1
$\begingroup$

I have the following $9$ number systems at hand and I am to determine which of them possess a particular property. I am having trouble understanding some of the subtleties between the questions and could use some verification/guidance. The goal is to use the information determined in parts $1$-$4$ to determine which number systems are fields. I should be able to make those determinations if someone can clear up my confusion on what's below.

$1) \mathbb N$, the natural numbers. For which we shall include $0$. $\{0,1,2,...\}$
$2) \mathbb Z$, the integers.
$3) \mathbb Q$, the rational numbers.
$4) \mathbb R$, the real numbers.
$5) \mathbb C$, the complex numbers.
$6) \{1, \zeta_{n}, \zeta_{n}^2,...,\zeta_{n}^{n-1}\}$ where $n$ is a positive integer. Note: $\zeta_n= \cos \frac{2\pi}{n}+i\sin\frac{2\pi}{n}$
$7) \{z\in \mathbb C$ such that $|z|=1\}$. This is the set of all complex numbers whose modulus is $1$. The modulus of a complex number, $a+bi$, is $|a+bi|=\sqrt{a^2+b^2}$
$8) \mathbb Z_n=\{[0]_n,[1]_n,...,[n-1]_n\}$, where $n$ is a positive integer. Where $[1]_n= 1\pmod n$.
$9) \mathbb H$, the Quaternions.

Part $1$: First I need to determine which of the $9$ number systems aren't closed under addition or multiplication.

I determined all are closed under multiplication and all are closed under addition except for $6)$ and $7)$. I used counterexamples to show this.

Part $2$: Next, I needed to identify the identity (additive or multiplicative) that goes with each operation.

Is this a trick question? Wouldn't it be the same as the first question? All of them have a multiplicative identity and all of them but $6)$ and $7)$ have additive identities?

Part $3$: For which of the operations is it true that every element has an inverse for $(1$-$9)$?

Since $0$ will never have a multiplicative inverse, none of $(1$-$9)$ will have have multiplicative inverses, correct?

The follow up to this is, determine when every nonzero element has a multiplicative inverse for $(1$-$9)$.

This part I am a bit confused on. How can we show when every element will have one? I know that for $\mathbb Z_n$ has a multiplicative inverse if and only if $n$ is prime. But, I can't seem to figure out how to apply that.

Part 4: Which of the operations in $(1$-$9)$ are not commutative? An operation $\times$ on a set $S$ is commutative if, given any $x,y\in S, x \times y=y \times x$.

$8) \mathbb H$, the Quaternions are certainly not commutative, are there any others? Systems $1$-$6$ and $8$ are very much commutative, right? $6)$ and $7)$ I believe are commutative but I am not certain.

This would leave me with $\mathbb Q, \mathbb R, \mathbb C,$ and $\mathbb Z_n$ (when $n$ is prime) as fields.

Any input would be great.

$\endgroup$

1 Answer 1

0
$\begingroup$

For multiplicative inverses in a ring/field, we only look at the elements of the ring/field that are NOT the 0 element (The additive identity). You probably are being asked in the first inverse question to show whether they have additive inverses though, because that is a necessary quality for a field to have. Of your sets which are closed under addition, all but $\mathbb N$ are.

If you want to show every nonzero has a multiplicative inverse, you need to take an arbitrary element of the set and show it's inverse. For $\mathbb Z _n$, when n is prime, you use Bezout's identity that if two integers a,b are relatively prime, then there esists two integers s,t such that as+bt=1. Use [a] as your element, and b=n. That should do it, but I'm a bit rusty on that part.

For commutativity, most of your sets are subsets of $\mathbb R$ or $\mathbb C$, so they immediately inherit commutativity and associativity from the parent set. You can quickly show that $\mathbb H$ isn't commutative from the fact that $ij\ne ji$ .

You are correct on your 4 fields

$\endgroup$
1
  • $\begingroup$ $\mathbb N$ is closed under addition. $\endgroup$
    – Vincent
    Sep 17, 2014 at 5:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .