Circular numbers as opposed to integers and Gaussian integers?

Apr 2010
118
6
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 \(\displaystyle Z\), which he contrasts with the regular integers \(\displaystyle Z\). In the second chapter he introduces "circular" numbers \(\displaystyle Z_n\), 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 \(\displaystyle Z_6\)?

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!
 

TheEmptySet

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The units are the elements of \(\displaystyle \mathbb{Z}_6\) 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
\(\displaystyle 1(5)=5\)
\(\displaystyle 2(5)=5=4\)
\(\displaystyle 3(5)=15=3\)
\(\displaystyle 4(5)=20=2\)
\(\displaystyle 5(5)=25=1\)
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
 
Apr 2010
118
6
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 \(\displaystyle n\) does \(\displaystyle \mathbb{Z}_n\) have zero divisors?
All non-primes.

2. What are the units in \(\displaystyle \mathbb{Z}_6\)? In \(\displaystyle \mathbb{Z}_7\)? In \(\displaystyle \mathbb{Z}_8\)?
\(\displaystyle \mathbb{Z}_6\) = 5
\(\displaystyle \mathbb{Z}_7\) = 2, 3, 4, 5, 6
\(\displaystyle \mathbb{Z}_8\) = 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 \(\displaystyle \mathbb{Z}_{\infty}\)?
\(\displaystyle \mathbb{Z}\). (I'm really not sure though.)

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

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

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

8. For which positive integers \(\displaystyle n\) less than 20 does the polynomial \(\displaystyle x^2+1\) have a root in \(\displaystyle \mathbb{Z}_n\)?
2, 5, 10, 13, 17.
 
Jul 2010
432
166
Vancouver
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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Apr 2010
118
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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, \(\displaystyle \mathbb{Z}_n\) is the set of all nonnegative integers less than n. \(\displaystyle \mathbb{Z}_{\infty}\) is then the set of all nonnegative integers less than \(\displaystyle \infty\), which includes every nonnegative integer. I don't know about the negative integers though.

Question 5: In \(\displaystyle \mathbb{Z}_3\), -1 is the same as 2. There is no integer \(\displaystyle a\) for which \(\displaystyle a^2=2\). Therefore, -1 has no square in \(\displaystyle \mathbb{Z}_3\). 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.
 
Jul 2010
432
166
Vancouver
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 \(\displaystyle \mathbb{Z}_n\). Those really are the integers 2,5,10 and 17. But those may not be all, right? x^2+1 has a root in \(\displaystyle \mathbb{Z}_n\) if \(\displaystyle x^2+1 \equiv 0 \pmod{n}\). This last statement is the same as \(\displaystyle n|x^2+1\), 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 \(\displaystyle \mathbb{Z}_n\).
 
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Apr 2010
118
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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.
 
Jul 2010
432
166
Vancouver
Hehe, I just ran them through a program and took a look at which numbers satisfied the relation.