ece108: add induction

docs/1b/ece108.md

@@ -1,6 +1,6 @@
 # ECE 108: Discrete Math 1
 
-An **axiom** is a defined core assumption held to be true.
+An **axiom** is a defined core assumption of the mathematical system held to be true without proof.
 
 !!! example
     True is not false.
@@ -246,10 +246,48 @@ An **odd number** is a multiple of two plus one.
 
 $$\boxed{n\text{ is odd}\iff\exists k\in\mathbb Z,n=2k+1}$$
 
-A number is **divisible** by another if it can be part of its product.
+A number is **divisible** by another $m|n$ if it can be part of its product.
 
 $$\boxed{n\text{ is divisible by } m\iff\exists k\in\mathbb Z,n=mk}$$
 
 A number is a **perfect square** if it is the square of an integer.
 
 $$n\text{ is a perfect square}\iff \exists k\in\mathbb Z,n=k^2$$
+
+### Induction
+
+Induction is a proof technique that can be used if the open sentence $P(n)$ depends on the parameter $n\in\mathbb N$. Because induction works in discrete steps, it generally cannot be applied domains of all real numbers.
+
+To do so, the following must be proven:
+
+- $P(1)$ must be true (the base case)
+- $P(k+1)$ must be true for all $P(k)$, assuming $P(k)$ is true (the inductive case)
+
+!!! warning
+    The statement **cannot** be assumed to be true, so one side must be derived into the other side.
+
+!!! tip "Proof"
+    This should more or less be exactly followed. For the statement $\forall n\in\mathbb Z,n!>2^n$:
+    
+    > We use mathematical induction on $n$, where $P(n)$ is the statement $n!>2^n$.
+    >
+    > **Base case**: Our base case is $P(4)$. Note that $4!=24>16=2^4$, so the base case holds.
+    >
+    > **Inductive step**: Let $k\geq 4$ for an arbitrary natural number and assume that $k!>2^k$. Multiplying by $k+1$ gives
+    >
+    > $$(k+1)k^2>(k+1)2^k$$
+    >
+    > By definition $(K=1)k!=(k+1)!$. Since $k\geq 4$, $k+1>2$ and thus $(k+1)2^k>2\cdot 2^k=2^{k+1}$. Putting this together gives
+    >
+    > $$(k+1)!>2^{k+1}$$
+    >
+    > Thus $P(k+1)$ is true and by the Principle of Mathematical Induction (POMI), $P(n)$ is true for all $n\geq 4$.
+    
+Induction can be applied to the whole set of integers by proving the following:
+
+- $P(0)$
+- if $i\geq 0, P(i)\implies P(i+1)$
+- if $i\leq 0, P(i)\implies P(i-1)$
+
+Alternatively, some steps can be skipped in **strong induction** by proving that if for $k\in\mathbb N$, $P(i)$ holds for all $i\leq k$, so $P(k+1)$ holds. In other words, by assuming that the statement is true for all values before $k$. If strong induction is true, regular induction must also be true, but not vice versa.
+