Prove that $n<3^n$ where $n \in \mathbb N$,
when $ n=1 $, I have proved it's true.
And assumed when $n = p$ , $p<3^p$ is true.
Can any body help me in showing that it is true for $n =p+1$
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Martin Sleziak
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Marlon Abeykoon
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1Related : http://math.stackexchange.com/questions/490983/prove-3n-ge-n3-by-induction and http://math.stackexchange.com/questions/434505/show-by-induction-that-n3-leq-3n-for-all-natural-numbers-n – lab bhattacharjee Dec 07 '13 at 14:27
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Ya but the correct answer is not selected by the person who have asked the question and in my question it has no power in n. – Marlon Abeykoon Dec 07 '13 at 14:33
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no power? $n=n^1$ – lab bhattacharjee Dec 07 '13 at 14:35
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$$3^p+1\lt 3^p+3^p+3^p=3^{P+1}$$ – Bumblebee Mar 29 '16 at 09:40
3 Answers
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We have as our inductive hypothesis (IH) that $\,p \lt 3^p$.
So we need to show $$p+1 \lt 3^{p+1} = 3 \cdot 3^p.$$
Can you see that $$p + 1 \quad \overset{IH}{\lt}\quad 3^p + 1 \quad \lt \quad 3\cdot 3^p \;= \;3^{p+1}, \quad \forall p \in \mathbb N$$
amWhy
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Great! but in last line you have mentioned $(3^p)+1 < 3*(3^p)$. How can we say like that directly. – Marlon Abeykoon Dec 07 '13 at 14:44
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it is false when $p=-1$ for example,but in natural numbers it is correct – dato datuashvili Dec 07 '13 at 15:09
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@Marlon Well, we can say that $3^p + 1 \leq 3^p + 3^p = 2\cdot 3^p \lt 3\cdot 3^p = 3^{p+1}$. – amWhy Dec 07 '13 at 15:12
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@amWhy why did you add $3^p$ to $3^p$?. If you add that to one side same should be done to the other right? and why did you put "smaller than or equal" sign? – Marlon Abeykoon Dec 07 '13 at 15:22
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1Not necessarily when we are working with inequalities: $3^p + \color{red}{1} \leq 3^p + \color{red}{3^p}$ because $1 \leq 3^p$, and for all $p > 0$, it happens to be true that $1 \lt 3^p$. This allows us then to say that $3^p + 1 \lt 3^p + 3^p \lt 3^p + 3^p + 3^p = 3\cdot 3^p = 3^{p+1}$. Note when $\lt$ is used, and note when $=$ is used. – amWhy Dec 07 '13 at 15:25
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If $n < 3^n$ then it follows that $$n+1 <3^n +1 < 3^n +(2\cdot 3^n)=3^{n+1}$$ here I used that $1<2\cdot 3^n$ for all $n\in \Bbb N$.
Slugger
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so we have proved that for $n=1$ ,$1<3$, what would be for $n+1$? we have $(n+1)<3^{n+1}$
which means that $(n+1)<3*3^n$ ,now clearly we have $3^p+1<3*3^p$, but also $3^p+1>p+1$ because we have used such thing that $p<3^p$, finally we conclude that $p+1<3*3^p$.
Marlon Abeykoon
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dato datuashvili
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