Prove $k^2>k+1$ by induction

Question

How would you prove that: $$n^2>n+1 \text{ for } n\ge2$$ using induction?

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Progress

The base is clear, and after that I have assumed $n=k$ and I am trying to prove $(k+1)^2>k+2$ , but I get stuck on proving $k^2>k+2$

Answer 1

$$n=2$$

$2^2 > 2 + 1$ (true)

$$n=k \implies n=k+1$$

Prove $(k+1)^2 > k + 2$ given $k^2 > k + 1$.

Try to make LHS of given look like LHS of conclusion:

$k^2 +2k + 1 > 3k + 2$

Can you finish? ^_^

Answer 2

Suppose $n^2 > n+1$. you want to show that $(n+1)^2 > (n+1)+1 = n+2$.

Start with $(n+1)^2 = n^2+2n+1 $ and go from there.

Answer 3

Let $P(n)$ be the statement $n^2 > n + 1$. The base case statement $P(2)$ is true as $2^2 = 4 > 3 = 2 + 1$.

Suppose $P(k)$. Then

$$(k+1)^2 = k^2 + 2k + 1 > (k + 1) + 2k + 1 \ \ \text{ by } P(k)$$

Thus $P(k) \Rightarrow (k+1)^2 > (k+1) + (2k + 1) > (k+1) + 1$ for any $k \geq 2$.

That is, $P(k) \Rightarrow (k+1)^2 > (k+1) + 1$ for any $k \geq 2$; in other words $$P(k) \Rightarrow P(k+1) \text{ for any } k \geq 2$$

Therefore by the Principle of Mathematical Induction, $P(n)$ is true for all $n \geq 2$.