Talk:High School Mathematics Extensions/Primes

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"In mathematics, only the most important (yet basic) theorems are coined fundamental. If you see the word fundamental as part of the naming of a theorem, it means that theorem is of extreme importance."

I am afraid that we have one exception: the Fundamental Theorem of Algebra.


Good point, the exception being that it's extremely hard, and hence not basic. But... by basic i dont mean easy, but fundamental. I'll fire up the thesaurus then

Do the multiplicative inverses have to be written as 1/k? Dysprosia 09:17, 6 Nov 2003 (UTC)

No, but I guess what i wanted to do was to illustrate the ideas using something they're familiar with. I guess a better notation will be x-1. I will adopt that notation in later sections. Xiaodai

The problem I have with it implies that you can divide in Zn which in fact you can't. That √(-1) notation threw me off too... Dysprosia 00:16, 29 Nov 2003 (UTC)

Okay, I fixed the grammar. I'm probably the only person who can listen to music, fix grammar, learn, and fix source code at the same time :-). Ugen64 01:03, 18 Dec 2003 (UTC)

Hey, thanks a thousand. Those fixes are exactly what this online-book needs!!! You are da man!Xiaodai AUSEDT 12.34 18 Dec 03

I might look stupid, but I don't understand what I'm supposed to notice about the values of the powers of two in modulo 11... can someone help me? Ugen64 20:45, 19 Dec 2003 (UTC)

In mod 11 arithmetic, 2 is called a generator. Let's list all the powers of 2 from 1 to 10:
2, 4, 8, 5, 10, 9, 7, 3, 6, 1.
The number 2 is called a generator, because its powers generate all the numbers from 1 to 10! That is, for every number x mod 11, we can find a number p, such that 2p = x. Utilizing this idea, calculating xn for big x and n is no longer a problem! Consider 67, notice 29 is 6, therefore 67 = 263 = 23 = 8.
Notice 6 is also a generator, can you find the other generators(Using the powers of 2 or 6)? Xiaodai 8:05, 20 Dec 03 Sydney
Which means that Zn is a cyclic group for some values of n :) Dysprosia 23:21, 19 Dec 2003 (UTC)
I also just happened to notice that the operations between numbers are mirrored on either side of the 10.
1+1=2+2=4+4=8-3=5+5=10
10-1=9-2=7-4=3+3=6-5=1

Contents

[edit] info -- Recursion

xn = x * xn-1 if n>0
why is n>0?
x0 = x * x0-1 = x/x = 1 this is not a special case it follows the same rules
x-1 = x * x-1-1 = x/x2 = 1/x ...
I'm not a math guy but n>0 seems not to be needed.
Pk2000 12:01, 3 Mar 2005 (UTC)
Very good point. Thanks for pointing that out. Will seriously consider making a point of that in the article. Thanks again Xiaodai 16:04, 3 Mar 2005 (UTC)
You need to point out then that x ≠ 0 if you do this. 0^0 is not defined. Dysprosia 22:18, 3 Mar 2005 (UTC)

[edit] One?

My mum tells me that when she did mathematics, 1 was considered a prime number. Was it? Is counting 1 as a non-prime more useful? Is her memory just at fault? *g* r3m0t 11:27, 23 Dec 2003 (UTC)

She may be might right, but that was long time ago. Now, one is not consider a prime. Xiaodai 15:00, 23 Dec 2003 (UTC)
...so why is one not considred a prime? I see the question being asked but I don't see it answered anywhere in the text. 209.67.181.254 (talk) 00:44, 3 May 2008 (UTC)
It is not considered a prime because it is more useful to do so. The fundamental theorem of arithmetic would have to be changed to say "every number can be expressed as a unique product of non-unitary primes" (where by non-unitary primes, I mean what we call a prime today), and the phrase "prime number" occurs much more often in mathematics than "prime number or 1". --Taejo (talk) 14:36, 9 June 2008 (UTC)

[edit] One chapter done

I consider this chapter done (a first draft at least)! I guess it's appropriate at this time to... clap. Thanks to all those who contributed to this 22 page beast! It was hard work. Xiaodai

*claps vigorously* By the way, you know you can just sign by typing ~~~~? r3m0t 23:43, 4 Jan 2004 (UTC)
Thanks for that i didn't know. Xiaodai 07:57, 5 Jan 2004 (UTC)

For problem 11(b), what exactly is your notation? Where you have \sqrt{-1}\equiv \frac{p-1}{2}!, you don't say what it's equivalent to. (There's no (mod n) clause.) Furthermore, should the "p=1 (mod 4)" equation be written as an equivalence instead?

I'm not entirely sure about these, so I didn't make the edit. Can someone confirm or deny? Grendelkhan 17:16, 9 Feb 2004 (UTC)

Thanks for the question. The editing is done. Where obvious I sometimes do use the two notations interchangably. But I'll change it in this case. Xiaodai 23:20, 9 Feb 2004 (UTC)

I do believe factorization is spelled wrong throughout the document.... Someone may like to fix that.

It's not "wrong". Some people spell it with an s and some with a z - they are variants, not misspellings. Dysprosia 04:59, 21 Apr 2004 (UTC)

It might be a temporary thing, but there appears to be a bug in the formula rendering - the example for marking 5 as a prime in the "Finding primes" section does not appear to render. I found that removing the terminating // on the last line of the matrix definition causes the matrix to render correctly again - however (bizarrely) it causes a visible rendering error (in preview mode at least) when applied to any of the other matrixes. Waveform 01:04, 30 May 2004 (UTC)

For the life of me, I can't figure out the fifth modulo 11 problem. I don't see how adding or subtracting 11 is ever going to get me a multiple of i. Infact, after the 5th question, the page starts to become a giant nonsequiter for me (I'm sure the page is correct, it's just not clear to me). The question is unclear -- I'm not sure if it means find x if (x mod 11)^2 = -2 or if x^2 mod 11 = -2.

I don't like the notation in the question. My interpretation of it is that it is asking you for all the numbers modulo 11, that when you square it, you get -2 = 9 (mod 11). So, x^2 = 9 normally you have 3 as a solution. But, 11 mod 11 is zero, so conceivably, 11+3 will also be a solution too, since we're basically "adding zero". So 3+11 = 14 = 3 (mod 11), and you can consider -3 = 8 to be a solution too (since (-3)^2 = 9 normally).

Then in the division example it says x=3/5. Ok, sure. Then x = 3 * (1/5). Alright, still with you. Then x = 3 * 3. Uh, where the hell did that come from? 3*3 = 9. 3/5 = 0.6. Then, to really confuse the hell out of me, 3 * 3 becomes 2!

Again, the notation (I've complained about it before, above). You can't really divide when talking about modular arithmetic because you can't always do it, and you don't get fractions. The multiplicative inverse of 5 mod 7 is the number you multiply 5 by modulo 7 to get 1. 3 * 5 = 1 (mod 7), so you can write "1/5" (which should really be written as 5-1) as 3. The rest should then be clear: get back to me if it isn't.

But I'm confused even before that. 1/5 = 0.2. 1/5 mod 7 is still 0.2. Why are we doing this funky stuff with x in the first place?

See above. 1/5 is supposed to be the multiplicative inverse, not a fraction. We only work with numbers {0, 1, 2, 3, 4} modulo 5, we don't have fractions or decimals or whatever. If we want to "divide" numbers, we look for numbers we multiply to get 1 (like if we had real numbers, 1/5 is supposed to be the number we multiply 5 by to get 1, modulo 7, 3 is supposed to be the number we multiply 5 by to get 1).

Can someone please clarify here or in the module itself what is going on? I can get a 4 on my AP calc examination but this makes no sense to me ;)

dataangel

As I mentioned, get back to me if anything I told you still doesn't make sense. Dysprosia 10:25, 13 Aug 2004 (UTC)
It seems that my notation (or my uni lecturer's notation) have confused quite a lot of ppl. Just think of "1/5" as the number when multiplied by 5 will give 1, dont think of it as 0.2. I know that "1/5" will be confusing to those with prior exposure to higher mathematics but this book is for high school student and so i want to do it in notations familiar to them. Maybe i should clarify it in the book. Xiaodai 14:18, 14 Aug 2004 (UTC)
I know you want to use notation familiar to high school students, but you really need to use a different notation. The two operations aren't the same, and you're going to confuse people more if you do, even that high schooler above. But it's up to you. Dysprosia 23:52, 14 Aug 2004 (UTC)
I think you can effectively use the "1/5" notation. What's wrong with thinking it is 0.2? What does 0.2 mean anyway? It only ever means 2/10 which makes just as much sense in modulo 11 arithmetic as it does in the rationals. You can "divide" in prime modulos because they are fields. The important thing to recognize (and communicate) is that the canonical form for all numbers in "modulo 11" arithmetic comes from the list {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. You get a lot of leeway for students to play around with the algebra and still get sensible answers, they just have to do the algebra correctly.
By the way, I have always liked the analogy of "clock" arithmetic as a way of explaining all the funny bits of Z12. So image a clock with only 11 hours on the face (strange day!). How many hours would you have to wait from "midnight" before the clock showed 2 o'clock? There are lots of equivalent answers. Now DIVIDE that many hours into 10 equal pieces. You get lots of different numerical answers for "how long", but they are all equivalent to 9 (mod 11). Which only means that if you work a 9 hour shift 5 times in a row starting at midnight, the clock will read 1. But if you work a 2/10 hour (do I dare call it 12 minute?) shift 5 times in a row starting at midnight the clock will still read 1. beej 13:51, 20 May 2005 (UTC)

5^99 mod 11? I'm probably not thinking through this properly/missing something (it's been a while for me since I did any of this stuff), but isn't that a little difficult to do, even with something like Euler's theorem for help? Dysprosia 12:40, 23 Aug 2004 (UTC)

I guess the question is a bit tricky. Just note that the inverse of 3 is 4. Xiaodai 07:55, 24 Aug 2004 (UTC)
I think what students are supposed to see is that 5 is a generator, just like all the other elements in modulo 11 arithmetic. If you can find which exponent i make 5i = 1, then you can do normal arithmetic on the exponent. In this case 510 = 1, so you figure 599 = 510 * 9 + 9= (510)9 * 59 = (1)9 * 59.

beej 13:51, 20 May 2005 (UTC)

Is the new project a good challenge or what? Need opinions. Xiaodai 10:59, 20 Nov 2004 (UTC)

hello everyone just reading through. the discription of what the modular arithmatic is was a little confusing so I fixed it up a bit but the rest of the article is a bit hard to fallow through the modular arithmatic section. I'm not exactly sure what the whole division and inversion section is saying. the whole section on modular arthmetic brings up some un answered questions. is there a real number version of modular arthimetic or is there only integers in modular arthimetic. from there I kind of get lost. good article though I like what I've read so far. --Stranger104 08:41, 16 Jun 2005 (UTC)

[edit] Revision 1

This chapter is the first chapter to be written. The material, despite others and my best effoects, remain somewhat chaotic and lacking focus. I have moved the much disputed (and confusing) part about division and inverses to Further Modular Arithmetic where it will received a rigorous treatment. I have also tried to replace the somehwhat colloquiall language/tone with a formal one in most parts. Removal of the personality section on Arnold Ross has been agonised over many times, but the final decision is to remove it. Xiaodai 07:21, 6 August 2005 (UTC)

  • Update*

Reorganised a bit, rewrote the inverses section after feedback receive, hope it's not as confusing as before, and that we won't get into arguments about technicalities this time. Xiaodai 08:05, 12 August 2005 (UTC)

what a pretentious and patronising piece of junk this is. Zhuo Jia Dai 12:08, 17 November 2006 (UTC)

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