# Circular numbers as opposed to integers and Gaussian integers?

#### Ragnarok

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

MHF Hall of Honor
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

• HallsofIvy and Ragnarok

#### Ragnarok

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.

#### Vlasev

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.

• Ragnarok

#### Ragnarok

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.

#### Vlasev

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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• Ragnarok

#### Ragnarok

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.

#### Vlasev

Hehe, I just ran them through a program and took a look at which numbers satisfied the relation.