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Math Help - Circular numbers as opposed to integers and Gaussian integers?

  1. #1
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    Circular numbers as opposed to integers and Gaussian integers?

    Hi, I have almost no experience with abstract algebra (just a bit with matrices), and to start I checked out the book Numbers and Symmetry by Bernard L. Johnston. The first section introduces the Gaussian integers Z[i], which he contrasts with the regular integers Z. In the second chapter he introduces "circular" numbers Z_n, which he explains as being the number line wrapped around a circle, basically. I'm familiar with basic modular arithmetic so I got that bit, but then he starts talking about zero divisors and units ("a number in a number system is a unit if every number in the system is a multiple of it"). The section is pretty brief and I didn't quite get a grip on it, but I can't find anything about it on the Internet. When I search for "circular numbers" I get a different definition involving squares. Is there another word for this type of thing?

    For example, one of the questions is:

    What are the units in Z_6?

    I couldn't find any units, but I am not sure if that is right. Could someone perhaps point me in a direction where I could read alternate explanations? That usually helps me greatly.

    Thanks!
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    The units are the elements of \mathbb{Z}_6 that have multiplicative inverses. The units of this ring are those that are relatively prime to 6. So 1 and 5 are the only units and they are their own inverses mod 6. What he means by "every number is a multiple of it" is that 5 is a generator of this group under addition.
    e.g
    1(5)=5
    2(5)=5=4
    3(5)=15=3
    4(5)=20=2
    5(5)=25=1
    Then the pattern repeats.

    I hope this helps. You may want to try the integers modulo n, or the multiplicative group of integers modulo n
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    Thank you so much for the explanation. I understand it better now.

    It's still hard to find information online though, so would anyone mind checking a few of my answers to the problem set?


    1. For which integers n does \mathbb{Z}_n have zero divisors?
    All non-primes.

    2. What are the units in \mathbb{Z}_6? In \mathbb{Z}_7? In \mathbb{Z}_8?
    \mathbb{Z}_6 = 5
    \mathbb{Z}_7 = 2, 3, 4, 5, 6
    \mathbb{Z}_8 = 3, 5, 7

    3. Can a unit be a zero divisor?
    No. (I think this is true, but I have no idea why.)

    4. What is \mathbb{Z}_{\infty}?
    \mathbb{Z}. (I'm really not sure though.)

    5. Does -1 have a square root in \mathbb{Z}_3?
    No.

    6. Does the polynomial x^2+1 have a root in \mathbb{Z}_5?
    Yes; it is 2.

    7. Does the polynomial x^2+1 have a root in \mathbb{Z}_{12}?
    No.

    8. For which positive integers n less than 20 does the polynomial x^2+1 have a root in \mathbb{Z}_n?
    2, 5, 10, 13, 17.
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    Question 1: Correct. Do you know why?

    Question 2: Almost correct. You are forgetting something in each line and it's important. Otherwise the units stated are good.

    Question 3: Correct. If you assume that a certain unit is a zero divisor, you may be able to reach a contradiction pretty quickly. You should check your definitions for both to see what you can come up with.

    Question 4: Correct. Can you think of a reason why, though. It has to do with the infinity.

    Q5: Correct, but can you explain why?

    Q6 and Q7: True.
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    Thank you so much for going to the trouble to check these! I am working on my own and I really, really appreciate it.


    Question 1: Well, to be a zero divisor a number has to multiply to zero - which is the same thing as n. Obviously prime numbers will not have integer divisors, so that rules them out. But any non-prime value for n will by definition have an integer divisor, and thus will have a zero divisor since n = 0.

    Question 2: Ah, I am forgetting 1. I wasn't sure if that "counted" or not.

    Question 3: Okay, I see. A unit is a number that, for one thing, multiplies to 1. So if x is a unit, then ax = 1. A zero divisor is a nonzero number that multiplies to 0. If x is also a zero divisor, then we have bx = 0. B cannot equal 0, so that must mean that x = 0. But that would make ax = 0, which is false. So there is a contradiction, and a number cannot be both a unit and a zero divisor. (I hope this is right.)

    Question 4: Well, \mathbb{Z}_n is the set of all nonnegative integers less than n. \mathbb{Z}_{\infty} is then the set of all nonnegative integers less than \infty, which includes every nonnegative integer. I don't know about the negative integers though.

    Question 5: In \mathbb{Z}_3, -1 is the same as 2. There is no integer a for which a^2=2. Therefore, -1 has no square in \mathbb{Z}_3. If I am thinking correctly, then the only values of n for which -1 would have a square root are those equal to perfect squares plus one (2, 5, 10, 17, etc.). I think this calls into question my answer to #8 above.
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    Question 3: you got the idea!

    Question 4: I think you are right. The non-negative integers will definitely be in the set. However, I'm not sure about the negatives either. Maybe someone else can help with this.

    Question 5: That's right!

    Now for question 8, here's what I think. If x^2+1 is indeed equal to n, then you definitely have a root in \mathbb{Z}_n. Those really are the integers 2,5,10 and 17. But those may not be all, right? x^2+1 has a root in \mathbb{Z}_n if x^2+1 \equiv 0 \pmod{n}. This last statement is the same as  n|x^2+1, that is n divides x^2+1.

    An exhaustive calculation shows that, indeed, 2,5,10,13 and 17 are the only n for which x^2+1 has a root in [LaTeX ERROR: Convert failed] .
    Last edited by Vlasev; August 2nd 2010 at 04:52 PM.
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    Thank you again! I see what you mean about question 8. (I'm sorry you had to go through each calculation! I did it mostly by rote but I assumed there would be a quicker way.)

    I really like this algebra. I may be posting more questions in the future. If anyone has any other comments I'd love to hear them as well.
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    Hehe, I just ran them through a program and took a look at which numbers satisfied the relation.
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