162 lines
5.3 KiB
Markdown
162 lines
5.3 KiB
Markdown
---
|
|
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)
|
|
|
|
|
|
|
|
|
|
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.
|