Wikipedia:Reference desk/Archives/Mathematics/2009 November 10

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November 10

Question About Primes Near nⁿ

I'm wondering primarily whether anyone has found a solution to nⁿ+2=Prime beyond n=3. Generally, has anyone taken up the question of the nearest prime values to nⁿ? Has the question appeared in the literature at all? I imagine it must have.Julzes (talk) 17:56, 10 November 2009 (UTC)

I've just run a little procedure to check. I got n = 737 and n = 1,349. I had to stop the program after that point because the numbers were getting so big that my computer nearly exploded! ~~ Dr Dec (Talk) ~~ 18:05, 10 November 2009 (UTC)

Can your computer actually tell if such a number is prime? RSA encryption is based on the fact that it's hard to factor a 400 digit number, even with a supercomputer. When n = 1349, you're talking about a number with over 1 million digits around 4000 digits. At best, you can do primality testing to have a good probability that such a number is prime. Isn't this right? I'm not claiming I know exactly. StatisticsMan (talk) 18:59, 10 November 2009 (UTC)

Factorization is harder than testing for primality. Taemyr (talk) 19:05, 10 November 2009 (UTC)

(e/c) First, primality testing can be done deterministically in polynomial time (though that does not necessarily mean that's what Dr Dec did, since the AKS algorithm is actually rather inefficient in practice). Factoring and testing primality are in a completely different ballpark. Second, 1349¹³⁴⁹ has only ~4000 or so decimal digits, not a million. — Emil J. 19:08, 10 November 2009 (UTC)

Okay, good points, especially about the number of digits. I did 3^1000 = 1000000, instead of 3 * 1000 = 3000. So, my bad. By deterministically, do you mean that you can tell 100% for sure that a number is prime, correct? So, it's not a probabilistic test? StatisticsMan (talk) 19:13, 10 November 2009 (UTC)

Yes, that is correct. But 3^1000 is not 1000000. -- Meni Rosenfeld (talk) 20:38, 10 November 2009 (UTC)

Yea, I agree. I realized after I logged off that I said it wrong. I'm not sure what I did honestly. But, I agree with around 4000 digits. StatisticsMan (talk) 20:55, 10 November 2009 (UTC)

I used Maple and the command isprime. For example isprime(2); returns true whereas isprime(4); returns false. I ran the very simple program:

• L := [ ]: (making an empty list)
  
• for i from 1 to 1000 do (setting the variable and number of cases)
  
• if isprime((2n+1)²ⁿ⁺¹ + 2) = true then L := [op(L),i]: fi: od: (checking all odd n, if it's prime add n to the list)
  

At the end my list L came out as [1,368,674] meaning that nⁿ + 2 is prime for n = 3, 737 and 1,349. ~~ Dr Dec (Talk) ~~ 19:26, 10 November 2009 (UTC)

P.S. I've asked Maple to factor 737⁷³⁷ + 2 and it seems to have either got lost or gone on strike. Either way, it's not very happy. As has been mentioned above: factorisation is much more difficult than primality testing. Hence the security of RSA encryption. ~~ Dr Dec (Talk) ~~ 19:35, 10 November 2009 (UTC)

According to MapleSoft's web page, isprime returns true if p is very probably prime. It does not say which algorithm it uses but gives references at the bottom. It says there is no known example of it being wrong and such a number would have to be hundreds of digits long. StatisticsMan (talk) 19:45, 10 November 2009 (UTC)

Mathematica confirms the results of User:Dr Dec. From the implementation notes:

• PrimeQ first tests for divisibility using small primes, then uses the Miller-Rabin strong pseudoprime test base 2 and base 3, and then uses a Lucas test.
• As of 1997, this procedure is known to be correct only for $n<10^{16}$, and it is conceivable that for larger n it could claim a composite number to be prime.
• The Primality Proving Package contains a much slower algorithm which has been proved correct for all n. It can return an explicit certificate of primality.

I can try to verify this using the last item if it is important to you. -- Meni Rosenfeld (talk) 20:38, 10 November 2009 (UTC)

This is interesting, but not terribly important. Thanks. I'm more interested in the general question of how to do a literature search. Is there a symbolic-string search engine that will pop out what's desired, or would one have to go through Mathematical Reviews' Number Theory subsection for all recent years?Julzes (talk) 23:17, 10 November 2009 (UTC)

oeis:A100407: "Numbers n such that n^n+2 is prime. 1, 3, 737, 1349." There are no others up to 3500 which corresponds to 12405 digits. It took me 12 minutes at 2.4 GHz to double check this limit with PrimeForm/GW 3.2.2. A search of n^n+2 [1] at http://www.primenumbers.net/prptop/searchform.php finds nothing. If any other terms had been found then they would probably have been submitted to that database of gigantic probable primes and turned up in the search. 1349^1349+2 is a 4223-digit probable prime. There is no known primality test which can quickly determine whether it is really prime, because the currently known quick tests require a large part of the factorization of p-1 or p+1 to be known. See more at http://primes.utm.edu/prove/index.html. The fastest algorithm to prove n^n+2 for large n is ECPP (Elliptic curve primality proving). The fastest publicly available ECPP implementation is Marcel Martin's Primo at http://www.ellipsa.eu/public/primo/primo.html. It might take around 1 GHz month to prove primality of 1349^1349+2 with Primo. Don't attempt any other public program like Mathematica's ProvablePrimeQ at this size. If you try a program and it claims prime in less than a GHz week then the program did not make a primality proof although it may have claimed to do so (some programs hide that they only make probable prime tests). oeis:A074966: NextPrime(n^n) - n^n. I easily found these OEIS sequences by computing the first few terms and search them in OEIS. PrimeHunter (talk) 00:42, 11 November 2009 (UTC)

I doubt Wolfram would outright lie about this - especially since it produces an Atkin-Goldwasser-Kilian-Morain certificate which I suppose can be easily checked. Anyway, I did run ProvablePrimeQ for 737 for several minutes and it did not terminate. -- Meni Rosenfeld (talk) 06:24, 11 November 2009 (UTC)

ProvablePrimeQ makes a primality proof but nobody uses it at sizes like 1349^1349+2 because it is too slow if it ever completes. Maybe it could handle 737^737+2 within a GHz week but I don't know. It is other programs which may give misleading messages about showing a number being prime. It was perhaps confusing that I mentioned this right after talking about ProvablePrimeQ. There are also many Mathematica users who have falsely thought that Mathematica's PrimeQ makes primality proofs. I don't have Mathematica but apparently some parts of the documentation can give the impression that PrimeQ shows primality. PrimeHunter (talk) 14:03, 11 November 2009 (UTC)

Indeed, the documentation for PrimeQ does not mention that it is probabilistic, but neither does it explicitly say otherwise. Also, that information is not exactly hidden for those who are aware of the issue and willing to look. -- Meni Rosenfeld (talk) 14:34, 11 November 2009 (UTC)

By the way - do you happen to have any data about how academic this discussion is? That is, was there ever a report that PrimeQ, or an equivalently accurate function, claimed a composite number was prime? -- Meni Rosenfeld (talk) 15:23, 11 November 2009 (UTC)

Thanks, PrimeHunter. I guess this is a question I might have tried a little harder to answer on my own, with my prior knowledge of OEIS. oeis:A074967: n^n-PrevPrime(n^n) also. There's still the question of literature, but a starting point for that is at OEIS anyway, for this specific problem at least.Julzes (talk) 01:18, 11 November 2009 (UTC)

One thing I noted is that oeis:A074966 has a strange quote in its comments. Asking whether there are any solutions of nⁿ+1=Prime beyond the three solutions n=1, 2 and 4; it is stated that any such prime is greater than 10³⁰⁰⁰⁰, according to Sierpinski. Now, this doesn't make a whole lot of sense. Any n that gives a prime for nⁿ+1 must be of the form 2^2^k, with the result being a (2^k+k)th Fermat number. Even if it wasn't known at the time of the quote that the 20th Fermat number is composite, this number is 315653 digits. At this date, not only the 20th but also the 37th Fermat number is known to be composite. The 70th appears to still be open (?), so the smallest possible prime would be this.Julzes (talk) 15:04, 11 November 2009 (UTC)

Infinite-time probability

This problem follows from a debate I was having with a friend in one of my classes. "If two people walk randomly around a room, given infinite time, will they always collide?" To simplify, if the chance of hitting in one minute is between 0 and 1 (say 0.01), what is the chance they will collide given an infinite number of minutes? As it happens, I said "1" and my friend said "almost 1".

Now, it would appear from our articles that the answer is that it will "almost surely" occur.

“ The difference between an event being almost sure and sure is the same as the subtle difference between something happening with probability 1 and happening always.

If an event is sure, then it will always happen, and no other event can possibly occur. If an event is almost sure, then other events are theoretically possible in a given sample space; however, as the cardinality of the sample space increases, the probability of any other event asymptotically converges toward zero. Thus, one can never definitively say for any sample space that other events will never occur, but can in the general case assume this to be true.

”

I'm afraid these paragraphs, as they apply to my problem, are a loss to me. Is it a certainty or not? (I realise that I probably technically won the debate here; the answer is 1. More interesting is the interpretation of that figure, it would seem.) - Jarry1250 ^{[Humorous? Discuss.]} 17:57, 10 November 2009 (UTC)

You can't actually do the experiment for an infinite amount of time, so we can't say what would actually happen. What we can say is that, as the length of time you do it for increases the chance of a collision tends towards one. In the limit as you do it for infinite amount of time we say the probability is one, but it is still possible that it wouldn't happen (both people could, purely by chance, walk round in small, non-intersecting circles for the whole time, for example). All that being said, you could both be wrong. It depends on what you mean by "walk randomly". Random walks are a much studied mathematical technique and different things happen in different situations. I think the best way to model your question is to consider the room as a grid and each person steps either north, east, south or west into the next square each second. You then need to decide on the distribution - are they more likely to keep walking in the same direction than they are to turn around and go backwards, for example? You can't calculate it based on the collision chance in a minute since each minute will be different depending on where they start that minute, and the minutes will be dependant on each other. If they move around really quickly (or the room is really small) then it is possible each minute would be approximately the same, but if they are going at a normal walking speed around a normal sized room, a minute is too short a time period. You may also be interested in Brownian motion, which is like the random walk I described but with an infinitely small grid (roughly speaking). --Tango (talk) 18:09, 10 November 2009 (UTC)

Thanks Tango - I understand the problem with modelling the exact scenario, to the point it doesn't really matter so much to me as the meta-issue. So the probability is 1, but they might not collide? Weird. - Jarry1250 ^{[Humorous? Discuss.]} 18:16, 10 November 2009 (UTC)

"Might" is such a strong word. The only thing separating "sure" from "almost sure" is that your sample space includes instantiations in which they don't collide. Tango's example of walking forever in circles matches the problem specification of walking inside the room. However, the probability that any of these instantiations will manifest is 0, and for any sensible notion of probability, this means that they will collide. -- Meni Rosenfeld (talk) 18:25, 10 November 2009 (UTC)

The notion of probability here is completely well-specified. What you seem to want is a different notion of possibility. If you declare probability-zero events to be impossible, you have to explain just how they get to be impossible, given that all initial segments of them are possible. I think that's contrary to the usual intuitive understanding of the modal operators. --Trovatore (talk) 00:15, 11 November 2009 (UTC)

For a reasonable mathematical model of the problem, the answer is : "The probability that they will collide at some point is exactly 1. They will almost surely collide at some point". The fact that these two statements do not contradict each other is what the paragraph you quote attempts to clarify. IMO, the distinction between "surely" and "almost surely" is too subtle to survive the translation to the physical world. -- Meni Rosenfeld (talk) 18:19, 10 November 2009 (UTC)

The distinction involves infinity - nothing involving infinity translates nicely to the physical world. --Tango (talk) 19:06, 10 November 2009 (UTC)

Think of "almost surely" as meaning something like this: pick a random real number in the interval [0,10] with uniform probability. If the random number is exactly π, they don't collide; otherwise, they do. Observations: 1) the probability of picking any specific number from an infinite collection is zero; 2) despite this, you can't exactly say the random number won't be π. The problem is "pick a random real number" is inherently imprecise. It's the same way with "run the experiment for an infinite amount of time". 69.228.171.150 (talk) 23:35, 10 November 2009 (UTC)

Why is nobody answering the question?

• Random walks in 2 dimensions are recurrent.
• Random walks in more than 2 dimensions are transient.

That means the probability of eventually coming within a distance ε of each other, no matter how small ε is, is 1, if they're in a plane. In 3-space, the probability is less than 1. (If it's a discrete random walk on integer points in 3 dimensions, with the usual assumptions (independence of steps, uniform distribution), then I think the probability they eventually meet again is something like 0.3.) I seem to recall that George Polya figured this out in about 1910 or something like that. Michael Hardy (talk) 04:18, 11 November 2009 (UTC)

If ε is finite, then ε^n represents a finite proportion of the total space in the room, no matter how many dimensions are involved, and thus the probability is 1. And if you make ε tend to 0 at the same time as the time tends to infinity, then the answer will depend on how the two are related.Ehrenkater (talk) 18:33, 11 November 2009 (UTC)

Since OP had already assumed recurrence I don't see how this is what the question was. Rather I read the OP question as dealing with the difference of "with probability 1" vs. "with surety". As such the answers given above is the proper answers to the question. Taemyr (talk) 05:25, 11 November 2009 (UTC)

Besides, the OP said they walked in a room, which I take to mean a bounded set. I think this guarantees recurrence no matter the dimension. -- Meni Rosenfeld (talk) 06:09, 11 November 2009 (UTC)

Thanks Michael for the answer, but yeah, I was kinda after the difference between "with probability 1" vs. "with surety" as it applied to this real life scenario. Sorry for not being more explicit. But anyhow, all useful information I will keep with me. - Jarry1250 ^{[Humorous? Discuss.]} 17:52, 11 November 2009 (UTC)

Set of integers

Resolved

Is there a simple name (e.g. weird number, prime number, even prime number, colossally abundant number) for this set of integers: 1, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53... ? (Note that this is essentially the set of odd prime numbers, except that 1 is not a prime number. These numbers all have the property that they cannot be divided by any number except 1 without becoming a non-integer number.) 68.116.36.179 (talk) 18:32, 10 November 2009 (UTC)

Some people used to include 1 in the primes but it just causes trouble. I'm not sure why you have excluded 2. Dmcq (talk) 18:37, 10 November 2009 (UTC)

Also a nitpick; 7 divided by 7 equals 1. Taemyr (talk) 18:54, 10 November 2009 (UTC)

Well, 2 is the only even prime number so being an odd prime is anything but special. If P denotes the set of prime numbers then your set is { { 1 } ∪ P } − { 2 }. So you seem to have substituted 2 for 1 in the set of primes. Why ought this new set have a special name? ~~ Dr Dec (Talk) ~~ 19:04, 10 November 2009 (UTC)

The phrase "odd prime" does come up quite often in number theory, since it is often desirable to exclude 2 (particularly when dealing with quadratics). Why anyone would want to include 1, though, I don't know. You could call that set "odd positive non-composite integers" if you wanted to, one being neither prime nor composite (and being unique among positive integers in having that property). --Tango (talk) 19:11, 10 November 2009 (UTC)

Thanks everyone, I think that that is indeed the closest name to what I wanted. I also saw it here: [2] (useful site, BTW). This site say they are the "odd noncomposite numbers". Re: the nitpick: OK, so I should have said "cannot be divided by any number except 1 or themselves without becoming a non-integer number." The reason I'm interested in having 1 in with the odd primes is because the number 1 shares some characteristics, but of course not all, with the other odd primes. I'm sure there are some applications for this - the reason I'm interested in this set is as a way to verify that a chemical equation cannot be reduced further. 68.116.36.179 (talk) 22:40, 10 November 2009 (UTC)

Then why are you excluding 2? I think you need to look at greatest common divisor - if the gcd is greater than 1, then you can cancel the common factor. --Tango (talk) 22:50, 10 November 2009 (UTC)

Oh yes, you're right. I failed to think about this very hard. How do you put the resolved checkmark on this thing? 66.178.144.142 (talk) 02:42, 11 November 2009 (UTC)

I have checked "resolved" for you. --PST 03:20, 11 November 2009 (UTC)

Thanks PST. 66.178.144.142 (talk) 18:06, 11 November 2009 (UTC)

Short division

I am trying to help my niece with her homework but I cannot remember doing short division at school and can't make any sense of any of the guides I have found. Can someone work through an example with me? Take 4 into 67 - I can get as far as getting to 16 and then I am stumped how I get to the .75? --Cameron Scott (talk) 19:38, 10 November 2009 (UTC)

If I remember well you do: 4 into 67 is 16 ,then you go on with the remainder 3=67-64. You put a 0 after 3 and a dot after 16, and go on this way: 4 into 30 gets 7 &c. Check short division also. --pma (talk) 19:45, 10 November 2009 (UTC)

What does "&C" mean? I couldn't make any sense of the short division article - it might as well have been in dutch. --Cameron Scott (talk) 19:52, 10 November 2009 (UTC)

&C is short for the borrowed latin phrase et cetera. 69.228.171.150 (talk) 20:22, 10 November 2009 (UTC)

Let's say that you want to divide 956 by 4. Well you write out

${\begin{matrix}4{\overline {)9\ 5\ 6}}\\\end{matrix}}$

Now how many times does 4 go into 9? Well, 4 goes into 9 twice with 1 left over, so we write a 2 above the 9 and put a little 1 in front of the 5 (like carrying in addition). So we have

${\begin{matrix}\!\!\!\!\!\!2\\4{\overline {)9\ {}^{1}5\ 6}}\\\end{matrix}}$

Next, how many times does 4 go into 15? Well, 4 goes into 15 thrice with 3 left over, so we write a 3 above the ¹5 and put a little 3 in front of the 6. So we have

${\begin{matrix}\!\!2\ \ 3\\4{\overline {)9\ {}^{1}5\ {}^{3}6}}\\\end{matrix}}$

Finally, how many times does 4 go into 36? Well, it goes 9 times with none left over. So we write a 9 above the ³6 and stop because the remainder was zero. This gives:

${\begin{matrix}\ \ 2\ \ 3\ \ 9\\4{\overline {)9\ {}^{1}5\ {}^{3}6}}\\\end{matrix}}$

If the remainder weren't zero then we could have carried on by considering

${\begin{matrix}4{\overline {)9\ 5\ 6\ .\ 0\ 0\ 0\ 0\ldots }}\\\end{matrix}}$

The net result is that 956/4 = 239. ~~ Dr Dec (Talk) ~~ 20:21, 10 November 2009 (UTC)

Thanks, that's a bit clearer. While I remember doing long division at school, I have no recollection of short division. Between three (!!!!) adults, we finally worked it out. --Cameron Scott (talk) 20:26, 10 November 2009 (UTC)

I dont understand why you are trying to do division by hand. The only reason I can think of for doing division by hand is to check that the calculator is giving me the right answer. And to check that the calculator is giving me the right answer, I have to learn doing multiplication by hand (for example: A divide B = C. Check by multiplying B by C). So I cannot think of any reason to do division by hand. 203.41.81.1 (talk) 21:08, 10 November 2009 (UTC)

Unfortunately at school they have an annoying habit of telling you to find the answer a certain way, even if it is ultimately pointless. - Jarry1250 ^{[Humorous? Discuss.]} 21:19, 10 November 2009 (UTC)

Indeed, she has to show her workings to indicate how she's done it. --Cameron Scott (talk) 21:33, 10 November 2009 (UTC)

Why's it "annoying" or "pointless"? There are things you can prove by manual division that a calculator can't. For example if you ask your calculator to calculate m/n and it returns the answer 0.123456789 then all you know is that m/n is 0.123456789 to nine decimal places (it might be 0.1234567891…, or 0.1234567886…, or infinitely many other numbers). If you do the division by hand you might notice that the remainders start to repeat after the 9 and so the actual number is 0.123456789. After practising the method for a while you get a feel for it and you can start to formulate some nice theorems. For example, what is the maximum possible period length of a repeating decimal (in the case of 13717421/111111111 = 0.123456789, the period length is nine). Well, if we divide m by n then there are n possible remainders and so m/n as a repeating decimal can have a period of at most n digits. Prove that with a calculator! ~~ Dr Dec (Talk) ~~ 22:04, 10 November 2009 (UTC)

A good apology for division by hand. But really? Blink? Staecker (talk) 00:50, 11 November 2009 (UTC)

Short division is not only a terrible excuse to name an obscure concept, but also meaningless and un-mathematical. As I have said, and shall continue to say, schools fail to give a proper mathematics education to their students these days (mathematicians do not find new ways for dividing one number into another - ultimately, this is non-mathematical). Proper mathematics occurs when students are asked to think deeply and creatively rather than memorize set methods to solve problems (as you will know, there is not shortage of open problems in mathematics, and new methods must be discovered to solve them - one cannot "copy" old methods). Additionally, many of these open problems are well beyond the comprehension of adults (who do not have university level mathematics education) - even formal mathematicians spend years on some of these problems (this is not to say that one needs mathematics education to do research - but it certainly helps). There is rarely a "formula" that can simply be applied (mathematics is not about formulaic approaches). The point is that although I would certainly discourage the teaching of "short division", I would encourage your niece to consider problems which require abstract thinking in relatively simple contexts. Although having an enormous amount of depth and breadth of knowledge is required before one can begin research in mathematics (nowadays), mathematical thinking can start at any point, and I would encourage your niece to consider this should she have the spare time. For instance, suppose I have a pair of numbers a and b. The GCD (Greatest Common Divisor) of a and b is the largest number g that divides both a and b. Examples include:

GCD(2, 5) = 1

GCD(3, 6) = 3

GCD(18, 45) = 9

Use the naive technique of long division to establish an algorithm by which one can determine the GCD of two numbers (the answer lies in Euclidean algorithm, but try to solve it without looking at the article). As a simpler problem, what is the relation between the least common multiple and the greatest common divisor? Lastly, invent your own questions during your investigation of the GCD - a mathematician is also someone who is able to pose good questions, and solve them. Nevetheless, these problems do not illustrate that mathematics is about basic arithmetic only - mathematics has hardly anything to do with basic arithmetic (it is the schools which give the wrong impression). The article mathematics should give a reasonable overview as to what mathematicians research. Hope this helps (and remember to ask your niece if she is interested in solving the problems I posed!). --PST 03:17, 11 November 2009 (UTC)

I'm fairly certain that the point of teaching methods of doing arithmetic is not to train students to be mathematicians. It's to teach them the basic skills that will presumably help them in everyday life. The vast majority will never need to think about actual problems in mathematics, but they may very well need to do arithmetic without having access to a calculator. I don't know how much justification there is for long division these days since if you can break out some paper you probably also have access to a computer, but that doesn't change the basic idea, which is that these are general life skills. It's silly to say that there's no place for that, even if it's not preparing anyone for "real math". Rckrone (talk) 06:36, 11 November 2009 (UTC)

That is the problem. The curriculum should be focused on training future mathematicians (as is done for other subjects. I am sure that many aspects of science or history have no purpose in "daily life". Why are these subjects taught at (primary) school?). Mathematics is the center of intellectual ability - it encapsulates all forms of thinking from abstract to practical. --PST 07:31, 11 November 2009 (UTC)

I agree that it's also important to expose students to real math, since clearly some of them need to end up becoming mathematicians or physicists or whatever. And the same can be said for most of the more specialized subjects taught in school. But there are some general skills that are important to everyone which include arithmetic, reading/writing, and maybe some basic history and civics. Encouraging future mathematicians doesn't need to come at the expense of establishing those skills for everyone else, and shouldn't. Rckrone (talk) 07:58, 11 November 2009 (UTC)

I completely agree with you. However, basic arithmetic appears to be taught for approximately 7 years at school. Certainly, it could be taught within the span of 5 years at the very most (but I feel that 3 years is enough). If children are too young in the earlier years of schooling to learn quickly, then introducing them to games like chess would be appropriate (whence they could have fun and improve intellectually). I know that what I am proposing is being implemented at some special programs these days, but the point is that the curriculum is not only too slow for many intelligent students, but also very ineffective (arithmetic need not be taught for over half of a student's schooling). Furthermore, weight should not be put on establishing "life skills" - there should be some balance between actual mathematical thinking and arithmetic. Training students to follow set methods to solve problems should be done first, but not at the cost of "dumbing them down" by discouraging creativity. --PST 09:37, 11 November 2009 (UTC)

mathematicians do not find new ways for dividing one number into another - ultimately, this is non-mathematical Certainly most mathematicians don't, but I don't think it could be called non-mathematical. The question of the best way to multiply two numbers (which is closely related to dividing) interested the likes of Andrey Kolmogorov, who conjectured that the "schoolchild" method cannot be improved upon, only to be proven wrong by Anatolii Karatsuba. --196.210.152.43 (talk) 20:20, 15 November 2009 (UTC)

What's funny is a lot of people still don't know arithmetic after learning it for 7 years. StatisticsMan (talk) 18:44, 12 November 2009 (UTC)

When I was 7, I was very fond of multiplications indeed, and I must say that my ability was very highly estimated by the class and by the teacher. Unfortunately, one day she started the divisions, and my favourite topic became suddenly kind of completely out. I felt that somehow unfair, and as a consequence, never managed to like divisions. --pma (talk) 20:58, 14 November 2009 (UTC)