Induction

Example #3: Returning to our example from slides 2 – 3, we'll now prove that $2^n - 1 > n^2$ for $n \ge 5$ using induction.

Proof:

1. Base case: For $n=5$, we confirm that $2^n - 1 =$$\; 2^5 - 1 =$$\; 32 - 1 =$$\; 31 >$$\; 25 =$$\; 5^2 =$$\; n^2$.
2. Inductive hypothesis: We assume that, for some $k \in \mathbb{P}$ for which $k \ge 5$, we have $2^k - 1 > k^2$.
3. Inductive step: We'll now prove that, for $k + 1$, we have $2^{k+1} - 1 > (k+1)^2$. Note that $2^{k+1} - 1 =$$\; 2 \cdot 2^k - 1 >$$\; 2 \cdot 2^k - 2 =$$\; 2(2^k - 1) >$$\; 2(k^2) \;$ [The latter inequality based on the inductive hypothesis]. Now, we need to show that $2(k^2) >$$\; (k+1)^2 =$$\; k^2 + 2k + 1$, or that $k^2 > 2k + 1$ in other words. Indeed, note that plugging any value $k \ge 5$ will make $k^2$ larger than $2k + 1$.

According to our proof by induction above, $2^n - 1 > n^2$ for all $n \ge 5$.◼