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That would require knowing what F, m, L, E, n, G all are defined to be.
After that square both sides, then make another substitution that will leave you a polynomial equation on just 1 function. Solve the polynomial (easier said than done - unless I made a mistake somewhere it doesn't have any rational roots), then take the inverse trig function to get the answers.
A nice proof. Geometric proofs are always lovely. But I have to disagree with Nehushtan's definition of "simple". You came up with that analytic geometry solution within a few hours of the original question. It took Bob days to produce the geometric proof. The analytic solution is straight-forward, while the geometric solution required experimentation and insight. It may not be as beautiful, but the analytic solution is definitely simpler.
Even though apparently no one else found the recursion formula as nice as I did, I am going to put another favorite discovery, this one more recent - I only realized this last year. (Both results were of course known well before. They were just discoveries for me.)
I had long heard that the Real numbers were the only complete ordered field, but knew only half of why this was so. Last year while studying the Surreals, I finally noticed the missing half. First, the easy part (okay - it takes more work to dot the i's and cross the t's, but it is a well-known construction.)
Theorem: Every ordered field contains the rational numbers.
(Some would say that should be "contains a copy of the rational numbers". But the rationals (and the reals) are defined only by their properties, so in some sense any copy of the rationals is the rationals.)
I'll only give an outline of the proof:
Let F be an ordered field. Then F contains 0 and 1. Define inductively a map f from the set Q of rationals into F by f(0) = 0[sub]F[/sub] and for all non-negative integers n, f(n + 1) = f(n) + 1[sub]F[/sub]. Note that since F is ordered, for all x in F, x + 1 > x, which allows us to prove inductively that if n < m, then f(n) < f(m), so f is 1-1. That f also obeys f(n +m) = f(n) + f(m), and f(nm) = f(n) f(m) are also straight-forward inductive proofs.
f can be expanded to negatives by f(-n) = -f(n), and to all rationals by f(n/m) = f(n) / f(m). It is easy to check that these are well-defined, and that the homomorphic and injective properties still hold. Just f injects the rationals into F.
QED
For ordered sets, the concept of "completeness" is defined in terms of the order. An ordered set is complete if and only if it possesses the supremum property: every non-empty subset that is bounded above has a least upper bound. That is, if A is a non-empty subset, and if there is some M such that for all x ∈ A, x < M, then there is some S such that:
A ≤ S (that is, for all x ∈ A, x ≤ S).
If for some y, A ≤ y, then S ≤ y.
define sup(A) = S.
Theorem: Any complete ordered field contains the real numbers.
One again, an outline proof: Every r in the set R of real numbers is the supremum of the rational numbers below it: r = sup{q ∈ Q : q < r}. So extend the injection f of Q into F to all of R by f(r) = sup{f(q) : q is rational and less than r}.
f is well-defined and still 1-1 and a homomorphism. Thus it injects R into F.
QED
That is one side: any complete ordered field must contain the real numbers. That is, it is an extension field of the real numbers. The other side of this theorem is:
Theorem: Any proper ordered extension field of the real numbers is not complete.
First we prove a lemma:
Lemma: there is at least one infinite element in a proper ordered extension field of the real numbers.
By "infinite", I mean greater than all real numbers. The field must have at least one element, x, which is not a real number (otherwise it wouldn't be proper). Since -x also cannot be real, we can assume x > 0. If x is greater than all real numbers, then we are done. So assume there is at least one real number greater than x. Define the set A = { r ∈ R : r < x }. Then 0 ∈ A and A is bounded above in the reals, so, since the reals are complete, there is a real number s = sup(A). Since s is real, s ≠ x. s could be either above or below x. But either way, there cannot be any other real number between s and x. For by definition, s ≥ A, which include all real numbers less than x, and since all real numbers above x are upper bounds for A, s ≤ all real numbers above x.
Define y = | x - s |. y > 0 since s ≠ x. And if 0 < r < y, then either s + r < x, or x < s - r. In either case, r cannot be real, as one of s + r or s - r is closer to x than s. Therefore there is no real number between y and 0. If t is any positive real number, y < 1/t. Which means also that t < 1/y. Hence 1/y is an infinite element of the field.
QED
Now assume that the proper ordered extension field is complete. Then by the lemma, R is bounded above. So R must have a supremum w. But w - 1 < w, which means that it cannot be an upper bound for R. So there exists some real number r such that w - 1 < r. But then w < r + 1, which is also real. This contradicts the definition of w. Therefore there cannot a complete proper ordered extension field of R.
QED
Okay. There are ways to combine some of the axioms into one, but the result is usually some odd statement that is hard to make sense of until you manage to derive the normal axioms from them. So normally, we go with the easily comprehended axioms, instead of the fewest possible.
I was just curious if you were refering to some such reduction.
Hi;
In Bengali, "Tuesday" is called "Mangal Baar"("Baar" means day in Bengali). And "Mangal" means Mars.
And similarly "Tues" is called Mars?
Half a year late, but: the Romans named the days of the week after their gods. All other cultures eventually adopted the Roman naming convention, but some of them translated the god names to the names of equivalent gods in their own culture.
The Germanic tribes did this. Their god of war was named "Tir" or "Tew", depending on the dialect. So they changed Mars to Tew, which by the time it reached English had morphed into "Tues". European languages generally use a variant of either Mars or Tew as the name for this day.
The Germanic tribes has no god equivalent to Saturn, though, so they kept the Roman name for the last day of the week. That is why in English we have Germanic gods for the first 6 days, but a Roman god for the 7th.
Snell's Law
Assume (1) is the air and (2) is the medium, then the left side of the equation for Medium B and Medium C is the same. Hence:
Happened to come across this old thread via a google search. Anyways, we were asked to prove this for a homework assignment. Obviously Cantor's proof is elegant and so it is widely used. For my proof, I constructed a 1-to-1 mapping with the natural numbers mapping to their reciprocal. Then I merely pointed out a real number in that interval (I used 2/3).
Does this properly prove that there are more reals than natural numbers or am I missing something?
I don't know if n00b is still around, but yes, if I understand your description of your proof correctly, then you have missed the whole idea.
It is not sufficient to show that there is a 1-1 mapping from the natural numbers into the reals that misses some numbers. After all, your mapping, and your example, were actually in the rational numbers. So if your reasoning were correct, there would be more rational numbers than naturals, which is false. In fact, by the same idea, a simple mapping would show that there are more natural numbers than natural numbers! The 1-1 mapping
What these mappings really show is that the natural numbers are infinite (the definition of "infinite set" is "a set which has a 1-1 mapping with a proper subset of itself"), and that the cardinality of the Reals and Rationals are both greater than or equal to the Naturals, which also follows from the simple fact that the Naturals are a subset of both.
By definition, two sets are the same size, or cardinality, if there is a 1-1 correspondence between them that includes every element of both sets. Cantor's proof shows that every mapping from Natural numbers into the Real numbers must miss at least one real number. Therefore the cardinality of the Real numbers cannot be equal to that of the Naturals. Combined with "greater than or equal to" already noted, we get that the cardinality of the Reals is strictly greater than that of the Naturals.
(I am amazed that list of useful symbols at the top of the page doesn't include pi!)
tau = 2 pi, the measure of a full circle. There are those who argue that tau would be a better choice than pi, but I haven't found that their arguments hold much water. For every calculation it simplifies, it makes another more complex.
Patternman, when you learn about fields or even before that, you'll prove basic things which you considered axioms using 7 statements which are actually axioms over the field of rational numbers. For now, just proving the theorems is what most people focus on. If you want to prove the "axioms" which aren't really axioms, then read the first three chapters of Principles of Mathematics by Oakley and Allendoerfer.
7 statements? What set of axioms are you using? The usual for fields are:
(1) Addition is commutative.
(2) Addition is associative.
(3) There exists an additive identity 0.
(4) Existance of additive inverses (opposites).
(5) Multiplication is commutative.
(6) Multiplication is associative.
(7) There exists a multiplicative identity 1.
(8) Existance of multiplicative inverses (except for 0).
(9) Distributivity of multiplication over addition.
Those specify an arbitrary field. To get the rational or real numbers in particular, you need to add more.
And they are axioms, or definitions (the two concepts are really the same). Oakley and Allendoerfer may build a model of the rational or real numbers and prove that these axioms hold for their model. But commonly in math we simply consider such things as primative elements defined by axioms, rather than tying them to a particular model. Models are used to guarantee that our axioms are not contradictory, and they sometimes can provide good insight into the objects of study, but we are studying the objects themselves, not ways to construct them.
The problem is to evaluate
As bob as already indicated
Hopefully, you should now (bob already used this fact once) that
So putting it together gives:
Now put in your limits of integration (where I can drop the "+ C", as it cancels out):
It still needs a bit more simplification, but hopefully this will straighten out where you went wrong.
Duplicate colors allowed, but not adjacent:
The middle band can have 1 of 5 colors. The top band can be any color except the one chosen for the middle, so 4 colors. The bottom band can have any color except the one chosen for the middle, so 4 colors again. That makes 5 * 4 * 4 combinations.
The calculation for no two colors the same is almost identical, but then the bottom band can't have the same color as either the middle or the top band, so 5 *4 * 3.
Forgive me, please, for taking this thread all the way back to the beginning, and also if someone else has pointed this out in those thousands of posts I have despaired of hunting through to check:
How would you judge this answer? And why?
The question itself and the replies given suggest this is wrong, but I would judge this answer correct. The following theorems are all easily proved:
The calculation shown follows from them.
After "[hide", type "=", followed immediately by the text of your choice and the closing bracket ']'.
Thanks!
It's been a while since been in a forum. It's embarrassing that I'd forgotten about the quote trick!
Nice!
Ah. This is obviously some strange usage of the word "reliable" that I hadn't previously been aware of.
What I would call these numbers is very, very badly biased. What do you have against the poor folks in town 1 that you think that the value of a vote by an individual in this town as compared to the value of a vote of an individual in town 3 should be in the same ratio as the population of the towns?? It's bad enough that even normally, town 3 has so much more population that essentially it controls the outcomes of all elections, with the other two towns making a difference only when the population in town 3 is close to equally divided. But now you want to devalue the votes in these small towns even more?
It doesn't need conditions to make it work. It is just a condition that some functions can satisfy. If we add a very common additional condition to it, namely that the function f be continuous, then the only functions that satisfy it (assuming f is a function from real numbers to real numbers) are f(x) = Ax, for constants A.
Without adding the continuity condition, then it is possible to break the real numbers into a collection of subsets, each of which is closed under addition, and which intersect only in that they all contain 0, so that on each of these sets f(x) = Ax for some A, but the value A differs from set to set. These sets are intermixed with each other in much the same way that the rationals and irrational intermix, so the graph of such a function doesn't look like anything you are used to, but rather it looks like a cloud of points.
A minor alternative that can make the calculus a little easier:
Question: it was pretty easy to guess how to hide the solution, but how do you get alternative text other "Hidden Text"?
Probably. If I recall, the proof was to examine numbers up to 169 by brute force. Above 169, you first subtracted 169 from it, apply the Lagrange 4 square theorem to write the result as a sum of 4 squares, some of which could be zero. Then just took noticing that 169 = 13[sup]2[/sup] = 12[sup]2[/sup] + 5[sup]2[/sup] = 12[sup]2[/sup] + 4[sup]2[/sup] + 3[sup]2[/sup] = 10[sup]2[/sup] + 7[sup]2[/sup] + 4[sup]2[/sup] + 2[sup]2[/sup]. So if N is the original number and N - 169 = A[sup]2[/sup] + B[sup]2[/sup] + C[sup]2[/sup] + D[sup]2[/sup], with A >= B >= C >= D, then:
If D > 0, then N = A[sup]2[/sup] + B[sup]2[/sup] + C[sup]2[/sup] + D[sup]2[/sup] + 13[sup]2[/sup].
If D = 0, C > 0, then N = A[sup]2[/sup] + B[sup]2[/sup] + C[sup]2[/sup] + 12[sup]2[/sup] + 5[sup]2[/sup].
If C = 0, B > 0, then N = A[sup]2[/sup] + B[sup]2[/sup] + 12[sup]2[/sup] + 4[sup]2[/sup] + 3[sup]2[/sup].
If B = 0, then N = A[sup]2[/sup] + 10[sup]2[/sup] + 7[sup]2[/sup] + 4[sup]2[/sup] + 2[sup]2[/sup].
(A cannot be 0).
The greatest tool in the mathematician's tool chest is abstraction. I cannot count all the times when I've done some long, horrific calculation or proof, and some time later came across an abstraction that made the whole thing almost trivial.
Example: prove that det(AB) = det(A)det(B). If you try to do that for higher dimensions by looking at the coordinate formula, you'll be gibbering before very long (2D isn't bad, 3 is a pain, 4 is awful, much higher, and you might as well throw the whole thing in). But if you develop the concept of vector spaces, and introduce the wedge product, then suddenly, the determinant pops up in such a way that the det(AB) = det(A)det(B) result is a trivial consequence.
I find analysis more interesting myself, but abstract algebra shows up in pretty much all other fields, so it is a very good thing to master.
Agnishom has it right, of course. But this only works for very symmetric shapes. No matter how you try to work it on ellipses and ellipsoids, for example, it fails.
Oops - just checked, and its the sum of 5 positive squares that is always possible after a certain point. I think 4 positive squares will always have exceptions (though if 0 is allowed, Lagrange's 4 square theorem says that natural number can be written as sum of 4 squares).
Leaving aside questions of infinite sets, I think the comparison caters is wanting is asymptotic density: For any set A, let A[sub]N[/sub] be the number values in it that are <= N. His question is, I think, how does |A[sub]N[/sub] π B[sub]N[/sub]| / N for some pair of sets A and B compare to the same ratio for other pairs of sets as N increases without bound? (If you haven't seen the notation before, the "absolute value" signs here actually mean cardinality, or set size).
Unfortunately, you can't extrapolate behavior from low numbers to high numbers (and any time you are testing by brute force, you are only looking at low numbers, no matter how high you go). The problem is, in the low numbers, there just isn't enough variation for some things to happen that otherwise would, while the tight quarters also forces other coincidences to occur that would otherwise be quite rare. As the numbers grow, there is more room to move around, as it were, and the behavior starts to change.
I haven't studied prime sets much myself, so I can't comment too deeply on them, but as a different example of what I am talking about: try writing numbers out as a sum of 4 positive squares. It obviously cannot be done for 1, 2, or 3. 4 = 1[sup]2[/sup] + 1[sup]2[/sup] + 1[sup]2[/sup] + 1[sup]2[/sup]. For 5 and 6, it is impossible again, but 7 = 2[sup]2[/sup] + 1[sup]2[/sup] + 1[sup]2[/sup] + 1[sup]2[/sup], and so on.
From the first 30 numbers, you would expect that occasional failures to be expressible as 4 squares would continually crop up. But in fact, there are only 12 of them, with the highest being somewhere in the 30s (I don't remember the exact value and am too lazy to look it up). So if you don't look far enough, you get the wrong about the behavior.
The problem is, you also generally have no idea about how far is far enough. For some problems 40 or 50 are enough to see what's going on. For others, even Ackerman's number may prove far too small. This is the shortcoming of experimental mathematics - it can only tell you about where you've looked. The behavior might change just beyond.
Just want to share one of my favorite bits of mathematics.
Consider the set
of all sequences satisfying the linear recursionDefine addition and scalar multiplication on V by:
Now suppose
Assuming that all the roots of the polynomial are unique, then we have the needed k independent sequences, so any sequence X in V is given by
For example, consider the a sequence that *just possibly* you may have heard of before:
, with .So the Fibonacci sequence F satisfies
for some A, B. To find these values, note thatSo we have Binet's formula:
.Perhaps part of the reason this appeals to me is that I was able to figure it out myself. I had seen Binet's formula many times, but never a derivation, so I finally worked out how to prove it. What is really nice is that it generalizes. Any linear recursion sequence has a similar formula (though there is a twist thrown in if the polynomial has roots of multiplicity greater than one).
As a challenge (I do know the answer), what has to change if the polynomial does have a double or higher order root? Experimenting with the recursion
can be instructive.Some examples, to help you realize what is going on:
Let f(x) = 2x
Then, for example: f(3 + 1) = f(4) = 8. f(3) = 6. f(1) = 2, and 2 + 6 = 8, so indeed f(3+1) = f(3) + f(1). More generally f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)
But now, let f(x) = x^2
Then f(3 + 1) = f(4) = 16, but f(3) = 9 and f(1) = 1, so f(3+1) ≠ f(3) + f(1)
Just because someone has studied functions satisfying an equation does not mean every function satisfies it. Very, very few functions satisfy this one. The ONLY functions you know that satisfy this equation are those of the form f(x) = Ax for some constant A. Even constant non-zero functions fail it.