Wikipedia:Reference desk/Archives/Mathematics/2007 October 2

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October 2

"Super" composites

Hi,
For no particular reason I recently thought to myself what would be the opposite of a prime number. Well I suppose the conventional answer would be a composite number. But what I then thought was what would really be an opposite would be a number which is divisible by all numbers below it - ie it has every number as a factor. Yes that is a bit of an abstract concept but I was wondering if such a thing had been described anywhere and if it had a particular name? An interesting thing about this "ideal" number would be that it would be impossible to calculate since there are infinite primes and by it's definition it would need to be divisible by all primes. And hence it could only ever be defined within a limited set.

Anyway I had a shot at calculating these numbers in progressive sets and this is what I got for the first 0-50:
0, 1
1, 1
2, 2
3, 6
4, 12
5, 60
6, 60
7, 420
8, 840
9, 2520
10, 2520
11, 27,720
12, 27,720
13, 360,360 (interesting no.!)
14, 360,360
15, 360,360
16, 720,720
17, 12,252,240
18, 12,252,240
19, 232,792,560
20, 232,792,560
21, 232,792,560
22, 232,792,560
23, 5,354,228,880
24, 5,354,228,880
25, 26,771,144,400
26, 26,771,144,400
27, 80,313,433,200
28, 80,313,433,200
29, 2,329,089,562,800
30, 2,329,089,562,800
31, 72,201,776,446,800
32, 144,403,552,893,600
33, 144,403,552,893,600
34, 144,403,552,893,600
35, 144,403,552,893,600
36, 144,403,552,893,600
37, 5,342,931,457,063,200
38, 5,342,931,457,063,200
39, 5,342,931,457,063,200
40, 5,342,931,457,063,200
41, 219,060,189,739,591,200
42, 219,060,189,739,591,200
43, 9,419,588,158,802,421,600
44, 9,419,588,158,802,421,600
45, 9,419,588,158,802,421,600
46, 9,419,588,158,802,421,600
47, 442,720,643,463,713,815,200
48, 442,720,643,463,713,815,200
49, 3,099,044,504,245,996,706,400
50, 3,099,044,504,245,996,706,400

and the rest up to 500.

Where the LHS shows the number at which all integers below it are factors of the RHS number. (ie 2520 is divisible by all numbers from 1-10) Now I know this in now way constitutes a formula or anything but I noticed the following whilst generating this list: you need to multiply "the number" each time you reach a prime; each time you reach a square number you have to multiply by the lowest prime factor of the square root of this number (eg for 16 you needed to multiply the previous figure by 2 which is the lowest prime factor of 4 and when I reached 25 I needed to multiply by 5); I haven't generated enough numbers yet (probably should have gone to 27) but judging from 8 it looks the same "square rule" applies for cubic numbers...
Anyway would appreciate any comments regarding this! --Fir0002 03:29, 2 October 2007 (UTC)

Higher numbers can be calculated using http://wims.unice.fr/wims/wims.cgi?module=tool/popup.en&search=lcm 91.94.58.53 (talk) 11:52, 3 April 2008 (UTC)

The closest thing I can think of is the Primorial, but that's not quite the same. It is a number that is the product of all primes below a given number. - Rainwarrior 03:39, 2 October 2007 (UTC)

Oh, there's a link in that article to Highly composite number, which is getting a lot closer to your idea. - Rainwarrior 03:40, 2 October 2007 (UTC)

And continuing on Superior highly composite number is even closer. - Rainwarrior 03:43, 2 October 2007 (UTC)

Thanks for your links Rainwarrior, nearly what I was after. But they were useful anyway coz it lead me to Perfect number which is pretty interesting too. --Fir0002 03:57, 2 October 2007 (UTC)

A simple way to define this sequence is that its nth term is the Least common multiple of the integers 1 through n. It's in the On-Line Encyclopedia of Integer Sequences at [1]. --tcsetattr (talk / contribs) 03:49, 2 October 2007 (UTC)

Thanks a lot tcsetattr!! --Fir0002 03:57, 2 October 2007 (UTC)

Note that this sequence has very little to do with your primary idea of "a number which is divisible by all numbers below it". This is clearly impossible (for numbers greater than 2), since n would never be divisible by $n-1$. -- Meni Rosenfeld (talk) 08:52, 2 October 2007 (UTC)

That's true but I already acknowledged that such a number would be impossible due to the infinite number of primes --Fir0002 10:35, 2 October 2007 (UTC)

On a similar note, n! has the greatest number of divisors (namely, 2^n-1) out of numbers less than n!. —Preceding unsigned comment added by 122.148.69.121 (talk) 11:02, 2 October 2007 (UTC)

No. n! has 2^n-1 divisors for n = 1 to 5, but less for all n > 5. For example, 6! has 30 divisors. And numbers with more divisors than all smaller numbers are called highly composite numbers. The only factorials among them are n! for n = 0 to 7. PrimeHunter 12:07, 2 October 2007 (UTC)

Did anyone else notice that the digital number of every super composite after 9 (including 9 too) is 9? Xhin Give Back Our Membership! 21:45, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

Wait, disregard that, I forgot that the digital number of any number that's multipled by nine is nine :( Xhin Give Back Our Membership! 21:46, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

You must be referring to digital root. PrimeHunter 00:59, 3 October 2007 (UTC)

Mathematics problem!Need help please!

The product and quotient rules give you the ability to symbolically calculate the derivative of a wide range of functions. However, many calculators and almost every computer algebra system (CAS) can do this work for you. Discuss why you should learn these basic rules anyway. —Preceding unsigned comment added by Nelsonkhoo (talk • contribs) 05:23, 2 October 2007 (UTC)

The most basic answer is that it's impractical to reach for a calculator every time you want to take a derivative. Students who fail to learn the product rule or quotient rule will slow at differentiating, and therefore slow at solving calculus problems. They will also find themselves unable to understand integration by parts, they will have difficulty solving differential equations using integrating factors, they will be surprised by the form of the multivariable chain rule, and so forth. Learning mathematics involves mastering each stage before moving on to the next, and students who don't learn these rules will find that future mathematics makes less sense than it would otherwise.

In addition, the idea behind the product rule is one of the most important ideas in calculus: when multiple quantities are changing, calculate the change due to each effect separately and then add. The small errors accrued by neglecting the interaction between different changes can safely be ignored. This idea is almost as important as the product rule itself, and any good student of calculus ought to be able to explain why the product rule works, in addition to merely having it memorized. Jim 06:28, 2 October 2007 (UTC)

It's not as if learning these rules is difficult. Besides, isn't it irresistible to know that, to first order, multiplication resembles addition? For, let λ be a nonzero "infinitesimal" quantity such that λ² = 0; then (1+λa)(1+λb) = 1+λ(a+b).

Actually, we needn't invent infinitesimals; we can use 2×2 matrices.

${\begin{aligned}\Lambda &={\begin{bmatrix}0&1\\0&0\end{bmatrix}}&\quad \Lambda ^{2}&={\begin{bmatrix}0&0\\0&0\end{bmatrix}}\\I&={\begin{bmatrix}1&0\\0&1\end{bmatrix}}&\quad I^{2}&=I\\(I+a\Lambda )(I+b\Lambda )&={\begin{bmatrix}1&a\\0&1\end{bmatrix}}{\begin{bmatrix}1&b\\0&1\end{bmatrix}}\\&={\begin{bmatrix}1&a+b\\0&1\end{bmatrix}}\\&=I+(a+b)\Lambda \end{aligned}}$

This fact is important to the use of homogeneous coordinates for, say, computer graphics; the geometric operation called translation is essentially addition, yet we can use matrix multiplication to do it. --KSmrq^T 07:12, 2 October 2007 (UTC)

One might equally ask, why learn to do basic arithmetic when calculators can do it; or why think about your homework when you can get people on WP:RD to do it for you.. AndrewWTaylor 10:18, 2 October 2007 (UTC)

Also, one can derive more complicated theorems from these rules that a calculator couldn't handle. – b_jonas 09:01, 3 October 2007 (UTC)

Simple: Because there are problems that not even Computer Algebra Systems can solve. Of course, you could write a patch of code for the CAS so that it can solve your problem, but then we come full circle: you need to be able to understand how to solve the base case (much more difficult) in order to implement it. Ultimately, you're going to have to do some math on your own.--Mostargue 05:42, 4 October 2007 (UTC)

Axiom of Choice

Can someone please clear my doubts regarding this equivalence of axiom of choice: Prove that (i) implies (ii) where:

(i) For every nonempty set whose elements are non empty sets there exists a choice function.

(ii) If $\{a_{i}\}$ is a family of nonempty sets indexed by a nonempty set I, then there exists a family $\{x_{i}\}$ with $i\in I$ such that $x_{i}\in a_{i}$ for each $i\in I$.

Here is the proof which is in my book:

Let A be a collection of disjoint sets. We have $A\subset P(\cup _{i\in I}a_{i})$. (P is the power set). By (i) there exists a choice function f on $P(\cup _{i\in I}a_{i})$. Let b be the image of A. Pick an element a $\in$ A, f(a) $\in a\cap b$ since f(a) $\in$ a. Let y $\in$ b where y $\neq$ f(a) thus we have y = f(a') where a' $\neq$ a, and thus y $\in$ a'. Since a and a' are disjoint y $\notin$ a. Thus the only element of b $\cap$ a is f(a).

My doubts are:

• Why do we need to prove this? Won't the choice function f give us the family {f(i)} anyway.
• How is the existence of A guaranteed?
• Can't we take A={a_i} and work through the proof.
• How does b $\cap$ a being f(a) guarantee that (ii) holds.

Thanks a lot. Cheers--Shahab 06:08, 2 October 2007 (UTC)

It seems that the main difference is between a family of sets (as in (ii)) and a set of sets (as in (i)). I feel this proof is very confusingly worded. Here's an attempt at a clearer version; perhaps it will be helpful. I've tried to stay as close to the structure of your proof as I could.

Let A = {a_i} be a collection of disjoint sets, and let U be the union $\cup _{i\in I}a_{i}$ (for universe, or perhaps union). Then since each a_i can be thought of as a subset of the union U, we can say that each a_i is an element of P(U); this lets us think of A as a subset of P(U). There exists a choice function f on P(U); by definition, this maps a subset $S\subset U$ to some element $f(S)\in U$ with the property that the element f(S) must lie within the set S: $f(S)\in S\subset U$.

Now let B be the image of A under f; since A is a subset of P(U), B will be a subset of U: B = {f(a_i)|a_i in A}. That is, for each of our original sets a_i, B contains the "chosen element" f(a_i). At this point we are done, since B is the desired family, but we don't know it yet -- we need to check that each f(a_i) is in a_i where it should be. This is the B $\cap$ a = f(a) that you asked about. That equation says that B contains exactly one element from a_i, and that element is f(a_i) -- this says exactly that B is the family that (ii) claims exists. The proof is a simple idea (I will use a₁ and a₂ for simplicity, but of course your index set I may not be the integers).

By definition of B, f(a₁) is in B, and since f is a choice function f(a₁) is in a₁. We want to show that f(a₁) is the only element of B that lies in a₁. So take some other element y of B. Since the elements of B were defined to be the f(a_i), it must be that y = f(a_i) for some i -- say for simplicity that y = f(a₂). Then since f is a choice function, we must have that y is an element of the set a₂. But a₁ and a₂ are disjoint, so y cannot be in a₁; thus f(a₁) is unique, as desired.

Finally, the existence of A is not in question; you are given a collection A to start with. You're correct that you could take A = {a_i} and work through the proof that way -- in fact, this is what the original proof is doing. Tesseran 07:12, 2 October 2007 (UTC)

Thanks. But how is A given to us. The only thing that is given is the family {a_i} which consists of nonempty sets not non empty disjoint sets. But why can't we simply use {f(i)} in place of {x_i}. The meaning of the choice function guarantees that f(i) is in a_i. --Shahab 07:48, 2 October 2007 (UTC)

Pls Help me on my maths

Wiki, Pls I need wide information on the following mathmatisc topics Logic, Binomial theorems, permutation & Combination —Preceding unsigned comment added by 62.56.133.78 (talk) 12:14, 2 October 2007 (UTC)

Try logic, binomial theorem, permutation and combination. Algebraist 13:23, 2 October 2007 (UTC)

If you have more specific questions, feel free to ask them here. Algebraist 13:23, 2 October 2007 (UTC)

Stats n00b question

Would the quartiles of the standard normal deviation simply be the values for z = 0.25 and z = 0.75? Probably not. I'm confused.--Sonjaaa 19:15, 2 October 2007 (UTC)

No, because z=0 defines the median. The quartiles will lie above and below that. You're looking for the points where 25% of the area under the bell curve. You can find a table in your textbook and also here [2]. Donald Hosek 20:11, 2 October 2007 (UTC)

In which textbook?  --Lambiam 09:35, 3 October 2007 (UTC)

The quartiles of a probability distribution are those values of z for which F(z) = 0.25, 0.50 (the median itself is also a quartile) and 0.75, where F denotes the cumulative distribution function of the distribution. The cumulative distribution function of the standard normal distribution is usually denoted by Φ. Because Φ(−0.6745) = 0.2500... and Φ(0.6745) = 0.7500..., the first and third quartiles are (about) at z = −0.6745 and z = 0.6745. They are each other's opposite because the standard normal distribution is symmetric about 0.  --Lambiam 09:35, 3 October 2007 (UTC)

women

there are only 3 types of women in this world, and the types can be plugged into the equation: y=(a+b)x. if someone knows the answer, this would help me immensly. --Etro —Preceding unsigned comment added by 151.196.108.226 (talk) 20:40, 2 October 2007 (UTC)

The function is actually |a-b|x, A = the amount she wants there to be equality, B = the amount she likes being a woman. X = the amount of anger she has over not being able to be both. One way or another, you end up with a negative number, although if one trait overpowers the other, that number will be enormous, and you consequentially will be screwed (figuratively, yes, but literally, no). Xhin Give Back Our Membership! 21:41, 2 October 2007 (UTC) —Preceding unsigned comment added by Xhin (talk • contribs)

What is a way to represent the infinity symbol with a non-unicode keybord?

that is, how can I represent infinity as a symbol or a couple of symbols with my keyboard, instead of using the obtuse 1/0 ? Much help appreciated ! Xhin Give Back Our Membership! 21:35, 2 October 2007 (UTC)

"oo" is commonly used. —Tamfang 21:59, 2 October 2007 (UTC)

In personal communications and informal settings, many mathematicians use \infty, from the TeX code for $\infty$. Tesseran 23:00, 2 October 2007 (UTC)

Wow, a Unicode keyboard. Wouldn't that be something! —Bromskloss 07:32, 3 October 2007 (UTC)

A Unicode keyboard with one key per character would be … large: Chinese characters, and Korean characters, and Arabic characters, and on and on — each with one key. The Chinese do not use such a keyboard even for "simplified Chinese".

Thus we enter the wonderful world of character encodings.

• For Web pages we can use a numeric entity, a string of ASCII characters of a special form, to obtain any Unicode codepoint. The encoding for the usual infinity character is "∞", producing "∞" (with the exact appearance depending on the typeface).
• This particular character also has an HTML named entity, "∞", making it easier to remember.
• Many browsers and text editors support UTF-8, so if an instance of the character can be found somewhere, copy and paste will work. As it happens, Wikipedia stores its pages using this encoding, which is a default Web standard, so "∞" works fine.
• The MathML standard for Web mathematics markup defines a long list of named entities; don't expect these to work with plain HTML markup.

Your options depend on your needs. For text email, the double-small-o works well; it is usually instantly recognized with no explanation. --KSmrq^T 08:57, 3 October 2007 (UTC)

In HTML one can use the character entities ∞ or ∞ for ∞. Works here on Wikipedia too. —Ilmari Karonen (talk) 08:35, 3 October 2007 (UTC)

On the MacOS U.S. (not U.S. Extended) keyboard setting, option-5 is '∞'. —Tamfang (talk) 04:40, 21 October 2013 (UTC)