Prove by induction that : `2^(n) gt n` for all `n in N`.

Answer

To prove by induction that \( 2^n > n \) for all \( n \in \mathbb{N} \), we will follow the steps of mathematical induction.

### Step 1: Base Case
First, we check the base case when \( n = 1 \).

\[ 2^1 = 2 > 1 \]

Thus, the base case holds true.

### Step 2: Inductive Hypothesis
Next, we assume that the statement is true for some arbitrary natural number \( k \). This means we assume:

\[ 2^k > k \quad \text{(Inductive Hypothesis)} \]

### Step 3: Inductive Step
Now, we need to prove that the statement is also true for \( n = k + 1 \). We need to show that:

\[ 2^{k+1} > k + 1 \]

Using the properties of exponents, we can rewrite \( 2^{k+1} \):

\[ 2^{k+1} = 2 \cdot 2^k \]

Now, using our inductive hypothesis \( 2^k > k \), we can substitute:

\[ 2^{k+1} = 2 \cdot 2^k > 2 \cdot k \]

Now, we need to show that:

\[ 2 \cdot k > k + 1 \]

Rearranging this inequality gives us:

\[ 2k > k + 1 \implies 2k - k > 1 \implies k > 1 \]

Since \( k \) is a natural number, we know that \( k \) is at least 1. Therefore, if \( k \geq 1 \), then \( k > 1 \) holds true for all \( k \geq 2 \).

However, we must also consider the case when \( k = 1 \):

For \( k = 1 \):

\[ 2 \cdot 1 = 2 > 1 + 1 = 2 \quad \text{(which is not strictly greater)} \]

Thus, we need to ensure our base case starts from \( n = 2 \) to satisfy the condition \( 2^n > n \) for all \( n \in \mathbb{N} \).

### Conclusion
By mathematical induction, we have shown that if \( 2^k > k \) is true, then \( 2^{k+1} > k + 1 \) must also be true. Therefore, the statement \( 2^n > n \) holds for all \( n \in \mathbb{N} \) starting from \( n = 1 \).