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Claim: $$\forall n \in \mathbb{N}, 2^n+1 \leq 3^n $$

Proof (Induction): Base: Let $$n = 1, 2^1 + 1 \leq 3$$so this is true. Inductive Step: Suppose $$n \geq 1$$ and assume inductive hypothesis holds for all values less than or equal to n. $$2^{n+1} + 1 = 2*2^n + 1$$recall $$2^n \leq 3^n - 1$$So, $$2(2^n) + 1 \leq 2(3^n -1) + 1$$Now, we want to show $$\forall n \in \mathbb{N}, 2(3^n) - 1 \leq 3^{n+1}$$ This inequality becomes $$-1 \leq 3^n$$ which is obviously true $\forall n \in \mathbb{N}$. Therefore, $$2^{n+1} + 1 \leq 2(3^n - 1) + 1 \leq 3^{n+1} \implies \forall n \in \mathbb{N}, 2^{n+1}+1 \leq 3^{n+1}$$ by induction. This is one of the first inequality induction proofs I have done, and usually in an inductive proof, you construct the right hand side using the left hand side. I did not do this here, I just pulled the $3^{n+1}$ out of nowhere and tested if the simpler inequality $2(3^n) - 1$ was always less than or equal to $3^{n+1}$. Is this rigorous enough? Thanks.

Bill Dubuque
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cbankr22
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    This seems fine, however it is recommended to state the overall claim in terms of $n$ but the induction hypothesis and induction step in terms of another variable such as $k$. If it could all be done in one line rather than having to split lines and use words between like you have it might look even more clean, but there is nothing inherently wrong with what you have done. – JMoravitz Mar 14 '23 at 16:45
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    Recommended reading: How to write a clear induction proof – JMoravitz Mar 14 '23 at 16:46

1 Answers1

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Your argument is correct.

There's another method of packaging up an inductive argument that may be easier to understand and that is proof by infinite descent.

Proof by infinite descent is equivalent to induction if you try to prove the inductive hypothesis for the previous positive integer and equivalent to strong induction if you try to prove the inductive hypothesis for some previous integer.

This is a form of proof by contradiction, though, so it is less elegant than a direct proof.

I know from personal experience that it's easy to fall into the bad habit of overusing proof by contradiction. They are, however, useful tools for checking your work, especially if you find them immediately intuitive.

Here's an example for your case.

Consider the claim $2^n + 1 \le 3^n$ for all positive integers $n$.

It is indeed true that $2^1 +1 \le 3^1$.

Suppose for contradiction that $w$ is the least positive integer such that $2^w + 1 > 3^w$.

  • $ 2^w + 1 > 3^w \;\;\;\;$ is our starting hypothesis.
  • $ 2^w \ge 3^w \;\;\;\;$ follows because no integers exist strictly between $a$ and $a+1$ for any $a$.
  • $ 2 \cdot 2^{w-1} \ge 3 \cdot 3^{w-1} \;\;\;\;$ follows immediately.
  • $ 2^{w-1} \ge 3^{w-1} \;\;\;\; $ follows because $2 < 3$ and hence $\frac{1}{2} > \frac{1}{3}$.
  • $ 2^{w-1} + 1 \ge 3^{w-1} \;\;\;\;$ as desired.
  • It is therefore absurd that $w$ is the smallest number for which $2^w + 1 > 3^w$.

Thus it holds that $2^n +1 \le 3^n$ for all positive integers $n$.

Greg Nisbet
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