The Rational Numbers
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1
2
3The rational numbers
The equation 2 x = 1 has no solution within the integers. (Why not?) One therefore
extends the number system Z Z one more time to allow for “fractions”. The resulting
number system is that of the rational numbers I Q. This note shows how this is done.
Theorem 1
Consider the relation R on the set Z Z
(Z Z
0 ) = (p, q) p, q
Z Z and q = 0
defined by (p
, q
)R(p
, q
) if and only if p
q
= p
q
. Then R is an equivalence
1
1
2
2
1
2
2
1
relation.
Exercises
1. Prove Theorem 1.
We denote the set of equivalence classes of R by I Q. The equivalence class [(p, q)] of
(p, q) will be denoted by p/q and called a rational number.
Be careful not to interpret the notation p/q as a division of integers until you have
completed the very last exercise of this note. At this point, the division of integers
does not exist. That is exactly why we introduce the rational numbers.
For example, 2/3 = [(2, 3)] = [( 10, 15)] =
10/
15.
2. Why do we not allow for rational numbers of the form p/0 ?
[Hint: examine your proof of Theorem 1 to answer this question. Show that the
relation R as defined above is not an equivalence relation on the set Z Z
Z Z.]
Definition
On I Q we define
(1) [(p
, q
)] + [(p
, q
)] = [(p
q
+ p
q
, q
q
)]
1
1
2
2
1
2
2
1
1
2
(2) [(p
, q
)] [(p
, q
)] = [(p
p
, q
q
)]
1
1
2
2
1
2
1
2
p
q
< p
q
in case
0 < q
q
1
2
2
1
1
2
(3) [(p
, q
)] < [(p
, q
)] if and only if
1
1
2
2
p
q
< p
q
in case
q
q
< 0
2
1
1
2
1
2
(4)
[(p, q)] = [( p, q)]
1
(5) [(p, q)]
= [(q, p)] if p = 0
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