f9211433f03c2cd1d35cdd441e4fc975e28f22.png)
so every element of is a linear combination of 
with coefficients in . To prove this just argue by contradiction using the maximality of and let 
as a set which has a weighted sum equal to . be a field and 
a root of an irreducible polynomial 
(i.e. is 's minimum polynomial). Then ![\[\text{dim}_A(A[\alpha]) = \deg P(x)\]](http://latex.artofproblemsolving.com/8/1/c/81c5228ce125574396850b37e3124387f5c09801.png)
Proof: This is not hard. Taking 
as our and then using the minimality of one derives ![$\text{dim}_A(A[\alpha]) \ge \deg P$](http://latex.artofproblemsolving.com/5/6/1/561a740237f86efeb2e1531223170d42a56967b0.png)
. To show ![$\text{dim}_A(A[\alpha]) \le \deg P$](http://latex.artofproblemsolving.com/b/4/d/b4d0549abf4276389ac3003eda4522f1e35c2b28.png)
, suppose it were greater and now consider 
items of ![$A[\alpha]$](http://latex.artofproblemsolving.com/8/1/b/81b0b888c3ff2d0a57ba4aca9435a7d1c9ecf118.png)
which have no weighted sum equal to . Note that every element of is a linear combination of , so write everything as "vectors" whose coordinates are the coefficients on the 
. Then using elimination similarly to how you would solve a system of equations, you can get a nonzero sum of these vectors equals , which contradicts our definition of dimension so we are done.
be fields such that 
. Then the following equality holds: ![\[\text{dim}_{A}(C) = \text{dim}_{B}(C) \cdot \text{dim}_A(B)\]](http://latex.artofproblemsolving.com/2/d/0/2d056808beb0f39f84ea7c30d79d26174ccaa91a.png)
Proof: Since I'm lazy, just look here.
with coefficients in 
, WLOG it is irreducible so it is the minimum polynomial of its roots. Then we want to show its roots lie in . Consider an arbitrary root 
of it. Then applying Lemma 2 we have ![$F = \mathbb{Q}[a_0,a_1,...,a_n]$](http://latex.artofproblemsolving.com/e/5/0/e50f580307aa63b0939698b6f48271ebd5913261.png)
satisfies 
is finite. Applying Lemma 2 again on 
, 
, ![$C = F[r]$](http://latex.artofproblemsolving.com/4/d/2/4d2cb9015d0ca459368b2827df378214705281de.png)
we get ![$\text{dim}_{\mathbb{Q}}(F[r])$](http://latex.artofproblemsolving.com/b/e/e/beef6eed5b9a2a2c3682428cc3fb9d69ff710ec4.png)
is finite and then thus as ![$\mathbb{Q}[r] \subset F[r]$](http://latex.artofproblemsolving.com/b/9/0/b907d5cc8043a8493a94d7d01ff58a89c981471b.png)
, it follows ![$\text{dim}_{\mathbb{Q}}(\mathbb{Q}[r])$](http://latex.artofproblemsolving.com/b/0/9/b0945a954052fc0b0048ced60e6b7058b38f6519.png)
is finite as well. It follows that lies in as well by applying Lemma 1 to yield is the root of some polynomial with rational coefficients and finite degree, so it follows is algebraically closed as desired. in " is difficult to answer elementarily. A very short way is to note that the algebraic closure is a subset of the algebraic closure of 
and so the result follows by the Fundamental Theorem of Algebra, but I think I'm misinterpreting what ThinkFlow means...
that you may find interesting. be fields such that . Then the following inequality holds: Proof: Since I'm lazy, just look here.
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. 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 ![$F[z]$](http://latex.artofproblemsolving.com/f/1/d/f1d93ad0a9c1b4dd8a334e90f0463dfe1742c1f6.png)
(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!) 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.
algebraic implies 
are algebraic) works just as well for any field extension of . If the following arguments are confusing, think of as and as . is a field extension of , then 
. 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. 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 .
. All these polynomial equations have coefficients in . The number 
means the 
-th root of the polynomial 
in ![$F[x]$](http://latex.artofproblemsolving.com/3/0/5/3058cfec73981ccfab32d12bb576d63b2bf97bf6.png)
that is assumed to satisfy.
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.
, 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. directly, instead of our way of writing in terms of the (and ). 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 . has an algebraic closure , unique up to -isomorphism.
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. 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![\[
a_{n-1}z^{n-1} + a_{n-2}z^{n-2} + \cdots +a_0,
\]](http://latex.artofproblemsolving.com/a/3/f/a3f7e8c04107275e8dd65c4b09f38bb8166e0c34.png)
'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 .
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 . is called the minimal polynomial of ., 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 ![$F'[z]$](http://latex.artofproblemsolving.com/2/c/9/2c90d38e2b502377c6358e642539536dc29eaa85.png)
, 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.-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.), , 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. is algebraically closed, is actually a genuine algebraic closure of . This is because any polynomial in ![$\mathbb{Q}[x]$](http://latex.artofproblemsolving.com/3/d/9/3d963b0224d7c8b822ad2df9d699fd043f0013e8.png)
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, 
. 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. 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., and their applications to solving number theory problems in the integers.
.
on a MathCounts-level forum
ruined the thread, as the abstract algebra improved it.
where 
represents the 
![\[ \frac{\lim_{x\to\infty}f(x+1)}{\lim_{x\to\infty}f(x)}=\frac{1+\sqrt5}{2} \]](http://latex.artofproblemsolving.com/5/9/e/59e3e925f71c3f0d9f76bb1325c2f241b3265768.png)
![\[\lim_{x\to\infty}\frac{f(x+1)}{f(x)} = \]](http://latex.artofproblemsolving.com/c/8/6/c86f4d50d61f57337d5e54a77947e806dc201f35.png)
![\[\lim_{x\to\infty}\frac{f(x) + f(x-1)}{f(x)} \]](http://latex.artofproblemsolving.com/d/5/5/d55bf0a2de64d2a9d4b5f426cd250e48df023879.png)
![\[\lim_{x\to\infty}\left( 1 + \frac{1}{\frac{f(x)}{f(x-1)}}\right)\]](http://latex.artofproblemsolving.com/4/1/0/410d5ac92b3ae5b7268b8b37eb2b3e8ef6cde969.png)
![\[\lim_{x\to\infty}\left( 1 + \frac{1}{\frac{f((x-1)+1)}{f(x-1)}}\right)\]](http://latex.artofproblemsolving.com/a/0/1/a01d4f06b1c8dfaeb9c4ca0f484186b9467b2569.png)
so we get![\[1 + \frac{1}{1 + \frac{1}{1+\dots}}\]](http://latex.artofproblemsolving.com/2/d/c/2dc75ddb2f1a36cdd8643504dd1b228912d17a3d.png)
can only be 1 or 0 modulo 4.
.
when the perimeter is 
and the other be 
,
or 
.
.
.
,
.
.
and 
.![\[\frac{a+b}{2} \ge \sqrt{ab}\]](http://latex.artofproblemsolving.com/b/d/5/bd546a96b9fa8b1d77afe09bfc85d1390263ed11.png)
![\[\left(\frac{a+b}{2}\right)^2 \ge ab\]](http://latex.artofproblemsolving.com/9/d/3/9d3a72a2d5b8f5ea54eb032e1fbbf79660a3be00.png)
![\[s^2 \ge ab\]](http://latex.artofproblemsolving.com/7/3/7/737c246f212778d91d5c5af512fa4cb2d28425b9.png)
candies and Jill has 
.
.
.
(we will omit the parentheses when possible, so e.g. 
means the same) is successor function, then for all natural 
we have
,
and 
,
,
.
,
is such 
This finishes the proof.