1
$\begingroup$

I am supposed to prove the following:

$$2^n+1\leq3^n\;\;(n\in\Bbb N)$$

Base case: $n=1$.

$$2^1+1\leq3^1$$

$$3\leq3$$

Now for induction:

$$2^{n+1}+1\leq 3^{n+1}$$

$$2\cdot2^n+1\leq3\cdot3^n$$

$$2^n+\frac{1}{2}\quad\leq\quad \frac{3}{2}\cdot3^n$$

$$2^n+\frac{1}{2}\quad\leq\quad2^n+1\quad\leq\quad3^n\quad\leq\quad\frac{3}{2}\cdot3^n$$

Is this a valid approach? The last step I took was a bit strange and I want to know if it was okay.

  • 3
    You started with what you're trying to prove. Instead, END with that. – 2017-02-23
  • 0
    Is this proof backwards? – 2017-02-23
  • 0
    Your induction step is backwards, yes. – 2017-02-23
  • 0
    $2^{n+1}+1\leq \dots \leq^{\dagger} \dots \leq 3^{n+1}$, should be able to be written all in one line where somewhere in the middle such as at $\dagger$ you invoke your induction hypothesis – 2017-02-23

2 Answers 2

2

Your almost correct. But let me reconstruct the proof.

We have to assume that $$2^n+1\leq 3^n.$$ Then multiplying by $2$, we get $$\big[2^{n+1}+2\big]\leq \big[2\cdot 3^n\big].$$ Thus, $$\begin{align} 2^{n+1}+1\quad&<\quad 2^{n+1}+2\\ &\leq\quad 2\cdot 3^n\\ &<\quad 3\cdot 3^n\\ &=\quad 3^{n+1}. \end{align} $$

2

Working backwards from what you're trying to prove with equalities can work but only if each line is an iff statement from the previous one. In this case it is not and therefore is not a valid proof.

Consider $3^{n+1} - (2^{n+1} + 1)$

$\Rightarrow 3*3^{n} - 2*2^{n} - 1$

$\Rightarrow 2(3^{n} - 2^{n} -1) + 3^{n}$

From our assumption that $3^{n} \geqslant 2^{n} + 1$ it is simple to see that:

$2(3^{n} - 2^{n} -1) + 3^{n} \geqslant 0$

$\Rightarrow 3^{n+1} \geqslant 2^{n+1} + 1$ as required.

  • 0
    How might I do that? – 2017-02-23
  • 1
    I edited my answer to include a solution – 2017-02-23