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# INTEGER AND RATIONAL NUMBERS

Theorem 7.10 Both the integers and the rational numbers are countable.

Proof By Theorem 4.4 we know that N × N is countable. there exists the
natural embedding of Z into N × N by identifying [a, b] with its canonical
representative (a-b, 0) or (0, b-a). Thus there exists a bijection from Z to
a subset of a countable set, hence Z is countable. Again by Theorem 4.4 we
know that Z × Z is countable, and there exists the natural embedding of Q
into Z×Z by identifying [a, b] with its canonical representative, (c, d) where
d is positive and minimal . Thus there exists a bijection from Q to a subset
of a countable set, hence Q is countable.

Let Z* represent the image of the embedding of Z into Q. Also let and represent respectively the images of the embeddings of the positive
and negative integers into the rationals .

The Archimedian Property

Theorem 7.11 Archimedian Property such that r < n.

Proof Let r = [a, b], if r ≤ 1, then r < 2 and we are done. If r > 1, then
without loss of generality, both a, and b are positive integers , and

b(a + 1) ≥ a + 1 > a
=> [a, b] < [a + 1, 1] ∈ .

Lemma For positive rational numbers r and s, if r > s, then r -1 < s-1.

Proof First we note that if r = [a, b], then r -1 = [b, a], this is immediate
by computing [a, b] · [b, a] = [ab, ab] = [1, 1]. Now let r = [a, b] and s = [c, d]. For any integer, a, the product of a with itself b times, where b is a
positive integer is denoted ab.

Exercise Prove that for any positive integer, n, there exists an integer of
the form 2m for some positive integer m, such that 2m > n. (Hint: use
induction).

Solution For n = 1, 1 < 2 = 21. Now assume n < 2m for some m, then Exercise Prove that for any positive rational number, q, there exists a
rational number of the form 2n, where 2n > q, and where n is the embedded
image of a positive integer.

Solution Let q = (a, b) where a and b are positive integers. We then have
(a, b) ≤ (a, 1) < (2n, 1) for some n.

The Division Algorithm

Theorem 7.12 The division algorithm If a and d are integers with d > 0,
then there exist unique integers q and r such that a = dq +r and 0 ≤ r < d.

Proof This result is a consequence of the well ordering property.
Let , and let , S' is thus
the embedded image of some subset of N, and thus if it is non-empty it must
have a least element .

If a ≥ 0, then let n = 0, and thus x = a - 0 = a ≥ 0. Thus a ∈ S'. If
a < 0, then let n = a, thus x = a-ad = a(1-d) ≥ 0. Thus a-ad ∈ S'. Thus
S' ≠ . Since S' ≠ , and is embedded image of a subset of N, S' has a least
element. Let the least element be r. Thus we have r = a - dq ≤ s and a = dq + r where r ≥ 0.

We now must show r < d. We have a - d(q + 1) = a - dq - d = r - d,
thus r - d ∈ S. Since r is the least element in S' and r - d < r we have
r - d < 0 => r < d. We thus have a = dq + r with 0 ≤ r < d.

Now we have to show that q and r are unique. Suppose and where and . Without loss of generality
we may assume . We thus have . We note that . Thus is a multiple of d
and non
-negative. We thus have and  . Thus ,
and thus .

Exercises

1. Verify that the relation (a, b) ≡ (c, d) a+d = b+c is an equivalence
relation on N × N, but not on α × α where α > ω.

2. Verify that the relation (x, y) ≡ (z,w) xw = yz is an equivalence
relation on Z × (Z - { 0 })

For any Integral Domain D show:

3. and where e is the additive identity .

4. u · (-u) = -u, where u is the multiplicative identity .

5. (-u) · (-u) = u, where u is the multiplicative identity .

6. If a, z ∈ D and a + z = a, then show z = e.

Solution to 3: a = a · u = a · (u + e) = a + a · e => a · e = e:

7. If v, a ∈ F, where F is a field and a · v = a, then show v = u.

8. Show that every Field is an Integral Domain.

Mathematical Induction is often used to prove certain identities. Exer-
cise 9 and its solution exemplifies how this is done. Exercise 10 is left
as practice.
9. Show that Solution to 9: Let i. thus 1 ∈ A.

ii. Assume m ∈ A, then Thus m + 1 ∈ A. Therefore by Mathematical Induction A = N,
and 10. Show that :

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