To prove the inequality $2^n > n$ for all integers $n \geq 1$ using mathematical induction, we proceed as follows:

Step 1: Base Case

We check if the inequality holds for $n = 1$.

$2^1 = 2 \quad \text{and} \quad 1 = 1.$

Clearly, $2 > 1$, so the base case holds.

Step 2: Inductive Hypothesis

Assume that the inequality $2^k > k$ holds for some arbitrary positive integer $k$. That is, we assume:

$2^k > k.$

Step 3: Inductive Step

We need to show that $2^{k+1} > k+1$, assuming the inductive hypothesis $2^k > k$.

Now, consider:

$2^{k+1} = 2 \cdot 2^k.$

Using the inductive hypothesis, we know that $2^k > k$. Thus, multiplying both sides of this inequality by 2 gives:

$2^{k+1} = 2 \cdot 2^k > 2 \cdot k.$

Therefore, we need to show that:

$2 \cdot k \geq k + 1.$

Simplifying the inequality:

$2k \geq k + 1 \quad \Longleftrightarrow \quad k \geq 1.$

This is true because $k \geq 1$ for all positive integers $k$. Thus, we have shown that:

$2^{k+1} > k + 1.$

Conclusion:

By mathematical induction, the inequality $2^n > n$ holds for all integers $n \geq 1$.

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Would you like further clarification on any part of this proof?

Here are five related questions:

1. How do we prove more complex inequalities using induction?
2. Can the same approach be used to prove $3^n > n^2$?
3. What is the significance of base cases in induction?
4. How do we choose an appropriate inductive hypothesis for different problems?
5. How does strong induction differ from regular induction?

Tip: Always make sure to clearly state the inductive hypothesis and how it is used in the inductive step.