Mathematics in English

Tetsuro Kitahara

August 9, 2001

1 集合論

直積と関係

If S and T are non empty sets, then we define the cartesian product of S and T to be the set of all ordered pairs of the form (s, t), where s ∈ S and t ∈ T. We denote the cartesian product of S and T by S × T. Any nonempty subset of S × T is called a relation between S and T. We will denote relations between S and T by capital Greek letters. EXAMPLE: If S and T are each equal to the set of real numbers then Φ = {(s, t) | s = t + 7} and Ψ = {(s, t) | s^{2} + t^{2} = 1} are both relations between S and T.

同値関係

An equivalence relation on a nonempty set S is a relation Φ satisfying the following three conditions:

(1) (reflexivity) If s ∈ S then (s, s) ∈ Φ;

(2) (symmetry) If (s, s′) ∈ Φ then (s′, s) ∈ Φ;

(3) (transitivity) If (s, s′) ∈ Φ and (s′, s′′) ∈ Φ then (s, s′′) ∈ Φ.

2 整数論

ペアノの公理

Giuseppe Peano formulated a set of five postulates. These postulates are the following:

(I) 1 is a natural number.

(II) To every natural number n there is assigned a natural number S(n), called the successor of n.

(III) If n ≠ m then S(n) ≠ S(m).

(IV) There is no natural number n satisfying S(n) = 1.

(V) If U is a set of natural numbers containing 1 and having the property that if n ∈ U then S(n) ∈ U then U equals the set of all natural numbers.

ペアノの公理の別表現

If P is a property of the natural numbers such that:

(i) 1 has property P;

(ii) If n has property P then so does n + 1;

then all natural numbers must have property P.

数学的帰納法

THEOREM: If n is a natural number then n^{2} − n is an even number. PROOF: If n = 1 then 1^{2} − 1 = 0, and this is an even number. Now assume that n is a natural number having the property that n^{2}−n is even. Then (n+1)^{2}−(n+1) = (n^{2} + 2n + 1) − (n + 1) = n^{2} + n = (n^{2} − n) + 2n. But n^{2} − n is an even number by assumption, and 2n is also an even number. Since the sum of two even numbers is again even, this prove that (n + 1)^{2} − (n + 1) is also an even number. Thus we have proven the step of the induction and hence the theorem.

3 代数学

2 項演算と代数構造

If S is a nonempty set then a function from S × S to S is called a binary operation on S. A nonempty set together with at least one binary operation satisfying certain conditions is called an algeberaic structure.

半群

A set S with a single binary operation ∗ defined on it is called a semigroup if and only if this operation is associative, i.e., if and only if a ∗ (b ∗ c) = (a ∗ b) ∗ c for any three elements a, b and c of S. EXAMPLE: The set Z of integers, together with the operation of addition, forms a semigroup. Ths same set, together with the operation of subtraction, does not form a semigroup.

モノイドと単位元

A semigroup (S, ∗) is called a monoid if and only if there exists an element e of S satisfying the condition that a ∗ e = e ∗ a = a for all elements a of S. Such an element is called an identity of the monoid.

単位元の一意性

Any semigroup has at most one identity. Indeed, if e and f are both identities of a semigroup (S, ∗), then e = e ∗ f = f. Therefore, any monoid has precisely one identity.

自己準同形

A function α from a monoid (S, ∗) to itself is said to be an endomorphism of the monoid if and only if the following two conditions are satisfied:

(1) α(a) ∗ α(b) = α(a ∗ b) for all a, b ∈ S;

(2) α(e) = e.

4 組合せ

階乗と組合せ

If n is a positive integer, we define the number n! (read “n factorial”) to be the product of all positive integer from 1 to n. By convention, we set 0! equal to 1. If k ≤ n are nonnegative integers, we define the number

(\begin{matrix}n \\ k \end{matrix}) (read “n choose k”) to equal \frac{n!}{k!(n − k)!}.

2 項定理

THEOREM: If n is any positive integer then

(x + y)^{n} = ∑_{k=0}^{n} (\begin{matrix}n \\ k \end{matrix}) x^{n−k}y^{k}.

PROOF: We will prove this theorem by induction on n. For n = 1, we have

x + y = (\begin{matrix}1 \\ 0 \end{matrix}) x + (\begin{matrix}1 \\ 1 \end{matrix}) y,

as desired. Therefore assume that n > 1 and that we have already established the equality

(x + y)^{n−1} = ∑_{k=0}^{n−1} (\begin{matrix}n − 1 \\ k \end{matrix}) x^{n−1−k}y^{k}.

We can then compute (x + y)^{n} as follows:

(x + y)^{n} = (x + y)(x + y)^{n−1}
= x(x + y)^{n−1} + y(x + y)^{n−1}
= x [∑_{k=0}^{n−1} x^{n−1−k}y^{k}] + y [∑_{k=0}^{n−1} x^{n−1−k}y^{k}]

= (\begin{matrix}n − 1 \\ 0 \end{matrix}) x^{n} + ∑_{k=1}^{n−1} [(\begin{matrix}n − 1 \\ k − 1 \end{matrix})

+ (\begin{matrix}n − 1 \\ k \end{matrix})] x^{n−k}y^{k} + (\begin{matrix}n − 1 \\ n − 1 \end{matrix}) y^{n}.

We have

(\begin{matrix}n − 1 \\ 0 \end{matrix}) = (\begin{matrix}n \\ 0 \end{matrix}) = (\begin{matrix}n − 1 \\ n − 1 \end{matrix}) = (\begin{matrix}n \\ n \end{matrix}) = 1

and

(\begin{matrix}n − 1 \\ k − 1 \end{matrix}) + (\begin{matrix}n − 1 \\ k \end{matrix}) = (\begin{matrix}n \\ k \end{matrix})

for all 0 < k < n, which proves that

(x + y)^{n} = ∑_{k=0}^{n} (\begin{matrix}n \\ k \end{matrix}) x^{n−k}y^{k}.

5 位相数学

距離関数

A distance function defined on a nonempty set S is a function δ from S × S to the set of nonnegative real numbers such that for all s_{1}, s_{2} and s_{3} in S the following conditions are satisfied:

(1) δ(s_{1}, s_{2}) = 0 if and only if s_{1} = s_{2};

(2) δ(s_{1}, s_{2}) = δ(s_{2}, s_{1});

(3) δ(s_{1}, s_{2}) = δ(s_{1}, s_{2}) + δ(s_{2}, s_{3}).

収束

If (S, δ) is a metric space, we say that a sequence {s_{i}} of points of S converges to a given point s^{∗} ∈ S if and only if, for every ε > 0, there exists a natural number n such that δ(s, s^{∗}) < ε for all i > n.