Number Systems - Proof

Question

Can someone explain the second part of this proof?

This is literally the very first example of the introduction to number systems in my textbook, and I'm failing to grasp it.

In the introduction chapter the definition of a number system is given as: “A set of numbers closed under operations of addition and multiplication”

The set of rational numbers is defined as: $$\mathbb{Q} = \left\{ \frac{k}{n} : k \in \mathbb{Z} , n \in \mathbb{N} \right\}$$

Following this I've assumed that I needed to proof that, $ k \in \mathbb{N}$ which it does since $\mathbb{N}$ is a subset of $\mathbb{Z}$. But I don't understand the statement “$k$ or $-k$ is in $\mathbb{N}$”. I mean, I understand its truth value, but how does it fit into the definition?

It would be great if someone could give me a step-by-step breakdown of the proof.

Example 1.1.1. Assuming that $\mathbb{N}$ and $\mathbb{Z}$ are number systems, prove that $\mathbb{Q}$ is a number system. Show, also, that if $q \in \mathbb{Q}$ with $q \neq 0$ then $1/q \in \mathbb{Q}$.

Solution. Let $p, q \in \mathbb{Q}$. Then $p = h/m$ and $q = k/n$ where $h,k \in \mathbb{Z}$ and $m,n \in \mathbb{N}$. So $$ p + q = \frac{nh + mk}{mn} \quad\text{and}\quad pq = \frac{hk}{mn}. $$ Since $\mathbb{N}$ and $\mathbb{Z}$ are closed under addition and multiplication and $\mathbb{N} \subseteq \mathbb{Z}$ we have $nh + mk \in \mathbb{Z}$, $hk \in \mathbb{Z}$ and $mn \in \mathbb{N}$. Hence $p + q \in \mathbb{Q}$ and $pq \in \mathbb{Q}$ as required.

Now let $q \in \mathbb{Q}$ with $q \neq 0$. Then $q = k/n$ where $k \in \mathbb{Z}$, $k \neq 0$ and $n \in \mathbb{N}$. Hence $1/q = n/k = (-n)/(-k) \in \mathbb{Q}$ since $n$ and $-n$ are in $\mathbb{Z}$ and $k$ or $-k$ is in $\mathbb{N}$.

(Original image here.)

Answer 1

The author is being succinct (maybe too much so). The definition of $\mathbb{Q}$ without symbols is that it's the set of all numbers which can be written as the quotient of an integer by a natural number.

We are given a nonzero $q \in \mathbb{Q}$, and we want to show $1/q \in \mathbb{Q}$. Since $q \in \mathbb{Q}$, we can write $q=n/k$ with $n\in\mathbb{Z}$, $n\neq 0$, and $k\in\mathbb{N}$. Then $1/q = k/n$. If $n>0$, then $n\in\mathbb{N}$, and so we have $1/q$ expressed as the quotient of an integer by a natural. If $n <0$, write $1/q$ as $(-k)/(-n)$ instead. Then $-n >0$, so $-n\in\mathbb{N}$, and again we can represent $1/q$ as the quotient of an integer by a natural. Either way, $1/q$ is seen to be an element of $\mathbb{Q}$.

Answer 2

We have $\frac 1q = \frac nk$ and we need to prove $\frac 1q =\frac nk \in \mathbb Q$

We know that $k \in \mathbb Z$ and that $k \ne 0$.

CASE 1: $k > 0$. Then $k \in \mathbb N$ and $n \in \mathbb N \subset \mathbb Z$ so by definition we have $\frac 1q = \frac kn \in \mathbb Q$.

That was easy.

CASE 2: $k < 0$. Well, in that case we are screwed.

We can't say $n \in \mathbb Z$ and $k \in \mathbb N$ because $k$ is NOT in $\mathbb N$. We can not conclude that $\frac 1q =\frac nk \in \mathbb Q$. The definition requires that $\frac 1q = \frac ab$ for some $a \in \mathbb Z$ and some $b \in \mathbb N$ and we just can't do that if our denominator is $k < 0$.

We must show a case where the denominator is natural.

Well, that's easy. If $k < 0$ then $-k > 0$ and so $-k$ will be a natural number.

So we need to express $\frac 1q = \frac nk = \frac {?????}{-k}$ where $-k \in \mathbb N$ and $????\in \mathbb Z$.

Well, that's easy. $\frac 1q = \frac nk = \frac {-n}{-k}$ and $-n \in \mathbb Z$ and $-k \in \mathbb N$. So that means $\frac 1q= \frac {-n}{-k}\in \mathbb Q$.

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It might have been easy/less tedious if we had done.

Lemma: By definition $\mathbb Q := \{\frac kn|k\in \mathbb Z; n \in \mathbb N\}$. The lemma proposes that $ \{\frac kn|k\in \mathbb Z; n \in \mathbb N\} = \{\frac kj|j, k\in \mathbb Z; j \ne 0\}$.

Pf: just the same stuff we did above. But lets make it more succinct.

If $x = \frac kj; j\in \mathbb Z; j \ne 0$ then $\frac kj = \frac {-k}{-j}$. Both $k, -k \in \mathbb Z$ and one or the other of $k$ or $-k$ is natural so $\frac kj = \frac {-k}{-j}$ will be an element of both sets. So the sets are equal.

(Does the language and usage make more sense now?)

Then we can simply claim: $\frac 1q = \frac nk$ and $n\in \mathbb N\subset \mathbb Z$ and $k \in \mathbb Z; k \ne 0$ so $\frac 1q \in \{\frac kn|k\in \mathbb Z; n \in \mathbb N\}=\{\frac kn|k\in \mathbb Z; n \in \mathbb N\}=\mathbb Q$.