Inequalities proof by math induction

Question

I have some troubles with inequalities proof by math induction.

Let's consider an example. Prove by induction that the following is true:

$$ N^2 - 3 * N \ge 0, where: N \ge 0$$

Base: $$0-0 \ge 0, true$$

Induction:

$$ (N+1)^2 - 3 * (N+1) \ge 0 $$ $$ N^2 + 2 * N + 1 - 3 * (N + 1) \ge 0 $$ $$ N^2 + 2 * N + 1 - 3 * N - 3 \ge 0 $$ $$ N^2 + 2 * N - 3 * N - 2 \ge 0 $$

let's rearrange terms and add brackets:

$$ [N^2 - 3 * N] + [2 * N - 2] \ge 0 $$

Is it right to say that (I don't understand why it is correct) Expression in left brackets greater or equal to 0 according to our assumption.

Let's prove expression in right brackets. Check it for 0: $$ 2 * N - 2 \ge 0$$ $$ 2 * 0 - 2 \ge 0$$ $$ 0 \ge 2:false$$

Therefore, the following expression is false:

$$ N^2 - 3 * N \ge 0, where: N \ge 0$$

Answer 1

It false because it false for $n=1$.

Id est, the statement $$P(n)\Rightarrow P(n+1)$$ is wrong for $n=0$.

Here $P(n)$: for any integer $n\geq0$ we have $n^2-3n\geq0.$

But for $n\geq3$ it's true by your work.

I like the following way.

For $n=3$ it's true.

Let $n^2-3n\geq0$ be true for any $n\geq3$.

Thus, $$(n+1)^2-3(n+1)=n^2-n-2=n^2-3n+2n-2\geq0$$ and we are done!

Answer 2

What you have done is show that the statement:

$$N^2-3N\ge0\;\;\text{ where}\;\;N\ge0$$

is false.

It is true for $N\in \mathbb{R}\setminus (0,3)$, or for just $N^2$ for example.