This is an interactive website to help you learn how to write an induction proof. You should try to prove the statement yourself, and use the information on this webpage to check your work or to get a hint if you need one.

We will prove an inequality. You will prove the similar inequalities in your homework. You can see that most of the proof is missing. Click on the box on the left to reveal the next step.

But before you click, try to write the proof yourself and figure out what will appear!

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Prove that \(3k+2 \le k^3\) for all \(k \in \mathbb N,\ k\ge 2\).

Proof: We can see that \(3(2)+2 \le 2^3\), so assume that \(3k+2 < k^3\) for some \(k\ge 2\). Note also that \(k\ge 2\) implies that \(3k^2 + 3k >6>2\). Continuing, we compute (start by clicking on the \(3(k+1)+2\))

\(3(k+1)+2 \)	\( = (3k+2)+3\)
*click*	\( = 3k + 2 + 3 \)
*click*	\( \le k^3 + 3\) (by the induction assumption)
*click*	\( = k^3 +2+1 \)
*click*	\( < k^3 + 3k^2 + 3k +1 \) (as noted above)
*click*	\( = (k+1)^3 \) as desired.

We can also begin with the induction assumption (after the basis step of course!) and continue with valid operations. We assume that \(3k+2 < k^3\) for some \(k\ge 2\), and obtain (click on the implications to see the next step)

\(3k+2 \le k^3 \Rightarrow \)	\( 3k+2+3 < k^3 + 3 \)
\(\Rightarrow\)	\( 3(k+1)+2 \le k^3 +2 + 1 \)
\(\Rightarrow\)	\( 3(k+1)+2 \le k^3 + 3k^2 + 3k +1 \) (as noted above)
\(\Rightarrow\)	\( 3(k+1)+2 \le (k+1)^3 \) as desired.

Thus \(3k+2 \le k^3\) for all \(k \in \mathbb N,\ k\ge 2\).