ma1a_1 - Mashiro's Blog

ma1a_1

2026-09-28

Lecture Notes

Math

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Note

Office hour Thursday 4 - 5pm 356 Linde hall

Sets. Wednesday - Monday.

Definition: Set#

A set is a collection $X$ of objects called elements.

$\rightarrow$ If $x$ belongs to $X$ , write $x \in X$

\rightarrow X = \{x_1, x_2, x_3, \ldots\}

$\varnothing$ : empty set. Set with no elements.

$\forall$ : for any.

$\exists$ : exists

Two sets A, B#

Union of A & B

A \cup B = \{x \mid x \in A \text{ or } x \in B\}

Intersection.

A \cap B = \{x \mid x \in A \text{ and } x \in B\}

If $C$ is subset of $A$ $\rightarrow$ $C \subseteq A$

$C \subset A$ $\rightarrow$ Proper subset.

Set Difference#

A \setminus B = \{x \mid x \in A \text{ and } x \notin B\}

Definition. The cardinality of finite set A#

$|A|$ or $\#A$ . is the number of elements in $A$ .

If $A, B$ are finite sets.

|A \cup B| = |A| + |B| - |A \cap B|

$\rightarrow$ The simplest case of. Inclusion exclusion principal.

Examples#

Set of natural numbers. $\mathbb{N}$ .

Set of Integers. $\mathbb{Z}$

Set of rational numbers.

\mathbb{Q} = \left\{\frac{a}{b} \mid a, b \in \mathbb{Z} \text{ and } b \ne 0\right\}

set of real numbers $\mathbb{R}$

Two Statements A, B#

A false Statement implies any Statement.

$A \rightarrow B$ . A implies B.

$A \leftrightarrow B$ . A is equivalent to B. A iff B.

$\neg A$ $\sim A$ . Negation of A.

fact. $(A \rightarrow B) \leftrightarrow (\sim B \rightarrow \sim A)$ . Contrapositive.

Methods of proofs#

1. Proof by direct verification#

Example. for all $x, y \in \mathbb{R}$ , we have.

x^2 + y^2 \ge 2xy.

Proof. $(x - y)^2 \ge 0$ .

x^2 - 2xy + y^2 \ge 0. \quad \square

2. Proof by contrapositive#

Example. $n^2$ is odd $\rightarrow$ $n$ is odd.

$\hookrightarrow$ proof $n$ is even $\rightarrow$ $n^2$ is even.

n = 2m.

n^2 = 4m^2. \quad \square

3. Proof by Contradiction#

Assume desired conclusion is false & deduce smth impossible.

Example. eq. $x^2 + y^2 = 1$ has no solution in positive integers $x$ & $y$

proof. Assume $x, y \in \mathbb{Z}^+$ such that $x^2 + y^2 = 1$ .

(x + y)(x - y) = 1.

So $x + y = x - y = 1$ or $-1$ .

\rightarrow y = -y = 0. \quad \square

Same result through direct verification.

If $x > y$ are positive integers. then $x^2 - y^2 \ge 3$ .

Proof. $x > y \rightarrow x \ge y + 1$ .

x^2 - y^2 \ge (y + 1)^2 - y^2 = 2y + 1 \ge 3. \quad \square

Theorem. There’s no rational number x satisfy $x^2 = 2$ #

Proof otherwise, Assume $x = \dfrac{a}{b}$ , $x^2 = 2$ , $a, b \in \mathbb{Z}$ , $b \ne 0$

\gcd(a, b) = 1.

$a^2 = 2b^2$ $\rightarrow$ $a^2$ is even $\rightarrow$ $a$ is even

assume. $a = 2a_1$

$4a_1^2 = 2b^2$ . $\rightarrow$ $b$ is even.

$\rightarrow$ 2 is a common divisor of 2. $\square$ .