# Lecture 007

## Universal Statements

### Proving Even and Odd

Claim: $(\forall m, n \in \mathbb{Z})(2m \neq 2n+1)$ Proof: AFSOC $(\exists m,n \in \mathbb{Z})(2m = 2n+1)$, then $m-n=\frac{1}{2}$. However, $m-n \in \mathbb{Z}$. There is a contradiction. $\Rightarrow\!\Leftarrow$

## Existential Statement

Claim: $(\exists x \in S)P(x)$ Direct Proof: proof by demonstration that some number cannot be in integer

### Disprove by Counter Example for Union 1

Claim: $A\nsubseteq B$ Proof: prove by showing $(\exists a \in A)$ s.t. $a\notin B$. Proof:

• $A=\{10n-15|n\in \mathbb{Z}\}$

• $B = \{5n|n\in \mathbb{Z}\}$

• let $b\in B$ and $b = 0$, $0=10n-15$ iff $n=\frac{3}{2}$

### Disprove by Counter Example for Union 2

Claim: $(A\cup C)\times (B\cup D) \subseteq (A\times B)\cup (C\times D)$ Disprove:

• disprove by giving counter example

### Non-Constructive Direct Proof

Claim: There exist 2 irrational numbers a and b such that $a^b \in \mathbb{Q}$ Proof: let $a=b=\sqrt{2}$

• case 1: $\sqrt{2}^{\sqrt{2}} \in \mathbb{Q}$ - satisfy claim.

• case 2: $\sqrt{2}^{\sqrt{2}} \notin \mathbb{Q}$, then $\sqrt{2}^{\sqrt{2}^\sqrt{2}}=\sqrt{2}^2 \in \mathbb{Q}$ where $\sqrt{2}^{\sqrt{2}} \notin \mathbb{Q}$ - satisfy claim.

### Indirect Proof

Using $\lnot (\exists x \in S)P(x) \equiv (\forall x \in S)\lnot P(x)$ Claim: Let $(n\in \mathbb{Z}^{+}) \land (y_1,...,y_n \in \mathbb{R})$, $(\exists i \in [n])(y_i \geq \sum_{j=1}^n \frac{y_j}{n})$ Proof: AFSOC $(\forall i \in [n])(y_i < \sum_{j=1}^n \frac{y_j}{n})$, then summing up all $y_i$, $\sum_{i=1}^n y_i < n\sum_{j=1}^n \frac{y_i}{n} = \sum_{i=j}^n y_j$, which contradict with $\sum_{i=1}^n y_i = \sum_{i=j}^n y_j$. $\therefore (\exists i \in [n])(y_i \geq \sum_{j=1}^n \frac{y_j}{n})$

Claim: $P \implies Q$ Proof:

1. Direct Proof: assume $P$, show $Q$ holds
2. Contrapositive: show $\lnot Q \implies \lnot P$
3. Contradiction: show $\lnot(P\implies Q) \equiv (P\land \lnot Q)$ by AFSOC $P \land \lnot Q$.

### Even Odd

Claim: $(\forall n \in \mathbb{Z})((\exists m\in \mathbb{Z})(n=2m)\implies (\exists l \in \mathbb{Z})(n^2 = 2l))$ Proof: let $n\in \mathbb{Z}$ be even, s.t. n=2m for some $m\in \mathbb{Z}$. Fix m. Then $x^2 = (2m)^2 = 4m^2 = 2(2m^2)$, and since (2m^2 \in \mathbb{Z}), $n^2$ is even.

TODO

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