Mathematical induction is a powerful technique used to prove statements about natural numbers. It involves two steps: the base case and the inductive step. In the base case, we show that the statement holds for the smallest value of n. In the inductive step, we assume that the statement holds for some arbitrary value of n, called the inductive hypothesis, and then prove that it also holds for n+1. By repeatedly applying the inductive step, we can then conclude that the statement holds for all natural numbers. In this article, we will utilize mathematical induction to prove that the statement 2^n ≤ 2n holds for all integers n ≥ 1.
Proving the Inequality Using Mathematical Induction
To prove that the inequality 2^n ≤ 2n holds for all integers n ≥ 1, we will employ the principle of mathematical induction.
Base Case
The base case involves demonstrating the truth of the statement for the smallest possible value of n, which is 1 in this case.
When n = 1, we have:
2^1 = 2 and 2(1) = 2
Since 2 ≤ 2, the inequality holds for n = 1.
Inductive Hypothesis
We assume that the inequality 2^k ≤ 2k holds for some arbitrary integer k ≥ 1. This assumption is known as the inductive hypothesis.
Inductive Step
The inductive step involves proving that the inequality also holds for n = k+1, assuming that it holds for n = k. In other words, we need to show that 2^(k+1) ≤ 2(k+1).
Starting with the left-hand side of the inequality, we can rewrite 2^(k+1) as 2 * 2^k.
Since we assumed that 2^k ≤ 2k in the inductive hypothesis, we can substitute this inequality into the expression:
2^(k+1) = 2 * 2^k ≤ 2 * 2k = 4k
Now, we need to show that 4k ≤ 2(k+1).
Expanding the right-hand side of the inequality:
2(k+1) = 2k + 2
Since k ≥ 1, we know that 2k ≥ 2. Therefore:
2k + 2 ≥ 2k + 2k = 4k
Combining the inequalities, we have:
2^(k+1) ≤ 4k ≤ 2(k+1)
Hence, we have successfully demonstrated that if the inequality holds for n = k, it also holds for n = k+1.
Conclusion
By proving the base case and the inductive step, we have successfully proven that the inequality 2^n ≤ 2n holds for all integers n ≥ 1. This proof demonstrates the effectiveness of mathematical induction in establishing the truth of statements about natural numbers. The process involves showing that the statement holds for the smallest value of n and then demonstrating that if it holds for some value of n, it also holds for the next value. This recursive approach allows us to conclude that the statement is true for all natural numbers.
