PCDVD數位科技討論區 - [時事]國中程度考題　科大生三分之一抱蛋

引用:

作者sofaly

國小六年都跟鄰居走路上下課小孩子在途中總很多是可以做很多話可以說

我朋友:你知道上大學學的是什麼嗎?
我:ㄝ..學什麼阿?
我朋友:學1+1=2
我:不會吧怎麼可能
我朋友:(振振有辭的說)但是他們的1+1=2跟我們學的不一樣
他們是1+1為什麼等於2
我:挖對厚!對什麼1+1=2
對不起重點是....讀大學的大大請解釋1+1為什麼等於2
解開我這個小小的疑問

證明１＋１＝２

We will proceed as follows: we define

0 = {}.

In order to define "1," we must fix a set with exactly one element;

thus

1 = {0}.

Continuing in fashion, we define

2 = {0,1},

3 = {0,1,2},

4 = {0,1,2,3}, etc.

The reader should note that 0 = {}, 1 = {{}}, 2 = {{},{{}}}, etc.

Our natural numbers are constructions beginning with the empty set.

The preceding definitions can be restarted, a little more precisely,

as follows. If A is a set, we define the successor of A to be the set

A^+, given by

A^+ = A ∪ {A}.

Thus, A^+ is obtained by adjoining to A exactly one new element,

namely the element A. Now we define

0 = {},

1 = 0^+,

2 = 1^+,

3 = 2^+, etc.

現在問題來了, 有一個 set 是包括所有 natural numbers 的嗎 ? (甚至問

一個 class). 這邊先定義一個名詞, 接著在引 A9, 我們就可以造出一個 set

包括所有的 natural numbers.

A set A is called a successor set if it has the following properties:

i) {} [- A.

ii) If X [- A, then X^+ [- A.

It is clear that any successor set necessarily includes all the natural

numbers. Motivated bt this observation, we introduce the following

important axiom.

A9 (Axiom of Infinity). There exist a successor set.

As we have noted, every successor set includes all the natural numbers;

thus it would make sense to define the "set of the natural numbera" to

be the smallest successor set. Now it is easy to verify that any

intersection of successor sets is a successor set; in particular, the

intersection of all the successor sets is a successor set (it is obviously

the smallest successor set). Thus, we are led naturally to the following

definition.

6.1 Definition By the set of the natural numbers we mean the intersection

of all the successor sets. The set of the natural numbers is designated by

the symbol ω; every element of ω is called a natural number.

6.2 Theorem For each n [- ω, n^+≠0.

Proof. By definition, n^+ = n ∪ {n}; thus n [- n^+ for each natural

number n; but 0 is the empty set, hence 0 cannot be n^+ for any n.

6.3 Theorem (Mathematical Induction). Let X be a subset of ω; suppose

X has the following properties:

i) 0 [- X.

ii) If n [- X, then n^+ [- X.

Then X = ω.

Proof. Conditions (i) and (ii) imply that X is a successor set. By 6.1

ω is a subset of every successor set; thus ω 包含於 X. But X 包含於 ω;

so X = ω.

6.4 Lemma Let m and natural numbers; if m [- n^+, then m [- n or m = n.

Proof. By definition, n^+ = n ∪ {n}; thus, if m [- n^+, then m [- n

or m [- {n}; but {n} is a singleton, so m [- {n} iff m = n.

6.5 Definition A set A is called transitive if, for such

x [- A, x 包含於 A.

6.6 Lemma Every natural number is a transitive set.

Proof. Let X be the set of all the elements of ω which

are transitive sets; we will prove, using mathematical induction

(Theorem 6.3), that X = ω; it will follow that every natural

number is a transitive set.

i) 0 [- X, for if 0 were not a transitive set, this would mean

that 存在 y [- 0 such that y is not a subset of 0; but this is

absurd, since 0 = {}.

ii) Now suppose that n [- X; we will show that n^+ is a transitive

set; that is, assuming that n is a transitive set, we will show

that n^+ is a transitive set. Let m [- n^+; by 6.4 m [- n

or m = n. If m [- n, then (because n is transitive) m 包含於 n;

but n 包含於 n^+, so m 包含於 n^+. If n = m, then (because n

包含於 n^+) m 包含於 n^+; thus in either case, m 包含於 n^+, so

n^+ [- X. It folloes by 6.3 that X = ω.

6.7 Theorem Let n and m be natural numbers. If n^+ = m^+, then n = m.

Proof. Suppose n^+ = m^+; now n [- n^+, hence n [- m^+;

thus by 6.4 n [- m or n = m. By the very same argument,

m [- n or m = n. If n = m, the theorem is proved. Now

suppose n≠m; then n [- m and m [- n. Thus by 6.5 and 6.6,

n 包含於 m and m 包含於 n, hence n = m.

6.8 Recursion Theorem

Let A be a set, c a fixed element of A, and f a function from

A to A. Then there exists a unique function γ: ω -> A such

that

I. γ(0) = c, and

II. γ(n^+) = f(γ(n)), 對任意的 n [- ω.

Proof. First, we will establish the existence of γ. It should

be carefully noted that γ is a set of ordered pairs which is a

function and satisfies Conditions I and II. More specifically,

γ is a subset of ω╳A with the following four properties:

1) 對任意的 n [- ω, 存在 x [- A s.t. (n,x) [- γ.

2) If (n,x_1) [- γ and (n,x_2) [- γ, then x_1 = x_2.

3) (0,c) [- γ.

4) If (n,x) [- γ, then (n^+,f(x)) [- γ.

Properties (1) and (2) express the fact that γ is a function from

ω to A, while properties (3) and (4) are clearly equivalent to

I and II. We will now construct a graph γ with these four properties.

Let

Λ = { G | G 包含於 ω╳A and G satisfies (3) and (4) };

Λ is nonempty, because ω╳A [- Λ. It is easy to see that any

intersection of elements of Λ is an element of Λ; in particular,

γ = ∩ G

G[-Λ

is an element of Λ. We proceed to show that γ is the function

we require.

By construction, γ satisfies (3) and (4), so it remains only to

show that (1) and (2) hold.

1) It will be shown by induction that domγ = ω, which clearly

implies (1). By (3), (0,c) [- γ; now suppose n [- domγ. Then

存在 x [- A 使得 (n,x) [-γ; by (4), then, (n^+,f(x)) [- γ,

so n^+ [- domγ. Thus, by Theorem 6.3 domγ = ω.

2) Let

N = { n [- ω | (n,x) [- γ for no more than one x [- A }.

It will be shown by induction that N = ω. To prove that 0 [- N,

we first assume the contrary; that is, we assume that (0,c) [- γ

and (0,d) [- γ where c≠d. Let γ^* = γ - {(0,d)}; certainly

γ^* satisfies (3); to show that γ^* satisfies (4), suppose that

(n,x) [- γ^*. Then (n,x) [- γ, so (n^+,f(x)) [- γ; but n^+≠0

(Theorem 6.2), so (n^+,f(x))≠(0,d), and consequently (n^+,f(x)) [-

γ^*. We conclude that γ^* satisfies (4), so γ^* [- Λ; but γ is

the intersection of all elements of Λ, so γ 包含於 γ^*. This is

impossible, hence 0 [- N. Next, we assume that n [- N and prove

that n^+ [- N. To do so, we first assume the contrary -- that is,

we suppose that (n,x) [- γ, (n^+,f(x)) [- γ, and (n^+,u) [- γ

where u≠f(x). Let γ^。 = γ - {(n^+,u)}; γ^。 satisfies (3) because

(n^+,u)≠(0,c) (indeed, n^+≠0 by Theorem 6.2). To show that γ^。

satisfies (4), suppose (m,v) [- γ^。; then (m,v) [- γ, so

(m^+,f(v)) [- γ. Now we consider two cases, according as

(a) m^+≠n^+ or (b) m^+ = n^+.

a) m^+≠n^+. Then (m^+,f(v))≠(n^+,u), so (m^+,f(v)) [- γ^。.

b) m^+ = n^+. Then m = n by 6.7, so (m,v) = (n,v); but n [- N,

so (n,x) [- γ for no more than one x [- A; it follows that v = x,

and so

(m^+,f(v)) = (n^+,f(x)) [- γ^。.

Thus, in either case (a) or (b), (m^+,f(v)) [- γ^。, thus, γ^。

satisfies Condition (4), so γ^。[- Λ. But γ is the intersection

of all the elements of Λ, so γ 包含於 γ^。; this is impossible,

so we conclude that n^+ [- N. Thus N = ω.

Finally, we will prove that γ is unique. Let γ and γ' be functions,

from ω to A which satisfy I and II. We will prove by induction that

γ = γ'. Let

M = { n [- ω | γ(n) = γ'(n) }.

Now γ(0) = c = γ'(0), so 0 [- M; next, suppose that n [- M. Then

γ(n^+) = f(γ(n)) = f(γ'(n)) = γ'(n^+),

hence n^+ [- M.

If m is a natural number, the recurion theorem guarantees the

existence of a unique function γ_m: ω -> ω defined by the

two Conditions

I. γ_m(0)=m,

II. γ_m(n^+) = [γ_m(n)]^+, 對任意的 n [- ω.

Addition of natural numbers is now defined as follows:

m + n = γ_m(n) for all m, n [- ω.

6.10 m + 0 = m,

m + n^+ = (m + n)^+.

6.11 Lemma n^+ = 1 + n, where 1 is defined to be 0^+

Proof. This can be proven by induction on n. If n = 0,

then we have

0^+ = 1 = 1 + 0

0^+ = 1 = 1 + 0

(this last equality follows from 6.10), hence the lemma holds

for n = 0. Now, assuming the lemma is true for n, let us show

that it holds for n^+:

1 + n^+ = (1 + n)^+ by 6.10

= (n^+)^+ by the hypothesis of induction.

把 n = 1 並且注意 2 = 1^+, 故 1 + 1 = 2.