|
[QUOTE=philmoore;217242]Nice catch! At least someone is paying attention (not me...)[/QUOTE]
Ah. [b]Total[/b] brain damage on my part. I had some examples in a file and I cut/pasted together some incorrect parts of the file. Very careless. Try: Units of Z/10Z are 1,3,7,9. 3*7 = 1 mod 10, so what is left is 1*9 = 9 mod 10. |
[quote=R.D. Silverman;217245]
Now consider the elements whose inverse is not different from themselves. Thus, a = a^-1 mod m. What can you say about the product of any two of these elements?[/quote] These numbers a which sutisfy the following are real unities right? The product of 1 and -1 is -1. |
[QUOTE=R.D. Silverman;217247]Ah. [b]Total[/b] brain damage on my part. I had some examples in a file
and I cut/pasted together some incorrect parts of the file. Very careless. Try: Units of Z/10Z are 1,3,7,9. 3*7 = 1 mod 10, so what is left is 1*9 = 9 mod 10.[/QUOTE] Sure, you gave that latter example in post #464 too. My point was that this is [b]not[/b] total brain damage. It's just a mistake. All of us make them sometimes, whether students or advanced practitioners. We shouldn't be berated so mercilessly when it happens. |
[QUOTE=blob100;217250]These numbers a which sutisfy the following are real unities right?
The product of 1 and -1 is -1.[/QUOTE] Don't invent terminology. What is a "real unity"? What do you mean by "satisfy the following" when you write 1* -1 = -1?? |
[QUOTE=jyb;217253]Sure, you gave that latter example in post #464 too. My point was that this is [b]not[/b] total brain damage. It's just a mistake. All of us make them sometimes, whether students or advanced practitioners. We shouldn't be berated so mercilessly when it happens.[/QUOTE]
I disagree. It was total brain damage on my part because I failed to check the data that I pasted into my message. This is such an oversight that "brain damage" is appropriate. |
[QUOTE=R.D. Silverman;217263]I disagree. It was total brain damage on my part because I failed
to check the data that I pasted into my message. This is such an oversight that "brain damage" is appropriate.[/QUOTE] Unless you are actually brain damaged, then I disagree that brain damage is appropriate (not to mention inaccurate). |
[quote=R.D. Silverman;217262]Don't invent terminology. What is a "real unity"?
What do you mean by "satisfy the following" when you write 1* -1 = -1??[/quote] When I said "sutisfy the following" I meant: (1) Sutisfiy: Now consider the elements whose inverse is not different from themselves. Thus, a = a^-1 mod m. *You asked what is the product of these numbers. (2)As I know, a unity is i=(-1)^(0.5) or 1 or -1. When I said "real unity" I meant 1 or -1. So, back to your question: Numbers that agree the following (1) are the numbers: 1,-1. Thier product is -1. |
[QUOTE=blob100;217269]When I said "sutisfy the following" I meant:
(1) Sutisfiy: Now consider the elements whose inverse is not different from themselves. Thus, a = a^-1 mod m. *You asked what is the product of these numbers. (2)As I know, a unity is i=(-1)^(0.5) or 1 or -1. [/QUOTE] Not 'unity', as in 'root of unity'. The word is 'unit'. I already defined this for you. And Shanks' book discusses them. A unit is an element of a ring that has a multiplicative inverse. [QUOTE] When I said "real unity" I meant 1 or -1. So, back to your question: Numbers that agree the following (1) are the numbers: 1,-1. Thier product is -1.[/QUOTE] No. We are NOT discussing 'unity' (or whatever you think 'unity' might be). I have said this multiple times: The units of the integers taken mod m, where m is an integer greater than 1 are those integers that have a multiplicative inverse mod m. The question is then: What is the product of all of the units taken mod m??? Even you gave examples showing that you know what units are. So why are you confused now? The product of two units does not have to be 1 or -1. Consider the integers mod 10. The units are 1,3,7,9. Note that 3*9 is 7 mod 10. This clearly isn't 1 or -1. On the other hand, 3 and 7 are inverses. |
[quote=R.D. Silverman;217272]Not 'unity', as in 'root of unity'. The word is 'unit'. I already defined this
for you. And Shanks' book discusses them. A unit is an element of a ring that has a multiplicative inverse. No. We are NOT discussing 'unity' (or whatever you think 'unity' might be). I have said this multiple times: The units of the integers taken mod m, where m is an integer greater than 1 are those integers that have a multiplicative inverse mod m. The question is then: What is the product of all of the units taken mod m??? Even you gave examples showing that you know what units are. So why are you confused now? The product of two units does not have to be 1 or -1. Consider the integers mod 10. The units are 1,3,7,9. Note that 3*9 is 7 mod 10. This clearly isn't 1 or -1. On the other hand, 3 and 7 are inverses.[/quote] I know the difference between unit and unity. Unity- i, 1, -1. Unit of a ring- For any given natural number n, g is a unit of n if and only if g is natural <n, (n,g)=1. I'm deffinietly not confused (maybe just because of not understanding what question am I needed to answer now), Your last question was: What is the product of natural numbers a which satisfy: a=a^(-1)(mod m). We easily mention that these numbers are equal to thier inverse. These may be 1,-1 (I called these "real unities" becuase these are real and unities, I'm sorry for calling these like that, becuase of wasting too much time for it). Can you please give me a hint how to explain why is d(m) modulo m is allways 1 or -1? I want again to show my work: If "a" which is d(m) modulo m (the residue) isn't 1 or -1, it may be devide d(m) or m. If it devide any of them, it must be deviding the other one too, which is out of the definition of d(m). My confusion come by not understanding the definition of inverse here: "On the other hand, 3 and 7 are inverses. " Inverses of what? Do you mean, a number b^(-1) for any real b? Is there another meaning here by saying inverse? |
[QUOTE=blob100;217274]
<snip> These may be 1,-1 (I called these "real unities" becuase these are real and unities, I'm sorry for calling these like that, becuase of wasting too much time for it). [/QUOTE] Stop inventing terminology. And your statement: "(I called these "real unities" becuase these are real and unities, ) is a misuse of existing terminolgy. 1 and -1 are units of Z. In fact, they are the only units. Stop inserting extraneous words into (i.e. the word 'real', as in 'real unities') your discussion because they make it appear that you are just confused. Elements of a ring that are equal to their own inverse do not have to be 1 or -1. Why do you think that they are? Consider m = (say) 24. 5 is its own inverse, yet 5 certainly is not 1 or -1. [QUOTE] Can you please give me a hint how to explain why is d(m) modulo m is allways 1 or -1? [/QUOTE] For about the (3rd? 4th? 5th?) time: PAIR THE UNITS WITH THEIR INVERSES. [QUOTE] I want again to show my work: If "a" which is d(m) modulo m (the residue) isn't 1 or -1, it may be devide d(m) or m. If it devide any of them, it must be deviding the other one too, which is out of the definition of d(m). My confusion come by not understanding the definition of inverse here: "On the other hand, 3 and 7 are inverses. " Inverses of what? Do you mean, a number b^(-1) for any real b? Is there another meaning here by saying inverse?[/QUOTE] Stop adding words!!!!! (as in 'real' b). Do not use the word real in a mathematical context unless you are in fact discussing the REAL NUMBERS. We are discussing the units of the integers taken mod m. These are the ONLY numbers under discussion. Why do you keep babbling about "real" numbers? In R, every number [b]except 0[/b] is a unit. I made it clear at the very beginning of this discussion what an inverse is. It is a MULTIPLICATIVE INVERSE. How can this possibly be unclear? The definition of a multiplicative inverse is basic, pre-algebra arithmetic. |
[quote=R.D. Silverman;217275]Stop inventing terminology.
And your statement: "(I called these "real unities" becuase these are real and unities, ) is a misuse of existing terminolgy. 1 and -1 are units of Z. In fact, they are the only units. Stop inserting extraneous words into (i.e. the word 'real', as in 'real unities') your discussion because they make it appear that you are just confused. Elements of a ring that are equal to their own inverse do not have to be 1 or -1. Why do you think that they are? Consider m = (say) 24. 5 is its own inverse, yet 5 certainly is not 1 or -1. For about the (3rd? 4th? 5th?) time: PAIR THE UNITS WITH THEIR INVERSES. Stop adding words!!!!! (as in 'real' b). Do not use the word real in a mathematical context unless you are in fact discussing the REAL NUMBERS. We are discussing the units of the integers taken mod m. These are the ONLY numbers under discussion. Why do you keep babbling about "real" numbers? In R, every number [B]except 0[/B] is a unit. I made it clear at the very beginning of this discussion what an inverse is. It is a MULTIPLICATIVE INVERSE. How can this possibly be unclear? The definition of a multiplicative inverse is basic, pre-algebra arithmetic.[/quote] The confusion came from not understanding what your talking about. You said inverse instead of multiplicative inverse. I'm not confused anymore. |
[QUOTE=blob100;217280]The confusion came from not understanding what your talking about.
You said inverse instead of multiplicative inverse. I'm not confused anymore.[/QUOTE] My apology. I thought that the word inverse was understood. |
I'll show the answer as a criterion:
If we have: (m-1)^2=1(mod m), and there is no g<m-1 prime to m such that: g^2=1(mod m) we have d(m)=-1(mod m). If we have such a g then d(m)=1(mod m), Where (m-1)^2=1(mod m) does not needed to exist (it can, but it can be not too). Examples: m=24. We have g={5,7,11,17,19} where g^2=1(mod 24). d(24)=1(mod 24). m=10 We have no such a g, but we have (10-1)^2=1(mod 10). d(10)=-1(mod 10). I found 24 as a really interesting number. It has just one quadratic residues which is one itself. Are there any other numbers of this kind? Are there infinitely many numbers of this form? |
[QUOTE=blob100;217298]I'll show the answer as a criterion:
If we have: (m-1)^2=1(mod m), and there is no g<m-1 prime to m such that: g^2=1(mod m) we have d(m)=-1(mod m). If we have such a g then d(m)=1(mod m), Where (m-1)^2=1(mod m) does not needed to exist (it can, but it can be not too). Examples: m=24. We have g={5,7,11,17,19} where g^2=1(mod 24). d(24)=1(mod 24). m=10 We have no such a g, but we have (10-1)^2=1(mod 10). d(10)=-1(mod 10). I found 24 as a really interesting number. It has just one quadratic residues which is one itself. Are there any other numbers of this kind? Are there infinitely many numbers of this form?[/QUOTE] Yes, and Yes. You need to answer the following: What do you get when you multiply two units, each of which is its own inverse.? What determines when the unit product is 1 and when it is -1.? |
[quote=R.D. Silverman;217300]Yes, and Yes.
You need to answer the following: What do you get when you multiply two units, each of which is its own inverse.? What determines when the unit product is 1 and when it is -1.?[/quote] For what the "Yes and Yes" claiming was directet? |
[quote=R.D. Silverman;217300]
You need to answer the following: What do you get when you multiply two units, each of which is its own inverse.? [/quote] When you say so, you mean: g1g2=b(mod m) We get b. And g1 and g2 each units of m with inverse g1 and g2 respectively. Example: m=24, g1=5 g2=7. g1g2=11(mod 24). b=11. What happens if there are no two units agree the terms (have thier own inverse)? I think the product of two units (that agree the terms) modulo m is another unit of that kind. We may call it a triangle: For m=16. 15, 9, 7 agree the terms. 15*7=9(mod 16) 15*9=7(mod 16) 7*9=15(mod 16) Isn't my criterion true? I'll repeat: a=1 if and only if there is no such g: g<(m-1) (g,m)=1 g^2=1(mod m), (but m-1 can agree the congruence (m-1)^2=1(mod m)). a=-1 if and only if there is such g, (m-1 can agree th congruence here too). |
[QUOTE=blob100;217335]For what the "Yes and Yes" claiming was directet?[/QUOTE]
To the two questions that immediately preceded the answer |
[QUOTE=blob100;217336]When you say so, you mean:
g1g2=b(mod m) We get b. And g1 and g2 each units of m with inverse g1 and g2 respectively. Example: m=24, g1=5 g2=7. g1g2=11(mod 24). b=11. .[/QUOTE] Despite my repeated admonshments you are still using some bad habits. g1, g2, and b are undefined. Stop using variables without defining them! Stop doing everything by numerical example, and start doing things algebraically. Numerical examples only HIDE what is going on. Working mod m: Let g1 and g2 be units that are their own inverse. Hence, g1 * g1 = 1 ; Thus elements that are their own inverse have order 2! g2 * g2 = 1 Whence (g1 * g2)^2 = 1, Thus the product of two units that are their own inverse is ALSO a unit that is its own inverse. The purpose of all of these exercizes is for you to discover relationships among variables that occur in modular arithmetic. STOP doing the numerical examples and START doing things algebraically. Now ask yourself: what happens if I multiply ALL the elements of order 2? |
[quote=R.D. Silverman;217353]Despite my repeated admonshments you are still using some bad habits.
g1, g2, and b are undefined. Stop using variables without defining them! Stop doing everything by numerical example, and start doing things algebraically. Numerical examples only HIDE what is going on. Working mod m: Let g1 and g2 be units that are their own inverse. Hence, g1 * g1 = 1 ; Thus elements that are their own inverse have order 2! g2 * g2 = 1 Whence (g1 * g2)^2 = 1, Thus the product of two units that are their own inverse is ALSO a unit that is its own inverse. The purpose of all of these exercizes is for you to discover relationships among variables that occur in modular arithmetic. STOP doing the numerical examples and START doing things algebraically. Now ask yourself: what happens if I multiply ALL the elements of order 2?[/quote] First of all, g1 and g2 were defined "each one is a unit of m with inverse g1 and g2 respectivly (g1's inverse is g1 and same with g2)" The multiply of all the elements of order 2 modulo m is +/-1. Look: you said: "Whence (g1 * g2)^2 = 1, Thus the product of two units that are their own inverse is ALSO a unit that is its own inverse." And I said: "What happens if there are no two units agree the terms (have thier own inverse)? I think the product of two units (that agree the terms) modulo m is another unit of that kind." Which is the same.. When I sain "agree the terms" I meant a unit which is it's own inverse. |
[QUOTE=blob100;217371]First of all, g1 and g2 were defined "each one is a unit of m with inverse g1 and g2 respectivly (g1's inverse is g1 and same with g2)"
[/QUOTE] This is not what you wrote. You wrote: "When you say so, you mean: g1g2=b(mod m) We get b." I could not find a definition of g1 or g2 in your prior posts either. [QUOTE] The multiply of all the elements of order 2 modulo m is +/-1. Look: you said: "Whence (g1 * g2)^2 = 1, Thus the product of two units that are their own inverse is ALSO a unit that is its own inverse." And I said: "What happens if there are no two units agree the terms (have thier own inverse)? [/QUOTE] For the last time, please stop inventing terminology (i.e. your expression "agree the terms"). It is clear that "have their own inverse" is well defined, so why invent terminology? It only adds confusion. [QUOTE] I think the product of two units (that agree the terms) modulo m is another unit of that kind." Which is the same.. When I sain "agree the terms" I meant a unit which is it's own inverse.[/QUOTE] With regard to this last comment of yours: STOP! Don't say "agree the terms" when you mean "unit equal to its own inverse". Try proving the following: (working mod m) If x1 and x2 are units with order e1, prove that (x1*x2) has order e1. Note that if x1 has order e1 and x2 has order e2, then the order of (x1 * x2) is unpredictable. So, let us return to the question: What determines when the product of the units is 1 and what determines when it is -1.?? For this question you may restrict m to be squarefree. (otherwise things can get complicated because of Hensel's Lemma [which we have not yet discussed]). |
just for laugh...
Euler power hippies? |
[QUOTE=R.D. Silverman;217375]This is not what you wrote. You wrote:
"When you say so, you mean: g1g2=b(mod m) We get b." I could not find a definition of g1 or g2 in your prior posts either. For the last time, please stop inventing terminology (i.e. your expression "agree the terms"). It is clear that "have their own inverse" is well defined, so why invent terminology? It only adds confusion. With regard to this last comment of yours: STOP! Don't say "agree the terms" when you mean "unit equal to its own inverse". Try proving the following: (working mod m) If x1 and x2 are units with order e1, prove that (x1*x2) has order e1. Note that if x1 has order e1 and x2 has order e2, then the order of (x1 * x2) is unpredictable. So, let us return to the question: What determines when the product of the units is 1 and what determines when it is -1.?? For this question you may restrict m to be squarefree. (otherwise things can get complicated because of Hensel's Lemma [which we have not yet discussed]).[/QUOTE] Actually, let's skip this last question. It gets us too deep into group theory and away from number theory. Instead, we simply observe, for example, that integers m that have only 2 elements of order 2 must have the product of all the units equal to -1. Why? Because the product of the units that don't have order 2 is 1 (we already established this) and the units of order 2 are 1 and m-1, and their product is -1. It should also be easy to see that when the number of elements of order 2 is a power of 2, then the product of all the units is +1. (units of order 2 are closed under multiplication; pair them. then pair their products. then pair those products etc.) |
[quote=R.D. Silverman;217383]Actually, let's skip this last question. It gets us too deep into group
theory and away from number theory. Instead, we simply observe, for example, that integers m that have only 2 elements of order 2 must have the product of all the units equal to -1. Why? Because the product of the units that don't have order 2 is 1 (we already established this) and the units of order 2 are 1 and m-1, and their product is -1. It should also be easy to see that when the number of elements of order 2 is a power of 2, then the product of all the units is +1. (units of order 2 are closed under multiplication; pair them. then pair their products. then pair those products etc.)[/quote] Let's discuss Hensel's lemma? |
[QUOTE=blob100;217503]Let's discuss Hensel's lemma?[/QUOTE]
Premature. |
[QUOTE=R.D. Silverman;217504]Premature.[/QUOTE]
BTW, you might actually want to try finishing the current problem. We have shown that the product of the units whose order is not 2 equals 1. What remains is to show: The product of units of order 2 is either 1 or -1. Determine when this product is 1 and when it is -1. This can actually be done fairly simply, without any group theory. Hint: (-1)^n = 1 or -1 depending on n mod 2. |
[quote=R.D. Silverman;217525]BTW, you might actually want to try finishing the current problem.
We have shown that the product of the units whose order is not 2 equals 1. What remains is to show: The product of units of order 2 is either 1 or -1. Determine when this product is 1 and when it is -1. This can actually be done fairly simply, without any group theory. Hint: (-1)^n = 1 or -1 depending on n mod 2.[/quote] When you say "the product of units of order 2 is either 1 or -1." you mean that for units a1, a2, a3...,ak of order 2 we have (let's say) a1a2=+/-1(mod m)? Or you mean: (a1a2)^2=+/-1(mod m)? For example (I know you don't want me to give examples, but it will improve my understanding): m=24. a1=5 and a2=7 (we would say it can be 11 and 13 too, but let's take 5,7). a1a2=35. 35-1=\=0(mod 24) 35+1=\=0(mod 24) But 35^2=1(mod 24). |
[QUOTE=blob100;217580]When you say "the product of units of order 2 is either 1 or -1." you mean that for units a1, a2, a3...,ak of order 2 we have (let's say) a1a2=+/-1(mod m)?
Or you mean: (a1a2)^2=+/-1(mod m)? [/QUOTE] For units a1,a2.....ak, of order 2, we have a1*a2*....*ak = 1 or -1 [/QUOTE] |
We see that every product of two units ai of order 2 is antoher units of order 2.
The product of these three units (the two and the third we got by the product) is +/-1. We may find K such "triangles" (set of three units discussed before). There may be k such triangles -1. The product of all units of order 2 will be denoted as g(m). g(m)=(-1)^k(mod m). |
[QUOTE=blob100;217629]We see that every product of two units ai of order 2 is antoher units of order 2.
The product of these three units (the two and the third we got by the product) is +/-1. [/QUOTE] This claim would need proof. And it is not true in general. .[/QUOTE] |
[quote=R.D. Silverman;217630]This claim would need proof. And it is not true in general.
.[/quote][/quote] What do you mean by: "not true in general"? You would just say: "False". What is the direction? |
[QUOTE=blob100;217634][/quote]
What do you mean by: "not true in general"? You would just say: "False". What is the direction?[/QUOTE] Given three units of order 2, a1, a2, a3, it is [b]sometimes[/b] true that a1*a2 *a3 = 1 or -1 mod m, but it is not always true. Given any two such units a1, a2, we can find a unit a3, such that a1 * a2 * a3 = 1 or -1, but it will not be true for (say) a1 * a2 * a4 where a4 is a unit of order 2 different from a3. It is true for [b]some[/b] triples of units, but not for all. |
[quote=R.D. Silverman;217641]What do you mean by: "not true in general"?
You would just say: "False". What is the direction?[/quote] Given three units of order 2, a1, a2, a3, it is [B]sometimes[/B] true that a1*a2 *a3 = 1 or -1 mod m, but it is not always true. Given any two such units a1, a2, we can find a unit a3, such that a1 * a2 * a3 = 1 or -1, but it will not be true for (say) a1 * a2 * a4 where a4 is a unit of order 2 different from a3. It is true for [B]some[/B] triples of units, but not for all.[/quote] OK, This is the same as I said: "We see that every product of two units ai of order 2 is antoher units of order 2. The product of these three units (the two and the third we got by the product) is +/-1." |
[QUOTE=blob100;217692]Given three units of order 2, a1, a2, a3, it is [B]sometimes[/B]
true that a1*a2 *a3 = 1 or -1 mod m, but it is not always true. Given any two such units a1, a2, we can find a unit a3, such that a1 * a2 * a3 = 1 or -1, but it will not be true for (say) a1 * a2 * a4 where a4 is a unit of order 2 different from a3. It is true for [B]some[/B] triples of units, but not for all OK, This is the same as I said: "We see that every product of two units ai of order 2 is antoher units of order 2. The product of these three units (the two and the third we got by the product) is +/-1."[/QUOTE] No. What you said and what I said are NOT the same thing. You said that the product of two units of order 2 is another unit of order 2. This is correct. You then say "The product of these three units...", but you do not SPECIFY the 3rd unit. You do say "the third we got by the product", but this phrase is meaningless gibberish. In any event, you can not prove the result by the mathod you are attempting. Consider: What if the total number of units of order 2 is not divisible by 3? Then one cannot match them up in the way you suggest. Perhaps this is just a language difficulty. Would someone else care to help me explain it? Apparently, my message is not getting through. |
[quote=R.D. Silverman;217693]
In any event, you can not prove the result by the mathod you are attempting. Consider: What if the total number of units of order 2 is not divisible by 3? Then one cannot match them up in the way you suggest. [/quote] I'm sorry for not expressing myself legally. I knew there are naturals m with a number of units of order 2 not divisible by 3. You need to agree there are sometimes "triangles" with a same unit. For example: a1a2=a3(mod m) a5a4=a1(mod m) Let's say A(a1,a2,a3) and B(a5,a4,a1) which both contain a1. The point is that every unit of order 2 can be located in a tripe of that kind. My propisition is that: For g1=a1a2a3=+/-1(mod m) The product of every such g we have: P=(-1)^k(mod m) Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m). |
Let me attempt to "translate". If I fail to restate you argument in a proper manner, then you should attempt to clarify the language.
[QUOTE=blob100;217695]I'm sorry for not expressing myself legally.[/QUOTE] Rather than "legally", "succinctly" would be a better choice. In mathematics, the only "law" that comes to mind is "The Law of Large Numbers". [QUOTE]I knew there are naturals m with a number of units of order 2 not divisible by 3. You need to agree there are sometimes "triangles" with a same unit.[/QUOTE] A "triangle" is a two-dimensional geometric shape. I think that you are trying to discuss some kind of 3-tupes. [QUOTE] For example: a1a2=a3(mod m) a5a4=a1(mod m) Let's say A(a1,a2,a3) and B(a5,a4,a1) which both contain a1.[/QUOTE] For an appropriate integer m, within {a}, the set of integers mod m, consider the set of 3-tuples of the form {a[SUB]i[/SUB], a[SUB]j[/SUB], a[SUB]k[/SUB]} where a[SUB]i[/SUB] * a[SUB]j[/SUB] = a[SUB]k[/SUB] (mod m). [QUOTE]The point is that every unit of order 2 can be located in a trip[B]l[/B]e of that kind.[/QUOTE] Let A = {a1,a2,a3} and B={a5,a4,a1} be such -tuples. (Here I am "lost" because there is nothing "special" about the triples, as defined. [B]Every[/B] "a" appears in many such triples (as the a[SUB]i[/SUB], a[SUB]j[/SUB], and/or a[SUB]k[/SUB]). [QUOTE] My prop[B]o[/B]sition is that: For g1=a1a2a3=+/-1(mod m) The product of every such g we have: P=(-1)^k(mod m) Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m).[/QUOTE] |
[quote=Wacky;217702]Let me attempt to "translate". If I fail to restate you argument in a proper manner, then you should attempt to clarify the language.
Rather than "legally", "succinctly" would be a better choice. In mathematics, the only "law" that comes to mind is "The Law of Large Numbers". A "triangle" is a two-dimensional geometric shape. I think that you are trying to discuss some kind of 3-tupes. For an appropriate integer m, within {a}, the set of integers mod m, consider the set of 3-tuples of the form {a[sub]i[/sub], a[sub]j[/sub], a[sub]k[/sub]} where a[sub]i[/sub] * a[sub]j[/sub] = a[sub]k[/sub] (mod m). Let A = {a1,a2,a3} and B={a5,a4,a1} be such -tuples. (Here I am "lost" because there is nothing "special" about the triples, as defined. [B]Every[/B] "a" appears in many such triples (as the a[sub]i[/sub], a[sub]j[/sub], and/or a[sub]k[/sub]).[/quote] Thank you verry much for teaching me how to build a mathematical argument.:bow: |
"My propisition is that:
For g1=a1a2a3=+/-1(mod m) The product of every such g we have: P=(-1)^k(mod m) Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m). " I'll try to write it better: We will denote the product of such tripe as gi. The product of the whole products gi of tripes will be denoted as P. P=g1g2g3...gk where we have k combinations for triples. P=(-1)^k(mod m), Where we assume k as the number of tripes g such that g=-1(mod m). We see a connection between P and g(m) is the product of the units of order 2.P is the product of the units of order 2 powers>/=1 and g(m) is the product of the units of order 2. |
[QUOTE=blob100;217707]"My propisition is that:
For g1=a1a2a3=+/-1(mod m) [/QUOTE] Whoa! Back up. What are you trying to say? How does this relate to the 3-tuples; or the -tuples "A" or "B"? And, just to make it clear, repeat, from the top, all of the clauses that you are using. I don't see "a4" or "a5". Why did you introduce them if you are not going to use them? Are you saying: For some/any/all a[SUB]1[/SUB], a[SUB]2[/SUB], let a[SUB]3[/SUB] = a[SUB]1[/SUB] * a[SUB]2[/SUB] then (g[SUB]1[/SUB] = a[SUB]1[/SUB] * a[SUB]2[/SUB] * a[SUB]3[/SUB]) == +/-1 (mod m) ?? Perhaps you meant to restrict the "a" in some way? Perhaps to units of order 2? If the relationship is meant to represent a number of such "g", then rather than using integer subscripts, using letters for the "independent" selections might make your intention more clear-- as in: For some/any/all a[SUB]i[/SUB], a[SUB]j[/SUB], let a[SUB]ij[/SUB] = a[SUB]i[/SUB] * a[SUB]j[/SUB] then (g[SUB]ij[/SUB] = a[SUB]i[/SUB] * a[SUB]j[/SUB] * a[SUB]ij[/SUB]) == +/-1 (mod m) [QUOTE]The product of every such g we have: P=(-1)^k(mod m) Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m). " I'll try to write it better: We will denote the product of such tripe as gi. The product of the whole products gi of tripes will be denoted as P. P=g1g2g3...gk where we have k combinations for triples. P=(-1)^k(mod m), Where we assume k as the number of tripes g such that g=-1(mod m). We see a connection between P and g(m) is the product of the units of order 2.P is the product of the units of order 2 powers>/=1 and g(m) is the product of the units of order 2.[/QUOTE] I think that I see where you are going. This argument relies on the "definition" of which -tuples you are considering. Therefore, you need to state, precisely, which g get included in the product P. In particular, how does that selection process relate to the a ? |
[quote=Wacky;217711]Whoa! Back up. What are you trying to say? How does this relate to the 3-tuples; or the -tuples "A" or "B"? And, just to make it clear, repeat, from the top, all of the clauses that you are using.
I don't see "a4" or "a5". Why did you introduce them if you are not going to use them? Are you saying: For some/any/all a[sub]1[/sub], a[sub]2[/sub], let a[sub]3[/sub] = a[sub]1[/sub] * a[sub]2[/sub] then (g[sub]1[/sub] = a[sub]1[/sub] * a[sub]2[/sub] * a[sub]3[/sub]) == +/-1 (mod m) ?? Perhaps you meant to restrict the "a" in some way? Perhaps to units of order 2? If the relationship is meant to represent a number of such "g", then rather than using integer subscripts, using letters for the "independent" selections might make your intention more clear-- as in: For some/any/all a[sub]i[/sub], a[sub]j[/sub], let a[sub]ij[/sub] = a[sub]i[/sub] * a[sub]j[/sub] then (g[sub]ij[/sub] = a[sub]i[/sub] * a[sub]j[/sub] * a[sub]ij[/sub]) == +/-1 (mod m) I think that I see where you are going. This argument relies on the "definition" of which -tuples you are considering. Therefore, you need to state, precisely, which g get included in the product P. In particular, how does that selection process relate to the a ?[/quote] OK, I understand everything was deffinietly not understandable. Again: We denote g as the product of ai, aj and av. Where ai*aj=av(mod m) for ai,j,v units of order 2. This is the specified definition: g=ai*aj*av. The product P contains (includes) every g as defined before. And the proposition is: P=(-1)^k(mod m), where k is the number of numbers g such that g=-1(mod m). This proposition is directed to the actual problem becuase of P product of the units of order 2 powers and we are on the actual problem we deal with the product of the units of order 2. |
[QUOTE=blob100;217718]OK, I understand everything was deffinietly not understandable.
Again: We denote g as the product of ai, aj and av. Where ai*aj=av(mod m) for ai,j,v units of order 2. This is the specified definition: g=ai*aj*av. The product P contains (includes) every g as defined before. [/QUOTE] No, it does NOT. What if the number of order-2 units is not a multiple of 3??? What if ai*aj*av = 1 (say), but when you multiply ax * ay, you get ai? You have already used ai in a different triplet. It is quite possible that aj * av = ax * ay. You can not prove the result by combining them as 3-tuples. I already told you this. |
[QUOTE=R.D. Silverman;217375]Try proving the following:
(working mod m) If x1 and x2 are units with order e1, prove that (x1*x2) has order e1. [/QUOTE] [QUOTE=R.D. Silverman;217383]Actually, let's skip this last question. It gets us too deep into group theory and away from number theory.[/QUOTE] Not to mention that it's false. What did you really mean here? |
[QUOTE=jyb;217742]Not to mention that it's false. What did you really mean here?[/QUOTE]
You can prove the result by showing that the product of the units in a cyclic group is -1, using the cyclic decomposition theorem to decompose the unit group as a product of cyclic sub-groups, then use the Sylow theorem to determine whether the number of such sub-groups is even or odd (the parity determines whether the product is -1 or +1). Yech. |
[QUOTE=R.D. Silverman;217746]You can prove the result by showing that the product of the units in a
cyclic group is -1, using the cyclic decomposition theorem to decompose the unit group as a product of cyclic sub-groups, then use the Sylow theorem to determine whether the number of such sub-groups is even or odd (the parity determines whether the product is -1 or +1). Yech.[/QUOTE] Note that the "last question" was not: If x1 and x2 are units with order e1, prove that (x1*x2) has order e1. The "last question" was to discover what determines when the product is +1 and when it is -1. However, it *can* be done without the heavy group-theory machinery. I am trying to lead Tomer through that now. |
[QUOTE=R.D. Silverman;217747]Note that the "last question" was not:
If x1 and x2 are units with order e1, prove that (x1*x2) has order e1. The "last question" was to discover what determines when the product is +1 and when it is -1. However, it *can* be done without the heavy group-theory machinery. I am trying to lead Tomer through that now.[/QUOTE] Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason? |
[QUOTE=jyb;217755]Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason?[/QUOTE]
e1 = 2 here. |
[QUOTE=R.D. Silverman;217781][QUOTE=jyb;217755]Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason?[/QUOTE]
e1 = 2 here.[/QUOTE] It's still false. Unless you stipulate that x1 != x2, of course. But even then, why did you generalize the problem from order 2 to order e1 if the point you were making only works for order 2? Particularly since you had already shown it to be true for order 2 in this post: [url]http://mersenneforum.org/showpost.php?p=217353&postcount=503[/url]. |
I'll back to the tuples:
A tuple g is defined as: The product of 3 specified distinct units of order 2. The specified 3 units are: Given 2 units of order 2 which thier product mod m is the third unit of order 2. Two tuples can have a common devisor. Example: m=24. 5*7=11(mod 24) With the numbers 5,7,11 we make the tuple 5*7*11. On the same time we have: 5*13=17(mod 24) A common devisor 5 is recieved by the two tuples (5*7*11 and 5*13*17). |
[QUOTE=jyb;217807]It's still false. Unless you stipulate that x1 != x2, of course. But even then, why did you generalize the problem from order 2 to order e1 if the point you were making only works for order 2? Particularly since you had already shown it to be true for order 2 in this post: [url]http://mersenneforum.org/showpost.php?p=217353&postcount=503[/url].[/QUOTE]
I thought that it had been stipulated that the question was about the product of two different units of order 2. If I failed to specify e1 = 2, it was an oversight on my part. |
[QUOTE=blob100;217810]I'll back to the tuples:
A tuple g is defined as: The product of 3 specified distinct units of order 2. The specified 3 units are: Given 2 units of order 2 which thier product mod m is the third unit of order 2. Two tuples can have a common devisor. [/QUOTE] Since all elements are units, it is meaningless to speak of a 'divisor'. Every unit has a multiplicative inverse, so every unit is a divisor of all other units. [QUOTE] Example: m=24. 5*7=11(mod 24) With the numbers 5,7,11 we make the tuple 5*7*11. On the same time we have: 5*13=17(mod 24) A common devisor 5 is recieved by the two tuples (5*7*11 and 5*13*17).[/QUOTE] This last statement is pure [b]NONSENSE[/b]. You are working with units mod m, and [b]any[/b] unit will be a "common divisor" of 5*13*17, because every unit has an inverse! Given any unit x0 then x0 divides 5*7*11 and it divides 5*13*17 mod 24. You are not working in Z. I am almost ready to give up on you because: (1) I told you to forget about 3-tuples of units, yet you persist with them. (2) I told you to [b]stop[/b] discussing mathematics from numerical example(s) and to use algebra, yet you still insist on trying to discuss an algebraic question by presenting numerical examples. Your problem: prove that the product of (all) units of order 2 is either 1 or -1. I already told you (and explained why) that your "3-tuple" approach will not work. Try something else. The proof is only a couple of lines and is actually quite easy. |
[quote=R.D. Silverman;217813]Since all elements are units, it is meaningless to speak of a 'divisor'.
Every unit has a multiplicative inverse, so every unit is a divisor of all other units. This last statement is pure [B]NONSENSE[/B]. You are working with units mod m, and [B]any[/B] unit will be a "common divisor" of 5*13*17, because every unit has an inverse! Given any unit x0 then x0 divides 5*7*11 and it divides 5*13*17 mod 24. You are not working in Z. I am almost ready to give up on you because: (1) I told you to forget about 3-tuples of units, yet you persist with them. (2) I told you to [B]stop[/B] discussing mathematics from numerical example(s) and to use algebra, yet you still insist on trying to discuss an algebraic question by presenting numerical examples. Your problem: prove that the product of (all) units of order 2 is either 1 or -1. I already told you (and explained why) that your "3-tuple" approach will not work. Try something else. The proof is only a couple of lines and is actually quite easy.[/quote] First of all, This last post had not titled with "This post is the approach..." It was meant to explain a missunderstand of of you. I know what is the problem I'm meant to solve. Why won't you give me the directon to solve the problem, I deffinietly don't understand how to attack this problem. You gave me the hint: (-1)^n. I tought this may be helpfull myself, the hint does not help me... Why did you on one of the posts concerned the product of the units of order 2 mod m? This is what gave me the idea of the tuples. |
If we have g(m) the product of the whole units of order 2,
We have (g(m))^2=1(mod m). (g(m)-1)(g(m)+1)=0(mod m) Which gives m is a devisor of g(m)-1 or g(m)+1, Which is our claim. I tought I firstly need to examine whether the residue is 1 or -1, That is why I used the tuples (I won't talk about them anymore as you wish). |
[QUOTE=blob100;217821]If we have g(m) the product of the whole units of order 2,
We have (g(m))^2=1(mod m). (g(m)-1)(g(m)+1)=0(mod m) Which gives m is a devisor of g(m)-1 or g(m)+1, [/QUOTE] Go to google. Look up "zero divisor". If a*b = 0 mod m, where m is composite, it need not be the case that either a or b is zero. e.g. 3*4 = 0 mod 12, but neither 3 not 4 is zero. I will add a hint. Pair the units of order 2 in an appropriate way, so that the product of each pair is -1. |
[quote=R.D. Silverman;217844]Go to google. Look up "zero divisor".
If a*b = 0 mod m, where m is composite, it need not be the case that either a or b is zero. e.g. 3*4 = 0 mod 12, but neither 3 not 4 is zero. I will add a hint. Pair the units of order 2 in an appropriate way, so that the product of each pair is -1.[/quote] First of all, your example is not able to be given: g(m) lowest value is m-1, that so, we may have 11<a,b<12 (where you gave a=3 and b=4). Secondly: About the hint: we can find (sometimes) that the number of units of order 2 isn't even. |
[QUOTE=blob100;217917]Secondly:
About the hint: we can find (sometimes) that the number of units of order 2 isn't even.[/QUOTE] Please find one. |
[QUOTE=blob100;217917]First of all, your example is not able to be given:
g(m) lowest value is m-1, that so, we may have 11<a,b<12 (where you gave a=3 and b=4). [/QUOTE] g(m) is the product of all units of order 2. What do you mean when you say that the lowest value of g(m) is m-1? You have proved [b]nothing[/b] so far about the value of g(m). So how can you assert that its "lowest value is m-1" [whatever you mean by that phrase].? Your expression "11 < a,b < 12" is total nonsense. There is no element strictly between 11 and 12. Furthermore, nothing you have done indicates in [b]any[/b] way that the value of g(m) might be bounded. Yet you [b]assert[/b] without proof that it is. Furthermore, I told you to stop using numerical examples! Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0 mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero divisor. It does not prove that g(m) = 1 or -1. My example showed a product that was 0, even though neither element was 0. Why is this hard to understand? [QUOTE] Secondly: About the hint: we can find (sometimes) that the number of units of order 2 isn't even.[/QUOTE] Oh, really? Show an example. Didn't we previously go through an exercize where we showed that the number of solutions to x^2 = a mod m was a POWER OF TWO???? Did you pull this latest assertion from your ass? (that the number might not be even). I am ready to give up on you. You keep making assertions without justification, you do not take advice about defining your variables, you do not take advice about not using unnecessary variables, you keep resorting to a single numerical example to make an assertion instead of using algebra. AND YOU DO NOT TAKE HINTS. |
[QUOTE=axn;217918]Please find one.[/QUOTE]
I'm ready to give up on him. We spent a lot of time and effort in the previous exercize to show that the number of solutions to x^2 = a mod m was a power of two, and then he turns around and asserts that the number of solutions might be odd!!!! I have tried to be patient, but he does not seem to listen to my advice. |
[QUOTE=R.D. Silverman;217920]g(m) is the product of all units of order 2. What do you mean when you say
that the lowest value of g(m) is m-1? You have proved [b]nothing[/b] so far about the value of g(m). So how can you assert that its "lowest value is m-1" [whatever you mean by that phrase].? Your expression "11 < a,b < 12" is total nonsense. There is no element strictly between 11 and 12. Furthermore, nothing you have done indicates in [b]any[/b] way that the value of g(m) might be bounded. Yet you [b]assert[/b] without proof that it is. Furthermore, I told you to stop using numerical examples! Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0 mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero divisor. It does not prove that g(m) = 1 or -1. My example showed a product that was 0, even though neither element was 0. Why is this hard to understand? Oh, really? Show an example. Didn't we previously go through an exercize where we showed that the number of solutions to x^2 = a mod m was a POWER OF TWO???? Did you pull this latest assertion from your ass? (that the number might not be even). I am ready to give up on you. You keep making assertions without justification, you do not take advice about defining your variables, you do not take advice about not using unnecessary variables, you keep resorting to a single numerical example to make an assertion instead of using algebra. AND YOU DO NOT TAKE HINTS.[/QUOTE] Slight correction. In the previous exercize m was only odd. This makes a slight difference. However, it does not affect the result that the number of solutions to x^2= a mod m must be even. This last result can be proved trivially using bonehead secondary school first year algebra. |
[quote=R.D. Silverman;217920]g(m) is the product of all units of order 2. What do you mean when you say
that the lowest value of g(m) is m-1? You have proved [B]nothing[/B] so far about the value of g(m). So how can you assert that its "lowest value is m-1" [whatever you mean by that phrase].? Your expression "11 < a,b < 12" is total nonsense. There is no element strictly between 11 and 12. Furthermore, nothing you have done indicates in [B]any[/B] way that the value of g(m) might be bounded. Yet you [B]assert[/B] without proof that it is. Furthermore, I told you to stop using numerical examples! Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0 mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero divisor. It does not prove that g(m) = 1 or -1. My example showed a product that was 0, even though neither element was 0. Why is this hard to understand? Oh, really? Show an example. Didn't we previously go through an exercize where we showed that the number of solutions to x^2 = a mod m was a POWER OF TWO???? Did you pull this latest assertion from your ass? (that the number might not be even). I am ready to give up on you. You keep making assertions without justification, you do not take advice about defining your variables, you do not take advice about not using unnecessary variables, you keep resorting to a single numerical example to make an assertion instead of using algebra. AND YOU DO NOT TAKE HINTS.[/quote] If you wish to give up, your welcome. The number of units of order 2 for 16 is 3 (which isn't a 2 power): 7,9,15. When I say: "11 < a,b < 12 (.....)" I realize you are going to read what is written in the (,). I meant, that for ab=0(mod 12), you gave 3=a and b=4. I didn't say anything about g(m)'s bounding. I didn't use any, but ANY numerical examples on the last posts, I STOPPED. "Did you pull this latest assertion from your ass? (that the number might not be even)." Is a really polite sentence. When I say the lowest value of g(m) may be m-1, I mean: m-1 is always a unit of order 2, and we can find examples for numbers m with m-1 the only unit of order 2, which gives, g(m)'s value may be >/=m-1, We may agree m-1 is always a unit of order 2 becuase: (m-1)^2=1(mod m) m^2-2m+1-1=0(mod m) m^2-2m=0(mod m). I have a question for you: Why do you teach me? (I'm fully serious). Tomer. |
[QUOTE=blob100;217935]If you wish to give up, your welcome.
The number of units of order 2 for 16 is 3 (which isn't a 2 power): 7,9,15.[/QUOTE] Don't forget 1 (ok, so technically the order is 1, not 2). The required pairing then would be (1,15), (7,9). What is the pattern here? |
[quote=axn;217940]Don't forget 1 (ok, so technically the order is 1, not 2). The required pairing then would be (1,15), (7,9). What is the pattern here?[/quote]
So why won't we say the number of units of order 2 is 2^n-1 (for specified n natural)? |
[QUOTE=blob100;217944]So why won't we say the number of units of order 2 is 2^n-1 (for specified n natural)?[/QUOTE]
Because, for the problem under consideration, we're not worried about order. We're only concerned about units that are their own inverses -- and 1 clearly qualifies. I think this whole order-2 discussion has sidetracked the focus of the problem. |
[quote=axn;217940]Don't forget 1 (ok, so technically the order is 1, not 2). The required pairing then would be (1,15), (7,9). What is the pattern here?[/quote]
Yes, I do see a pattern: 15+1=16 7+9=16. |
[quote=axn;217948]Because, for the problem under consideration, we're not worried about order. We're only concerned about units that are their own inverses -- and 1 clearly qualifies. I think this whole order-2 discussion has sidetracked the focus of the problem.[/quote]
I see. |
[QUOTE=blob100;217950]Yes, I do see a pattern:
15+1=16 7+9=16.[/QUOTE] What is the product of each pair (mod 16, of course)? What conclusions can you draw from this observation? |
[QUOTE=blob100;217935]If you wish to give up, your welcome.
The number of units of order 2 for 16 is 3 (which isn't a 2 power): 7,9,15. [/QUOTE] NO! You should know the definition of element of order 2 by now. You are missing an element. And you STILL haven't learned from the previous exercize, which was: How many solutions are there to x^2 = a mod m??? And you still assert that there are only 3 solutions for m = 16. Don't you believe what you have already learned? [QUOTE] When I say: "11 < a,b < 12 (.....)" I realize you are going to read what is written in the (,). [/QUOTE] As is everyone else reading this thread. [QUOTE] I meant, that for ab=0(mod 12), you gave 3=a and b=4. I didn't say anything about g(m)'s bounding. [/QUOTE] You said 11 < a,b<12. This sure looks as if you imposing bounds on g to me! The question was about a*b = 0 mod m implying that a = 0 or b = 0. In this case a = g(m) - 1 and b = g(m) + 1. [QUOTE] I didn't use any, but ANY numerical examples on the last posts, I STOPPED. "Did you pull this latest assertion from your ass? (that the number might not be even)." Is a really polite sentence. [/QUOTE] It was [b] intended [/b] as a rebuke. [QUOTE] When I say the lowest value of g(m) may be m-1, I mean: m-1 is always a unit of order 2, [/QUOTE] This last statement is clear. But you did not connect it in any way to the value of g(m). [QUOTE] [and we can find examples for numbers m with m-1 the only unit of order 2, [/QUOTE] I don't know which universe you are living in, but in the one I live in, there are always at least two elements of order 2. Please show your examples. [QUOTE] I have a question for you: Why do you teach me? (I'm fully serious). Tomer.[/QUOTE] I'm still hopeful that you might actually learn something. |
[quote=axn;217956]What is the product of each pair (mod 16, of course)?
What conclusions can you draw from this observation?[/quote] 1) -1. The product between a unit of order 2 (denoted as g) and m-g mod m is -1. From here, we continue to: If there are 2k pairs (defined before as: two units of order 2 where thier product is -1 and thier sum is m), the product of [B]all[/B] the units of order 2 mod m is (-1)^k. |
[quote=R.D. Silverman;217958]
I don't know which universe you are living in, but in the one I live in, there are always at least two elements of order 2. Please show your examples. [/quote] It is the same as for 16. I was not actually including 1 as a unit of order 2 becuase it is of order 1. We may say there are 4 units of order 2 for 16 (including 1): 1, 7, 9, 15. An example for a number m from my universe I'm living on is 10 which his units of order 2 are 9 (not including 1), and 2 (including 1). I guess I need to include 1, but it isn't defined as a number of order 2. |
Good night mersenneforum.org.
|
[QUOTE=blob100;217962]
I guess I need to include 1, but it isn't defined as a number of order 2.[/QUOTE] No? Elements that are their own multiplicative inverse are elements of order 2. i.e. x = x^-1 mod m --> x^2 = 1 mod m. Isn't 1 a solution to x^2 = 1 mod m??? |
[QUOTE=axn;217940]Don't forget 1 (ok, so technically the order is 1, not 2). The required pairing then would be (1,15), (7,9). What is the pattern here?[/QUOTE]
1 is usually considered to have order 2. |
[QUOTE=blob100;217950]Yes, I do see a pattern:
15+1=16 7+9=16.[/QUOTE] Please, please, stop with the numerical examples!! Use algebra. The following should have been immediately obvious to anyone who has taken first year algebra: If x0 is a solution to x^2 = 1 mod m, then SO IS -x0!!!! The product : x0 * -x0 = -1!!!! Pair each element of order 2 with its additive inverse, then multiply the pairs together. If the number of pairs is odd, then the product is -1. Why was this so hard????? |
[QUOTE=R.D. Silverman;217958]I don't know which universe you are living in, but in the one I live in,
there are always at least two elements of order 2.[/QUOTE] [QUOTE=R.D. Silverman;217975]No? Elements that are their own multiplicative inverse are elements of order 2. i.e. x = x^-1 mod m --> x^2 = 1 mod m. Isn't 1 a solution to x^2 = 1 mod m???[/QUOTE] [QUOTE=R.D. Silverman;217976]1 is usually considered to have order 2.[/QUOTE] Okay, now I'm really curious. In a multitude of texts on number theory and algebra, I have [B]never[/B] seen a definition of element order that agrees with this. I have always seen it defined such that the order of the group identity is 1. So when you say it's "usually considered to have order 2", can you point me to the people/texts who are usually doing this considering? Or to put it another way, just which universe [B]are[/B] you living in? :grin: |
[QUOTE=jyb;217978]Okay, now I'm really curious. In a multitude of texts on number theory and algebra, I have [B]never[/B] seen a definition of element order that agrees with this. I have always seen it defined such that the order of the group identity is 1. So when you say it's "usually considered to have order 2", can you point me to the people/texts who are usually doing this considering? Or to put it another way, just which universe [B]are[/B] you living in? :grin:[/QUOTE]
You may be (probably/almost certainly are) correct with regard to what appears in books. Certainly the common definition of "order" is the smallest n such that a^n = 1, and for a = 1, n=1 is certainly the smallest. But I have seen/heard this notational issue debated vigorously during a West Coast Number Theory Conf. back in the late 80's. One of the Lenstra's was discussing some open problems related to improving the efficieny of the (then very new) APR-CL test. It was in the context not of groups, but of cyclotomic rings. It involved elements of prime order. Lenstra had tossed a set onto a screen. The set was elements of order p, but presented in terms of solutions to x^p = 1. Since 1 is a solution, Lenstra had presented it as part of the set of elements of order p. A debate ensued as to whether this was proper. The general consensus reached was that in the context of [b]prime power[/b] order, it was OK to include 1, in order to have to make an exceptional case for the element 1 when stating a general result. But the consensus was also that it is highly context sensitive. I also heard a similar discussion when receiving a lecture on finite fields from Birkhoff back when I was an undergrad (and yes, we used his book). I also remember a separate discussion about whether 1 could be considered as the order of a finite field with 1 element (and indeed whether such a thing existed) [usually considered the 'trivial' field; whether it is a field depends on whether one requires the convention that the multiplicative identity must be different from the additive. ]. Certainly there are many instances when 1 is tossed into a set to keep from having it singled out as a special case in some theorem. 1 is excluded from being prime, so we [b]don't[/b] have to create a special case in the FTA. I agree that if it were not for the [b]context[/b] of the current problem, considering 1 to have order 2 would be considered abuse of notation. I agree that the discussion might have been clearer if we had just stuck to "elements that are their own inverse". But I deliberately introduced x^2 = 1, precisely because I was hoping that a discussion of zero divisors would arise as a result. I will accept the Lang comment. (Do you know it?) Lang was well known for telling people at lectures "Your argument is fine, but your notation sucks". |
[QUOTE=R.D. Silverman;217992]You may be (probably/almost certainly are) correct with regard to what
appears in books. Certainly the common definition of "order" is the smallest n such that a^n = 1, and for a = 1, n=1 is certainly the smallest. But I have seen/heard this notational issue debated vigorously during a West Coast Number Theory Conf. back in the late 80's. One of the Lenstra's was discussing some open problems related to improving the efficieny of the (then very new) APR-CL test. It was in the context not of groups, but of cyclotomic rings. It involved elements of prime order. Lenstra had tossed a set onto a screen. The set was elements of order p, but presented in terms of solutions to x^p = 1. Since 1 is a solution, Lenstra had presented it as part of the set of elements of order p. A debate ensued as to whether this was proper. The general consensus reached was that in the context of [b]prime power[/b] order, it was OK to include 1, in order to have to make an exceptional case for the element 1 when stating a general result. But the consensus was also that it is highly context sensitive.[/QUOTE] Fair enough, but even then the most you can really say is that the order of 1 will sometimes be considered to be p, for a prime p that is dependent on context. It hardly seems fair to extend that to saying that it is usually considered to be 2, as you did above. And yes, I understand that the context you were interested in here was certainly the case p = 2, but your statement about the "usual meaning" implies that it is context-independent. (Or at least, I certainly took it that way, even though I knew what point you were really trying to make.) [QUOTE=R.D. Silverman;217992]I also heard a similar discussion when receiving a lecture on finite fields from Birkhoff back when I was an undergrad (and yes, we used his book). I also remember a separate discussion about whether 1 could be considered as the order of a finite field with 1 element (and indeed whether such a thing existed) [usually considered the 'trivial' field; whether it is a field depends on whether one requires the convention that the multiplicative identity must be different from the additive. ]. [/QUOTE] As an aside, here's perhaps another case where I would balk at your description. I have [B]always[/B] seen fields defined as having 1 != 0. Certainly the trivial [B]ring[/B] is well-defined and useful, but fields always have separate additive and multiplicative identities. And this isn't just arbitrary, it's an important point. For example, there's a theorem that R/I is a field iff I is a maximal ideal of R (R an arbitrary ring). But that only holds if fields are required to have differing identities. [QUOTE=R.D. Silverman;217992]Certainly there are many instances when 1 is tossed into a set to keep from having it singled out as a special case in some theorem. 1 is excluded from being prime, so we [b]don't[/b] have to create a special case in the FTA. I agree that if it were not for the [b]context[/b] of the current problem, considering 1 to have order 2 would be considered abuse of notation. I agree that the discussion might have been clearer if we had just stuck to "elements that are their own inverse". But I deliberately introduced x^2 = 1, precisely because I was hoping that a discussion of zero divisors would arise as a result. I will accept the Lang comment. (Do you know it?) Lang was well known for telling people at lectures "Your argument is fine, but your notation sucks".[/QUOTE] I wasn't familiar with the Lang comment, but I get the point. In particular, in this case the question of the order of the elements wasn't really of interest. What was much more important was simply whether or not an element was its own inverse (as you say above). And if your audience here had been a bunch of people who could be expected to get that distinction, then it wouldn't much matter. But your audience was a novice who is trying to learn this material, and when asked for an example of an m for which there are an odd number of elements with order 2, he gave you one, and he was correct, according to the precise definition (the letter of the law, if not the spirit which you intended). And your response was "NO! You should know the definition of element of order 2 by now." But if he had diligently looked up that definition [B]anywhere[/B], he would have reason to think he was correct. In light of that, doesn't your rebuke of him seem a little unwarranted? All of which, I guess, is to say don't be so quick to give up on Tomer. (I know, I know; you're thinking that the 500+ posts so far make this anything but "quick". :grin:) But he's trying to learn, and it's worth recognizing that not all of his difficulties have been entirely his fault. |
You, and others here asked me to give this "numerical example",
I gave it and showed the structure I found. Further more, I generalized my claim: The number of pairs (pairs of units of order 2 that so the product of each one is -1) is 2k and the product of the whole pairs mod m is (-1)^k. I said it (on post 552) and you left it as it is. |
[quote=blob100;218001]You, and others here asked me to give this "numerical example",
I gave it and showed the structure I found. Further more, I generalized my claim: The number of pairs (pairs of units of order 2 that so the product of each one is -1) is 2k and the product of the whole pairs mod m is (-1)^k. I said it (on post 552) and you left it as it is.[/quote] It is easy to prove: a will define the units of order 2 (a1, a2, a3,..., a2k will be simillarily defined). There are k pairs of units a{a1,a2,a3,...a2k} ai and aj such that aiaj=-1(mod m). We may say that the product of the whole such pairs defined before is (-1)^k. If k is odd we may say the residue is -1 and if it is even the residue is +1. |
Tomer,
Few months ago, I thought they were new and posted some of my conjectures. In this forum, the members responded in a professional manner. So, my advice is "Give a try", who knows what you had in your mind might be a new concept. Good Luck, Sastry Karra |
[quote=spkarra;218284]Tomer,
Few months ago, I thought they were new and posted some of my conjectures. In this forum, the members responded in a professional manner. So, my advice is "Give a try", who knows what you had in your mind might be a new concept. Good Luck, Sastry Karra[/quote] Hello Karra. I can't understand what do you talk about. My conjectures were trivially false and unreadable. Tomer. |
The fifth problem (1):
V1=(x1,y1), V2=(x2,y2). C is the angle between the two vectors. A is the angle between V1 and X. B is the angle between V2 and X. cosC=cos(|A-B|)=cosAcosB+sinAsinB= (x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G. G is an easy way to show the whole phrase. We try to prove: |V1||V2|cosC=x1x2+y1y2. ((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|. cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2. |
[QUOTE=blob100;218448]The fifth problem (1):
V1=(x1,y1), V2=(x2,y2). C is the angle between the two vectors. A is the angle between V1 and X. B is the angle between V2 and X. cosC=cos(|A-B|)=cosAcosB+sinAsinB= (x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G. G is an easy way to show the whole phrase. We try to prove: |V1||V2|cosC=x1x2+y1y2. ((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|. cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2.[/QUOTE] Please make this a little more readable..... There are several ways to prove the result. This may be one of them, but your notation is all "jammed together". BTW, This is a very important result. |
[quote=R.D. Silverman;218520]Please make this a little more readable.....
There are several ways to prove the result. This may be one of them, but your notation is all "jammed together". BTW, This is a very important result.[/quote] Yes, I know this result is important, And my proof is not readable, The problem I found aiming my writing was that geometical pictures can't be drawn here, and more then it, the phrases aren't written with the popular mathematical form (as you, me and everyone would write on a sheet of paper, I mean, not as on the computer, forum). I am able to write it on a word page and send it to you by e-mail. I can try explaining what is written. Thanks Tomer. |
[quote=blob100;218448]The fifth problem (1):
V1=(x1,y1), V2=(x2,y2). C is the angle between the two vectors. A is the angle between V1 and X. B is the angle between V2 and X. cosC=cos(|A-B|)=cosAcosB+sinAsinB= (x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))=G. G is an easy way to show the whole phrase. We try to prove: |V1||V2|cosC=x1x2+y1y2. ((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|. cosC|V1||V2|=G|V1||V2|=(x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))(((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2)))=x1x2+y1y2.[/quote] |V1|=(x1,y1), |V2|=(x2,y2). C is the angle between the two vectors (you denoted it as theta). A is the angle between V1 and X. B is the angle between V2 and X. cosC=cos(|A-B|)=cosAcosB+sinAsinB= (x2/((x2^2+y2^2)^(1/2)))(x1/((x1^2+y1^2)^(1/2)))+(y2/((x2^2+y2^2)^(1/2)))(y1/((x1^2+y1^2)^(1/2)))= (x2/|V2|)(x1/|V1|)+(y1/|V1|)(y2/|V2|)=G. G is an easy way to show the whole phrase. _______________________________ We try to prove: |V1||V2|cosC=x1x2+y1y2. ((x1^2+y1^2)^(1/2))((x2^2+y2^2)^(1/2))=|V1||V2|. _______________________________ cosC|V1||V2|=G|V1||V2|=((x2/|V2|)(x1/|V1|)+(y1/|V1|)(y2/|V2|))|V1||V2|= x1x2+y1y2. |
[QUOTE=blob100;218539]Yes, I know this result is important,
And my proof is not readable, The problem I found aiming my writing was that geometical pictures can't be drawn here, and more then it, the phrases aren't written with the popular mathematical form (as you, me and everyone would write on a sheet of paper, I mean, not as on the computer, forum). I am able to write it on a word page and send it to you by e-mail. I can try explaining what is written. Thanks Tomer.[/QUOTE] | | | | /(x1, y1) | / | / .(x2,y2) | / . | / . | / . |/a . b ---------------------------------- Let L1 = length of the segment labelled (x1,y1) L2 = '' (x2, y2) Then cos(b) = x2/L2 cos(a + b) = x1/L1 What then is cos(a)? |
[QUOTE=R.D. Silverman;218543]
[CODE]| | | | /(x1, y1) | / | / .(x2,y2) | / . | / . | / . |/a . b ----------------------------------[/CODE] [/QUOTE] Looks better with code tag |
[quote=R.D. Silverman;218543]|
| | | /(x1, y1) | / | / .(x2,y2) | / . | / . | / . |/a . b ---------------------------------- Let L1 = length of the segment labelled (x1,y1) L2 = '' (x2, y2) Then cos(b) = x2/L2 cos(a + b) = x1/L1 What then is cos(a)?[/quote] I can't understand what is drawn here. I think, after some changes I gave, you will understand easily what I wrote there. |
[quote=axn;218545]Looks better with code tag[/quote]
1) cos(a)=cos(a+b)cos(b)+sin(a+b)sin(b). sin(a+b)=x1/|V1| sin(b)=x2/|V2| cos(a+b)=y1/|V1| cos(b)=y2/|V2| 2) cos(a)=(x1/|V1|)(x2/|V2|)+(y1/|V1|)(y2/|V2|). If we product cos(a) with |V1||V2| we may get by (2): cos(a)|V1||V2|=x1x2+y1y2. |
[QUOTE=blob100;218552]1) cos(a)=cos(a+b)cos(b)+sin(a+b)sin(b).
sin(a+b)=x1/|V1| sin(b)=x2/|V2| cos(a+b)=y1/|V1| cos(b)=y2/|V2| 2) cos(a)=(x1/|V1|)(x2/|V2|)+(y1/|V1|)(y2/|V2|). If we product cos(a) with |V1||V2| we may get by (2): cos(a)|V1||V2|=x1x2+y1y2.[/QUOTE] This works nicely. |
[quote=R.D. Silverman;218558]This works nicely.[/quote]
Thanks. |
[QUOTE=blob100;218552]
<snip> .[/QUOTE] How about some exercizes on polynomial algebra? Throughout these exercizes let f(x) = a_n x^n + a_{n-1} x^(n-1) + .... a_0. Suppose that its roots are x_0, x_1, .... x_n. a_i are integers. (1) Find the coefficients of a polynomial whose roots are 1/x_0, 1/x_1, ... 1/x_n. Is this polynomial unique? (2) Express the coefficient of x^k as a function of the roots. (3) Prove the remainder theorem: The remainder left when f(x) is divided by (x-r) equals f(r). (4) Prove the rational roots theorem: If alpha = a/b is a rational root of f(x) then b divides a_n and a divides a_0. (5) Prove that any polynomial of odd degree must have a real root. This is easy with Calculus. Prove it without the Intermediate Value Thm. (6) Prove that if a_n = 1, (known as a monic polynomial), then any rational root must be an integer. (7) Solve x^4 + x^3 + x^2 + x + 1 = 0. Hint: use the result from problem (1). Note that this problem can be solved without knowing how to solve quartics in general. (8) Suppose all of the coefficients of f(x) equal 1. Prove that the absolute value of all of the roots must equal 1. |
[quote=R.D. Silverman;218562]How about some exercizes on polynomial algebra?
Throughout these exercizes let f(x) = a_n x^n + a_{n-1} x^(n-1) + .... a_0. Suppose that its roots are x_0, x_1, .... x_n. a_i are integers. (1) Find the coefficients of a polynomial whose roots are 1/x_0, 1/x_1, ... 1/x_n. Is this polynomial unique? (2) Express the coefficient of x^k as a function of the roots. (3) Prove the remainder theorem: The remainder left when f(x) is divided by (x-r) equals f(r). (4) Prove the rational roots theorem: If alpha = a/b is a rational root of f(x) then b divides a_n and a divides a_0. (5) Prove that any polynomial of odd degree must have a real root. This is easy with Calculus. Prove it without the Intermediate Value Thm. (6) Prove that if a_n = 1, (known as a monic polynomial), then any rational root must be an integer. (7) Solve x^4 + x^3 + x^2 + x + 1 = 0. Hint: use the result from problem (1). Note that this problem can be solved without knowing how to solve quartics in general. (8) Suppose all of the coefficients of f(x) equal 1. Prove that the absolute value of all of the roots must equal 1.[/quote] As you said, my skills are quiet losy in this area, will you please give me a name of a book I can read on the topic? |
[QUOTE=blob100;218565]As you said, my skills are quiet losy in this area, will you please give me a name of a book I can read on the topic?[/QUOTE]
This is basic pre-calculus algebra. You should have seen this stuff in 2nd/3rd year secondary school algebra. The Schaum Outline Series should cover this. I can also recommend a superb high-school level textbook: Introductory Modern Analysis that is surely long out of print. This is the text I used. |
[QUOTE=R.D. Silverman;218571]This is basic pre-calculus algebra. You should have seen this stuff in
2nd/3rd year secondary school algebra. The Schaum Outline Series should cover this. I can also recommend a superb high-school level textbook: Introductory Modern Analysis that is surely long out of print. This is the text I used.[/QUOTE] Here it is: [url]http://www.amazon.com/Modern-Introductory-Analysis-Mary-Dolciani/dp/0395350484[/url] |
[quote=R.D. Silverman;218571]This is basic pre-calculus algebra. You should have seen this stuff in
2nd/3rd year secondary school algebra. The Schaum Outline Series should cover this. I can also recommend a superb high-school level textbook: Introductory Modern Analysis that is surely long out of print. This is the text I used.[/quote] I have seen these stuff on these years of secondary school, but it does not say these exersizes are on the same level with the secondary school's level (there are many levels for everything, you can teach a theory on a verry deep way and on the same time the opposite). I am giong to get "Undergraduate Algebra" by Serge Lang in few weeks and I took from TAU's library "A course of pure mathematics" by G.H hardy, Are these discussing the area of polynomials as I needed for these exercises? |
[quote=R.D. Silverman;218562]How about some exercizes on polynomial algebra?
Throughout these exercizes let f(x) = a_n x^n + a_{n-1} x^(n-1) + .... a_0. Suppose that its roots are x_0, x_1, .... x_n. a_i are integers. (1) Find the coefficients of a polynomial whose roots are 1/x_0, 1/x_1, ... 1/x_n. Is this polynomial unique? (2) Express the coefficient of x^k as a function of the roots. .[/quote] The first question isn't understandable. The roots 1/x_0, 1/x_1, ... can be shown as roots of the form x_i instead of 1/x_i (every number have an inverse). The second question is concerning the famous Newton's Binom. |
[quote=R.D. Silverman;218571]This is basic pre-calculus algebra. You should have seen this stuff in
2nd/3rd year secondary school algebra. The Schaum Outline Series should cover this. I can also recommend a superb high-school level textbook: Introductory Modern Analysis that is surely long out of print. This is the text I used.[/quote] I don't have the whole Schaum Outline series, I have: Advanced Infinitesimal arithmetic and complex variables. |
[quote=blob100;218575]I have seen these stuff on these years of secondary school, but it does not say these exersizes are on the same level with the secondary school's level (there are many levels for everything, you can teach a theory on a verry deep way and on the same time the opposite).
I am giong to get "Undergraduate Algebra" by Serge Lang in few weeks and I took from TAU's library "A course of pure mathematics" by G.H hardy, Are these discussing the area of polynomials as I needed for these exercises?[/quote] Stay away from Lang's books!! They are terrific [B]references[/B] if you know the material, but make horrible textbooks. Also, this book is about modern algebra: boolean algebras, monoids, groups, rings, fields. It will assume that you [b]already[/b] know the polynomial results that are given by my exercizes. I mean no offense, but you keep trying to run before you can walk. Master the secondary school stuff first, then move on. |
[QUOTE=blob100;218576]The first question isn't understandable.
The roots 1/x_0, 1/x_1, ... can be shown as roots of the form x_i instead of 1/x_i (every number have an inverse). The second question is concerning the famous Newton's Binom.[/QUOTE] Huh? The first question is clear. You are given a polynomial and its roots and asked to derive the polynomial whose roots are the reciprocals of the given roots. Why isn't this understandable????? The second question is somewhat related to the binomial theorem. |
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