--- type: theoretical --- ## Proving Equivalences To prove that two statements $P$ and $Q$ are equivalent ($P \iff Q$): 1. ***Chain of Equivalences***: Show that $P(x_1, \ldots, x_n) \iff \ldots \iff Q(x_1, \ldots, x_n)$, transforming $P$ into $Q$ step by step. 2. ***Bi-conditional***[^10] ***Decomposition***[^1]: Use the fact that: $$ (P \iff Q) \iff ((P \implies Q) \land (Q \implies P)). $$ - Prove $P \implies Q$ (first direction). - Prove $Q \implies P$ (second direction). --- ## Proving Universal Statements To prove a statement of the form $\forall x\, P(x)$: 1. Proof by Exhaustion[^5]: - If the domain of $x$ is finite, verify $P(x)$ for each possible value. 2. Proof by Universal Generalization[^6]: - Let $c$ be an arbitrary element of the domain. - Prove $P(c)$. - Conclude that $\forall x\, P(x)$ holds. **Example**: - For all integers $n$, if $n$ is even, then $n^2$ is even. - ***Proof***: 1. Let $n$ be an arbitrary integer. 2. Assume $n$ is even. Then $n = 2k$ for some integer $k$. 3. Compute $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, which is even. --- ## Proving Existential Statements To prove a statement of the form $\exists x\, P(x)$: 1. Constructive Proof[^7]: Find a specific $c$ such that $P(c)$ is true. 2. Non-constructive Proof[^8]: - Assume $\forall x\, \neg P(x)$ (negation of existence). - Derive a contradiction. - Conclude that $\exists x\, P(x)$ must be true. --- ## Proof by Contraposition Instead of proving $P \implies Q$, prove its contrapositive[^3]: $$ \neg Q \implies \neg P $$ **Example**: - For all $n \in \mathbb{N}$, if $n^2$ is odd, then $n$ is odd. - ***Proof***: 1. Prove the contrapositive: If $n$ is even, then $n^2$ is even. 2. Assume $n = 2k$. Then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, which is even. --- ## Proof by Contradiction To prove $P$: 1. Assume $\neg P$. 2. Derive a contradiction[^4]. 3. Conclude that $P$ must be true. **Example**: - $\sqrt{2}$ is irrational. - **Proof**: 1. Assume $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{p}{q}$, where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\frac{p}{q}$ is in lowest terms. 2. Square both sides: $2 = \frac{p^2}{q^2}$, so $2q^2 = p^2$. 3. $p^2$ is even, so $p$ is even ($p = 2k$). 4. Substitute $p = 2k$: $2q^2 = (2k)^2 = 4k^2$, so $q^2 = 2k^2$. 5. $q^2$ is even, so $q$ is even. 6. Contradiction: $p$ and $q$ cannot both be even since $\frac{p}{q}$ is in lowest terms. 7. Conclude $\sqrt{2}$ is irrational. --- ## Mathematical Induction [^2] Generalized in [Proofs (FP)](Proofs.md) ![induction](induction.png) To prove a statement of the form $\forall n \geq n_0, P(n)$: 1. *Base Case*: Prove $P(n_0)$ is true. 2. *Inductive Hypothesis* - $IH$: Assume $P(k)$ is true for some $k \geq n_0$. 3. *Inductive Step*: Prove $P(k + 1)$ is true using $P(k)$. **Example**: - $1 + 2 + \dots + n = \frac{n(n + 1)}{2}$ for all $n \geq 1$. - ***Proof***: 1. Base Case: For $n = 1$, LHS = $1$, RHS = $\frac{1(1 + 1)}{2} = 1$. Holds true. 2. $IH$: Assume $1 + 2 + \dots + k = \frac{k(k + 1)}{2}$. 3. Inductive Step: Show $1 + 2 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2}$. $$ \text{LHS} = (1 + 2 + \dots + k) + (k + 1) = \frac{k(k + 1)}{2} + (k + 1). $$ Simplify: $$ \frac{k(k + 1)}{2} + (k + 1) = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2}. $$ Matches RHS. --- ## Strong[^9] Mathematical Induction To prove $\forall n \geq n_0, P(n)$: 1. *Base Cases*: Prove $P(n_0), P(n_0 + 1), \ldots, P(n_0 + m)$ are true. 2. *Inductive Hypothesis*: Assume $P(i)$ is true for all $n_0 \leq i \leq k$. 3. *Inductive Step*: Prove $P(k + 1)$ is true using the assumption $P(i)$ for all $i \leq k$. **Example**: - Every $n \geq 2$ can be factored into primes. - **Proof**: 1. *Base Case*: $n = 2$ is a prime. 2. $IH$: Assume every $n \leq k$ can be factored into primes. 3. *Inductive Step*: For $n = k + 1$: - If $k + 1$ is prime, done. - If composite, $k + 1 = a \cdot b$, where $2 \leq a, b \leq k$. - By hypothesis, $a$ and $b$ can be factored into primes. - Combine the prime factors of $a$ and $b$ to get $k + 1$'s factorization. [^1]: Breaking a complex problem into smaller, simpler parts to solve each one step by step [^2]: Proof that works by showing it works for the smallest case, then assuming it works for one number and proving it works for the next [^3]: Instead of directly proving "If A, then B," you prove "If not B, then not A," which means the same thing logically. [^4]: A way of proving something is true by assuming it is false and showing this leads to a logical impossibility, which, as we know, really messes with everything. [^5]: Checking every possible case individually to prove something is true for all of them. [^6]: Showing something is true for all cases by proving it for an arbitrary or random example. Similar to formal proof for introducing the $\forall$ quantifier [^7]: Directly finding an example to show something exists or is true. [^8]: Proving something exists or is true without giving a specific example, usually by ruling out the possibility of it not being true. [^9]: Like induction, but you assume everything is true up to a certain point to prove the next case [^10]: A statement where both directions are true, like "A if and only if B," meaning A leads to B, and B leads to A.