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How can we form an idea of the result of the operation of dividing a by b with a and b integers and b not equal to zero?

Peter
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    can you be more precise? – user42912 Feb 01 '14 at 22:57
  • "You have $a$ apples and $b$ persons. How many apples does each person get?" – Arthur Feb 01 '14 at 22:58
  • if the quantity of apples for each person isn't a whole number of apples then how do i answer "how many apples?"? – Peter Feb 02 '14 at 00:24
  • (1) Three apples between two people involves cutting one in half. The two people get one and a half apples each. (2) Three people get two thirds of an apple each from two apples. Cut both apples into thirds to get six pieces and give two pieces to each person. Rational numbers can be used to describe situations that integers cannot. – Geoff Pointer Feb 02 '14 at 20:47

3 Answers3

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Let $(a,b),(c,d)$ be two ordered pairs with $a,b,c,d\in\mathbf{Z}$ and $bd\neq 0$. Define the equivalence relation $(a,b)\sim (c,d)$ if $ad=bc$. Then, we can define the rationals $\mathbf{Q}$ to be the set of equivalence classes with $+$ and $\times$ such that $$\begin{align*}(a,b)+(c,d) &=(ad+bc,bd) \\ (a,b)\times (c,d) &= (ac,bd).\end{align*}$$ Now just show this construction satisfies field axioms. Hope this answers your question.

tc1729
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What are the rational numbers? The first thing you need to get rid of is your conception that a rational number $\frac{a}{b}$ means a divided by b. While this is true in practice, that's not how we define the rational numbers.

Take the set of integers, $\mathbb{Z}$. Take the set of nonzero integers, $\mathbb{Z}\setminus\{0\}$.

Now take the cartesian product of these sets: $\mathbb{Z} \times \mathbb{Z}\setminus\{0\}$, which is the set of all ordered pairs $(m,n)$ such that $m,n \in \mathbb{Z}, n \neq 0$. We may also write $\mathbb{Z} \times \mathbb{Z}\setminus\{0\} = \{(m,n) \mid m,n \in \mathbb{Z}, n \neq 0 \}$.

Then you write that two pairs $(m_1,n_1),(m_2,n_2)$ are equivalent, denoted $(m_1,n_1)\sim(m_2,n_2)$, iff $m_1n_2 - m_2n_1 = 0$. This relation $\sim$ is an equivalence relation, and allows us to say that fractions like $\frac12$ and $\frac24$ are equivalent.

Then $\mathbb{Q}$, the set of rational numbers, is the set of all equivalence classes given by $\sim$ on the set $\mathbb{Z} \times \mathbb{Z}\setminus\{0\}$. So, formally, if you denote the class of fractions equivalent to $\frac12$ by $C\left(\frac12\right)$, you can think of $\mathbb{Q}$ like this:

$$\mathbb{Q} = \{C\left(\frac12\right),C\left(\frac13\right),\ldots,C\left(\frac57\right),\ldots,C\left(\frac{11}{19}\right)\} $$

Where you note that every equivalence class of fractions is represented by its irreducible form (when the denominator and numerator are coprime).

Newb
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  • I've skipped defining the rationals as a field (show that they satisfy the field axioms), and their being ordered (refer to the integers and their axioms of order), because I was only interested in defining the rationals as a set of numbers (and I'm also a little lazy. You can easily find proofs that $\mathbb{Q}$ is an ordered field on M.SE or the internet, generally.) – Newb Feb 01 '14 at 23:06
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Not sure if I understand your question right, but you can just take an axiomatic approach: First of all rational numbers are the set

$\mathbb Q = \{ \frac a b : a \in \mathbb Z, b \in \mathbb N \}$

and we define the two operations $+$ and $\cdot$ by

$+ : \mathbb Q \times \mathbb Q \to \mathbb Q: (\frac a b, \frac c d) \mapsto \frac {ad + bc}{bd} $

$\cdot : \mathbb Q \times \mathbb Q \to \mathbb Q: (\frac a b, \frac c d) \mapsto \frac {ac}{bd} $

Having defined this just by axioms one can prove that $(\mathbb Q, +, \cdot)$ is a field any many other properties. More over this definition is compatible with addition and multiplication in $\mathbb Z$.

Or did you want some "deeper" insight into the nature of rational numbers?

Thekwasti
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