CMSC250

def fun(a):
  if a ==0:
    return 1
  return 2 * fun(a-1)

Claim: fun(\(n\)) = \(2^n\)

How to prove?

Induction: Proof methodology where we show that $\forall n \in \mathbb{N}, P(n)$

Prove $P(x)$ is true because $P(n\le x)$ is true

More specifically: $P(0) \land P(0) \Rightarrow P(1) \land P(1) \Rightarrow P(2) \land \dots$

Basically a way we can prove recursion works

Dominoes, or ladders

There are always two things

• Base Case
• Inductive Step

There are always two things

• I will place a lego brick on the ground
• For any brick, starting with the first one, if that brick is placed, a brick will be placed next to it.

What can be concluded?

Weak Induction: A way we can prove $\forall x \in \mathbb{Z}^{\ge n_0}, P(x)$ because $P(n_0) \land P(x) \Rightarrow P(x+1)$ and $n_0 \in \mathbb{Z}^{\ge 0}$

Weak Induction: A way we can prove $\forall x \in \mathbb{Z}^{\ge n_0}, P(x)$ because $P(n_0) \land P(x) \Rightarrow P(x+1)$ and $n_0 \in \mathbb{Z}^{\ge 0}$

We want to prove 2 things

• Base Case: $P(n_0)$
• Inductive Step: $\forall k \ge n_0, P(k) \Rightarrow P(k+1)$

Term: Inductive Hypothesis: for some arbitrary $ k \ge n_0, P(k)$

• Base Case: $P(n_0)$
• Inductive Step: $\forall k \ge n_0, P(k) \Rightarrow P(k+1)$

def fun(a):
  if a ==0:
    return 1
  return 2 * fun(a-1)

\[ f(x) = \begin{cases} 1 & x = 0 \\ 2f(x-1) & x \ge 1 \end{cases} \]

\[ f(x) = \begin{cases} 1 & x = 0 \\ 2f(x-1) & x \ge 1 \end{cases} \]

$\forall x \in \mathbb{Z}^{\ge 0},f(x) = 2^x$

• Base Case: $f(0) = 1 = 2^0$
• Inductive Hypothesis: for some arbitrary $k \ge 0, f(k) = 2^k$
• Inductive step: $f(k) = 2^k \Rightarrow f(k+1) = 2^{k+1}$

$\forall n \in \mathbb{Z}^{\ge 3}, 2n+1 < 2^n$

• Base Case: $2(3)+1 < 2^3 = 2(3) + 1 = 7 < 2^3 = 8 $
• Inductive Hypothesis: Suppose for some arbitrary $k \ge 3, 2k+1 < 2^k)$
• Inductive step: $2k+1 < 2^k \Rightarrow 2(k+1)+1 < 2^{k+1}$

$\forall n \in \mathbb{Z}^{\ge 1},\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

$P(x) = \sum_{i=1}^{n} i = \frac{n(n=1)}{2}, n_0 = 1$

• Base Case: $P(1) = \sum_{i=1}^{1} i = \frac{1(1+1)}{2} \Rightarrow 1 = \frac{2}{2} = 1 $
• Inductive Hypothesis: Suppose $k \ge 1, P(k)$
• Inductive step: $P(k) \Rightarrow P(k+1)$

\[ f(x) = \begin{cases} 3 & n = 0 \\ 5(f(x-1)) + 8 & n \ge 1 \end{cases} \]

$\forall x \in \mathbb{Z}^{\ge 0},f(x) \equiv 3 (\text{ mod }4)$

• Base Case: $f(0) = 3, 3 \equiv 3 (\text{ mod } 4)$
• Inductive Hypothesis: Suppose for some arbitrary $k \ge 0, f(k) \equiv 3 (\text{mod } 4)$
• Inductive step: $f(k) \equiv 3 (\text{mod } 4) \Rightarrow f(k+1) \equiv 3 (\text{mod }4)$

$\forall n \in \mathbb{Z}^{\ge 4},2^n - n^2 \ge 0$

• Base Case: $2^4 - 4^2 \ge 0 = 16 - 16 = 0 \ge 0$
• Inductive Hypothesis: Suppose for some arbitrary $k \ge 4, 2^k - k^2 \ge 0$
• Inductive step: $2^k - k^2 \ge 0 \Rightarrow 2^{(k+1)} - (k+1)^2 \ge 0$