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  • Brother Daniel
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Re: less-than-uncool preschool
Reply #50
We have now seen that for any m, n ∈ ℕ, at most one of the following three statements is true:
1.  m = n,
2.  m < n,
3.  m > n.

In fact, exactly one of those three statements is true.  (This is called the law of trichotomy.)

Proof:
Let A = {p ∈ ℕ | 1 ≤ p}.
Clearly A ⊆ ℕ.
1 ∈ A, trivially.
Suppose nA.  If n = 1, then n* = 1 + 1 > 1, so n* ∈ A.  On the other hand, if n > 1, then 1 + m = n for some m ∈ ℕ, so n* = (1 + m)* = 1 + m*, so again n* ∈ A.
So A = ℕ by induction.
Now let m ∈ ℕ.
Let B = {p ∈ ℕ | p < m or p = m or p > m}.
We have already shown that 1 ∈ B.
Suppose nB.
Case 1.  If n < m, then either m = n + 1 = n*, in which case n* ∈ B, or m = n + q for some q ∈ ℕ where q > 1, in which case q = 1 + r = r* for some r ∈ ℕ, so m = n + r* = (n + r)* = (r + n)* = r + n*, so n* < m, so again n* ∈ B.
Case 2.  If n = m, then n* = m + 1, so n* > m, so again n* ∈ B.
Case 3.  If n > m, then n = m + q for some q ∈ ℕ, so n* = (m + q)* = m + q*, so yet again n* > m and n* ∈ B.
In all three cases, n* ∈ B.
Thus B = ℕ by induction.
So we have n < m or n = m or n > m for all n ∈ ℕ, given our choice of m.  But m was arbitrary, so we're done.


[edited to add:  But there's probably a simple one-line proof that a smarter person would have written.]

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Re: less-than-uncool preschool
Reply #51
Another useful result:  For any m, n, p ∈ ℕ, if m > n, then mp > np.

Proof:  If m > n then there is a q ∈ ℕ such that m = n + q.
Then mp = (n + q) ∙ p = (np) + (qp).
Let r = qp.  Then we have a natural number r such that mp = (np) + r.  Done.

So we have another "cancellation law":
pm = pn implies m = n, for all m, n, p ∈ ℕ.

Proof:  Let m, n, p ∈ ℕ.
If m > n, then pm > pn, as proved above.
Similarly, if m < n, then pm < pn.
So we can have pm = pn only if m = n.  Done.

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Re: less-than-uncool preschool
Reply #52
Here's the "well-ordering principle for ℕ":

If A ⊆ ℕ and A ≠ ∅, then there is a pA such that pn for all nA.  In other words, every nonempty subset of ℕ has a "first element".

Proof:  Let A be a subset of ℕ and suppose that A does not have a first element.
Let B = {q ∈ ℕ | q < n for all nA}.
Clearly B ⊆ ℕ.
We can't have both q < n and q = n, as we have already seen.  So BA = ∅.
1 ∉ A, for otherwise 1 would be the first element of A, and A has no such thing.  So 1 < n for all nA, and thus 1 ∈ B.
Now suppose that pB.  Then p < n for all nA, so p* ≤ n for all nA.  But that means that p* ∉ A, for otherwise it would be the first element of A, and A has no such thing.  So p* < n for all nA, and thus p* ∈ B.
Thus B = ℕ by induction.
But then A ⋂ ℕ = ∅.  Since A ⊆ ℕ, it follows that A = ∅.
We got there by assuming that A ⊆ ℕ and that A has no first element.  Thus, if A ⊆ ℕ and A ≠ ∅, then A has a first element.  Done.

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Re: less-than-uncool preschool
Reply #53
Following convention, let's allow ourselves to drop the "∙" when talking about multiplication of numbers, and simply jam the arguments of the multiplication operation together.  So if m, n ∈ ℕ, then we can write mn to mean mn.

An exception arises when we have two or more explicit digits in a row.  For example, 23 does not mean 2 ∙ 3; rather, it means "twenty-three".  As noted earlier, a string of explicit digits should be treated as a single symbol.  Of course, we can always use brackets, and thus (for example) say 2(3) to mean 2 ∙ 3.

Another convention:  Let's make multiplication have a higher priority than addition, so that when we have an expression that mixes addition and multiplication, we treat the multiplications as bracketed.  Thus m + np means m + (np), not (m + n)p.

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Re: less-than-uncool preschool
Reply #54
Way back at the start of the thread, I alluded to "even" numbers.  An "even" natural number is any natural number that is equal to 2n for some n ∈ ℕ.  An "odd" natural number is any natural number that isn't "even".

1 is odd.  Because if n ∈ ℕ, then n ≥ 1 and so 2n ≥ 2 and 2 > 1.

For any n ∈ ℕ, 2n + 1 is odd.  Because if m ∈ ℕ and mn, then 2m ≤ 2n and 2n < 2n + 1, so 2m < 2n + 1.  On the other hand, if m ∈ ℕ and m > n, then we have p ∈ ℕ such that m = n + p, so 2m = 2n + 2p, and we cannot have 2n + 2p = 2n + 1, or else we would have 2p = 1 by the cancellation law.  And as we have already seen, we can't have that because 1 is odd.

Other than 1, every odd natural number is equal to 2n + 1 for some n ∈ ℕ.

Proof:  Let A = {1} ⋃ {2n | n ∈ ℕ} ⋃ {2n + 1 | n ∈ ℕ}.
Obviously 1 ∈ A.
Suppose pA.
Case 1.  If p = 1, then p* = 2 = 2 ∙ 1.
Case 2.  If p = 2n for some n ∈ ℕ, then p* = 2n + 1.
Case 3.  If p = 2n + 1 for some n ∈ ℕ, then p* = 2n + 1 + 1 = 2n + 2 = 2(n + 1) = 2(n*).
In all three cases we have p* ∈ A.  So A = ℕ by induction.
We have already shown that ({1} ⋃ {2n + 1 | n ∈ ℕ}) ⋂ {2n | n ∈ ℕ} = ∅.
It follows that ℕ \ {2n | n ∈ ℕ} = {1} ⋃ {2n + 1 | n ∈ ℕ}.
So this latter union is precisely the set of all odd natural numbers.  Done.

The product of two even numbers is even, because (2m)(2n) = 2(2mn).

The product of two odd numbers is odd.  In the non-trivial case where neither of the two odd numbers is equal to 1, this result holds because (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1.

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Re: less-than-uncool preschool
Reply #55
There's a tremendous amount of stuff we could do with the natural numbers alone.  I haven't touched on divisibility, prime numbers, GCF, LCM, or any of that jazz.  And I don't intend to, in this thread.  Instead, I want to extend our set of numbers.

Consider the equation a = b + x.  Given natural numbers a and b, we'd like to be able to solve for x.  But very often there is no solution within ℕ.  So let's extend our system of numbers in such a way that an equation of this form will always have a solution.  We need to deal with a set of numbers called the "integers".  I'll use ℤ to denote the set of integers.

By "extending" from ℕ to ℤ, I mean that we'll end up considering ℕ as a subset of ℤ, and we'll have addition and multiplication operations on ℤ that are consistent with those we've already defined for ℕ.

But I don't want to add any further assumptions.  So I will present a way to construct ℤ from ℕ.

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Re: less-than-uncool preschool
Reply #56
Consider the set ℕ × ℕ of all ordered pairs of natural numbers.  I'm going to define a relation on this set, and I'll use the "∼" notation for relatedness according to this relation.

Let m, n, p, q ∈ ℕ.  Let's say that
(m, n) ∼ (p, q)
whenever m + q = n + p.

This relation is reflexive:  (m, n) ∼ (m, n) because m + n = n + m.

This relation is symmetric:  If (m, n) ∼ (p, q), then m + q = n + p, so p + n = q + m, so (p, q) ∼ (m, n).

This relation is transitive:  If (m, n) ∼ (p, q) and (p, q) ∼ (r, s), then m + q = n + p and p + s = q + r, so m + q + p + s = n + p + q + r, so p + q + m + s = p + q + n + r, so by the law of cancellation (for addition), we have m + s = n + r, so (m, n) ∼ (r, s).

So we have an equivalence relation.

Now consider the set of equivalence classes that we can derive from this relation.

Let's treat these equivalence classes as numbers in their own right.  For a suitable choice of how to define the operations of addition and multiplication on these equivalence classes, they turn out to have all the familiar properties of the integers.  So let's go ahead and define an "integer" as one of these equivalence classes.

I'll use the notation m - n as a shorthand for the equivalence class [(m, n)].  (Peeking ahead toward stuff I haven't yet defined, we might think of this equivalence class as the "difference" resulting from subtracting n from m.)

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Re: less-than-uncool preschool
Reply #57
Let's define addition of integers as follows.

Let m, n, p, q ∈ ℕ.  Then
(m - n) + (p - q) = (m + p) - (n + q).

But wait.  We've defined an operation on equivalence classes in terms of the constituent parts of some representatives of those equivalence classes.  What if m - n = m' - n' and p - q = p' - q' for some m', n', p', q' ∈ ℕ?  Will we get the same answer if we write the same equivalence class a different way?  We'd better, or else our definition of addition is crap.

Fortunately, it works.  To say that m - n = m' - n' and p - q = p' - q' is to say that (m, n) ∼ (m', n') and (p, q) ∼ (p', q'), or in other words, m + n' = n + m' and p + q' = q + p'.

Therefore, m' + p' + n + q = m' + n + p' + q = n' + m + q' + p = n' + q' + m + p,
so (m' + p', n' + q') ∼ (m + p, n + q),
so (m' + p') - (n' + q') = (m + p) - (n + q),
so (m' - n') + (p' - q') = (m - n) + (p - q).

So our addition of integers is well-defined.

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Re: less-than-uncool preschool
Reply #58
Now let's define multiplication of integers.

Let m, n, p, q ∈ ℕ.  Then
(m - n) ∙ (p - q) = (mp + nq) - (mq + np).

Again, just as with addition, we have defined an operation on equivalence classes in terms of the constituent parts of some representatives of those equivalence classes.  So we need to check that our definition works.  Again, suppose that (m - n) = (m' - n') and (p - q) = (p' - q') for some m', n', p', q' ∈ ℕ.  So again, m + n' = n + m' and p + q' = q + p'.

Now mp + nq + m'q' + n'p' + mq' + np' = m(p + q') + n(p' + q) + m'q' + n'p'
= m(p' + q) + n(p + q') + m'q' + n'p'
= (m + n')p' + (m' + n)q' + mq + np
= (m' + n)p' + (m + n')q' + mq + np
= mq + np + m'p' + n'q' + mq' + np'.

So by the cancellation law (for addition), we have mp + nq + m'q' + n'p' = mq + np + m'p' + n'q'.
So (mp + nq, mq + np) ∼ (m'p' + n'q', m'q' + n'p'),
so (mp + nq) - (mq + np) = (m'p' + n'q') - (m'q' + n'p'),
so (m' - n') ∙ (p' - q') = (m - n) ∙ (p - q).

So our multiplication of integers is well-defined.


[edited to tidy up a bit]
  • Last Edit: July 26, 2016, 10:23:57 AM by Brother Daniel

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Re: less-than-uncool preschool
Reply #59
Let's take a closer look at these equivalence classes m - n.  We know from the law of trichotomy that either m = n or m > n or m < n.

Consider the case m = n.  The equivalence class m - m is just the set {(q, q) | q ∈ ℕ}.

Proof:  Let (a, b) ∈ m - m.
Then (a, b) ∼ (m, m), so a + m = b + m, so (by the cancellation law for addition) a = b, so (a, b) ∈ {(q, q) | q ∈ ℕ}.
Conversely, let (a, b) ∈ {(q, q) | q ∈ ℕ}.
Then a = b, so a + m = b + m, so (a, b) ∼ (m, m), so (a, b) ∈ m - m.  Done.

Recall that we can name an equivalence class by using any representative thereof.  So m - m = 1 - 1.

Now consider the case m > n.  In this case, there is a unique p ∈ ℕ such that m = n + p.  The equivalence class m - n is just the set {(q + p, q) | q ∈ ℕ}.

Proof:  Let (a, b) ∈ m - n.
Then a + n = b + m = b + p + n, so (by the cancellation law for addition) a = b + p, so (a, b) ∈ {(q + p, q) | q ∈ ℕ}.
Conversely, let (a, b) ∈ {(q + p, q) | q ∈ ℕ}.
Then a = b + p, so a + n = b + p + n = b + m, so (a, b) ∼ (m, n), so (a, b) ∈ m - n.  Done.

Given p ∈ ℕ such that m = n + p, we can write the equivalence class m - n as (1 + p) - 1.

Now consider the case m < n.  In this case, there is a unique p ∈ ℕ such that n = m + p.  The equivalence class m - n is just the set {(q, q + p) | q ∈ ℕ}.

The proof for this case is just like the proof for the previous case, with trivial changes.

Given p ∈ ℕ such that n = m + p, we can write the equivalence class m - n as 1 - (1 + p).

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Re: less-than-uncool preschool
Reply #60
So every integer falls nicely into one of these three categories:
There's 1 - 1.
For each p ∈ ℕ, there's (1 + p) - 1.  Let's call these integers "positive".
For each p ∈ ℕ, there's 1 - (1 + p).  Let's call these integers "negative".

Let's focus on the positive integers for a moment.  We can define a function z : ℕ → ℤ, where z(p) = (1 + p) - 1 for all p ∈ ℕ.  Clearly this function is an injection, and its range is equal to the set of positive integers (as defined above).

Let's see what happens if we add or multiply two positive integers.

((1 + p) - 1) + ((1 + q) - 1) = (1 + p + 1 + q) - (1 + 1) = (1 + (p + q)) - 1.
So z(p) + z(q) = z(p + q).

((1 + p) - 1) ∙ ((1 + q) - 1) = ((1 + p)(1 + q) + 1) - (1 + p + 1 + q) = (1 + p + q + pq + 1) - (1 + p + q + 1) = (1 + pq) - 1.
So z(p) ∙ z(q) = z(pq).

So the positive integers behave exactly like the natural numbers, with regard to addition and multiplication.

So let's identify the positive integers with the natural numbers.

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Re: less-than-uncool preschool
Reply #61
Now the integers can be named.

The positive integer (1 + p) - 1 will henceforth be given the same name as the natural number p.  We will have no need to distinguish (for example) between the integer 7 and the natural number 7.

The integer 1 - 1 will be given the name 0 (read "zero").  This integer, uniquely, is neither positive nor negative.

I'll get back to the negative integers soon.

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Re: less-than-uncool preschool
Reply #62
Addition and multiplication of integers have many of the properties that we already saw for addition and multiplication of natural numbers.

In particular:  Addition is associative and commutative, multiplication is associative and commutative, and multiplication distributes over addition.  These properties for addition and multiplication in ℤ follow easily from the same properties for addition and multiplication in ℕ.

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Re: less-than-uncool preschool
Reply #63
Let m, n ∈ ℕ.  Then (m - n) + (1 - 1) = (m + 1) - (n + 1) = m - n.
In other words, a + 0 = a, for all a ∈ ℤ.
So we can call 0 the "additive identity".

Notice also that (m - n) + (n - m) = (m + n) - (m + n) = 1 - 1.
So the integer n - m is the "additive inverse" of m - n.

For any a ∈ ℤ, we can write -a for the additive inverse of a.
a + -a = 0, for all a ∈ ℤ.

Clearly -0 = 0, and -(-a) = a for all a ∈ ℤ.  If a is positive then -a is negative.  If a is negative then -a is positive.

We can use this "-" notation to give the negative integers names.
So the additive inverse of 3 is called -3, for example.

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Re: less-than-uncool preschool
Reply #64
Ignoring multiplication for now:  The set ℤ, taken together with the addition operation, conforms to the definition of a "group", given earlier.  Here, addition plays the role of the group operation.  Addition is associative, as a group operation must be.  We have an (additive) identity, zero.  And for every integer, there is an (additive) inverse.

Our addition operation is also commutative.  So ℤ (with addition) is not only a group, it is a "commutative group" (also known as an "Abelian group").

I'm not going to dwell on this here, but quite a lot follows from it.  Group theory, as I indicated a while back, is a very big area of study.

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Re: less-than-uncool preschool
Reply #65
Now let's define another binary operation on ℤ, called "subtraction", and denoted with the symbol "-":
a - b = a + -b.

The result of subtracting a number from another number is called the "difference" between those numbers.

This definition of subtraction is consistent with the notation we have been using so far to denote equivalence classes in ℕ × ℕ.

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Re: less-than-uncool preschool
Reply #66
It's easy to see that we have a cancellation law for addition again:
a + b = a + c implies b = c, for all a, b, c ∈ ℤ.

Proof:  Let a, b, c ∈ ℤ.  Suppose that a + b = a + c.
Then b = (a + b) - a = (a + c) - a = c.  Done.

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Re: less-than-uncool preschool
Reply #67
Let m, n ∈ ℕ.  Then (m - n) ∙ (2 - 1) = (2m + n) - (m + 2n) = m - n.
In other words, a ∙ 1 = a, for all a ∈ ℤ.
So we can call 1 the "multiplicative identity".

Notice also that (m - n) ∙ (1 - 2) = (m + 2n) - (2m + n) = n - m = -(m - n).
So a ∙ (-1) = -a, for all a ∈ ℤ.

Also, (m - n) ∙ (1 - 1) = (m + n) - (m + n) = 1 - 1.
So a ∙ 0 = 0, for all a ∈ ℤ.

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Re: less-than-uncool preschool
Reply #68
Let p, q ∈ ℕ.

We have already seen that ((1 + p) - 1) ∙ ((1 + q) - 1) = (1 + pq) - 1.

By going through the same steps, we can see that
((1 + p) - 1) ∙ (1 - (1 + q)) = 1 - (1 + pq),
(1 - (1 + p)) ∙ (1 - (1 + q)) = (1 + pq) - 1, and
(1 - (1 + p)) ∙ ((1 + q) - 1) = 1 - (1 + pq).

But now we've covered every possible case of multiplication of two nonzero integers.  The product of two positive integers is positive, as is the product of two negative integers.  The product of a positive integer and a negative integer is negative.

In particular, notice that the product is never zero in any of those cases.  So if a product of two integers is zero, then at least one of those two integers must be zero.  This fact is sometimes called the "product law of arithmetic".

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Re: less-than-uncool preschool
Reply #69
We have a cancellation law for multiplication again, but now we have to qualify it:
ab = ac implies b = c, for all a, b, c ∈ ℤ with a ≠ 0.

Proof:  Let a, b, c ∈ ℤ.  Suppose that ab = ac and that a ≠ 0.
Then a(b - c) = ab - ac = 0.  Since a ≠ 0, we must have b - c = 0 by the product law of arithmetic.  So b = c.  Done.

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Re: less-than-uncool preschool
Reply #70
We can extend the "<" relation (and thus the ">" and "≤" and "≥" relations) to ℤ very easily:

Let a, b ∈ ℤ.
To say that a < b means that b - a is positive.

It's easy to see that this relation is transitive, and that it is consistent with the way we previously defined it for natural numbers.  The law of trichotomy is still valid, as well.  But we no longer have a well-ordering principle; ℤ does not have a "least element", and a nonempty subset of ℤ can easily fail to have a "least element".

Notice that -a - -b = b - a.  So if a < b, then -b < -a.

So ... < -3 < -2 < -1 < 0 < 1 < 2 < 3 < ...

A couple of useful things worth noticing:
If a < b, then a + c < b + c for any c ∈ ℤ.
If a < b and c > 0, then ac < bc.
I think of these properties (along with transitivity) as expressing how any "<" relation worthy of the name ought to behave.

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Re: less-than-uncool preschool
Reply #71
We can extend the idea of "even" and "odd" to integers as well.  Let a ∈ ℤ.  If a = 2b for some b ∈ ℤ, then a is said to be "even"; otherwise, it is said to be "odd".  a is odd if and only if it is equal to 2b + 1 for some b ∈ ℤ.

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Re: less-than-uncool preschool
Reply #72
A "group" was defined earlier, as a kind of "structure" (my word again).  As already noted, ℤ with addition qualifies as a group.  Considering that we also have a multiplication operation, it is perhaps not surprising that there are other kinds of "structures" that demand two binary operations, by analogy to our addition and multiplication, and that ℤ with both of these operations conforms to the definition of one of these "structures".

In particular, a "ring" is a set S, taken together with an "addition" operation and a "multiplication" operation, such that
(1) S together with the addition operation forms a commutative group;
(2) the multiplication operation is associative; and
(3) multiplication distributes over addition (both ways).

So ℤ, taken together with the addition and multiplication operations already defined, is a ring.

It's even better than that.  Since the multiplication operation is commutative, ℤ is a "commutative ring".  Since we have a multiplicative identity, ℤ is a "ring with identity".  Having the product law of arithmetic in place is also considered to be of significance.  There's probably a name for that.  When you have a ring with all three of these properties - i.e., a commutative ring with identity, for which the product law of arithmetic also holds - I think that's called an "integral domain".

Again, we're touching on a big area of study that I won't get into.

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Re: less-than-uncool preschool
Reply #73
So we have extended our study of numbers from ℕ to ℤ.  From this point on, we have no need to unpack the integers.  That is, we have no need to think of an integer as an equivalence class of ordered pairs of natural numbers.  We can simply take the properties of the integers as given.

Working in ℤ, we now will always have a solution to the equation a = b + x.  Given integers a and b, we can find x easily; the solution is x = a - b.  Yay.

But what about the equation a = bx?  Very often, there is no solution within ℤ.  It would be nice, perhaps, to extend our system of numbers in such a way that an equation of this form will always have a solution.

But wait.  That's unlikely to happen.  Consider the case of b = 0.  We know that for any x ∈ ℤ, 0x = 0.  Thus any possible value of x is a solution if a = 0, and there is no solution if a ≠ 0.  Intuitively, 0 will have that same property in any extension of ℤ.

So let's set a slightly more modest goal:  Let's extend our system of numbers so that an equation of the form a = bx will always have a solution, provided that b ≠ 0.  It often fails to have a solution in ℤ.  We need to deal with a set of numbers called the "rational numbers".  I'll use ℚ to denote the set of rational numbers.

By "extending" from ℤ to ℚ, I mean that we'll end up considering ℤ as a subset of ℚ, and we'll have addition and multiplication (and subtraction) operations on ℚ that are consistent with those we've already defined for ℤ.

But again, I don't want to add any further assumptions.  So I will present a way to construct ℚ from ℤ.  Again, I'll take the approach of considering equivalence classes of ordered pairs.

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Re: less-than-uncool preschool
Reply #74
Consider the set ℤ × (ℤ \ {0}) - that is, the set of all ordered pairs of integers, excluding those in which the second coordinate is zero.  I'm going to define a relation on this set, and again I'll use the "∼" notation for relatedness according to this relation.

Let a, b, c, d ∈ ℤ, where b ≠ 0 and d ≠ 0.  Let's say that
(a, b) ∼ (c, d)
whenever ad = bc.

This relation is reflexive:  (a, b) ∼ (a, b) because ab = ba.

This relation is symmetric:  If (a, b) ∼ (c, d), then ad = bc, so cb = da, so (c, d) ∼ (a, b).

This relation is transitive:  If (a, b) ∼ (c, d) and (c, d) ∼ (e, f), then ad = bc and cf = de, and we have two cases.  If c = 0, then bc = cf = 0, so ad = de = 0.  But d ≠ 0, so the product law of arithmetic tells us that a = e = 0, and we have af = be = 0.  If c ≠ 0, then we have adcf = bcde, so by the law of cancellation (for multiplication), taking into account that cd ≠ 0, we again have af = be.  Either way, (a, b) ∼ (e, f).

So we have an equivalence relation.

Now consider the set of equivalence classes that we can derive from this relation.

Let's treat these equivalence classes as numbers in their own right.  For a suitable choice of how to define the operations of addition and multiplication on these equivalence classes, they turn out to have all the familiar properties of the rational numbers.  So let's go ahead and define a "rational number" as one of these equivalence classes.

I'll use the notation a/b as a shorthand for the equivalence class [(a, b)].  (Peeking ahead toward stuff I haven't yet defined, we might think of this equivalence class as the "quotient" resulting from dividing a by b.)

In this notation a/b, the number before the "/" is the "numerator", while the number after the "/" is the "denominator".