# Proof by Induction (concerning $3^n\ge1+2n$)

I've been able to follow the idea and steps of induction so far but I've hit a road block in understanding one of the examples in a text book. This is what the book says p.97:

Prove: $3^n \geq1 + 2n$

Skipping past the base case and assuming it's true, the books inductive step is as follows:

Show: $3^{n+1} = 1 + 2(1+n)$

LHS $= 3\cdot 3^n$
LHS $\geq 3(1+2n)$ [by assumption]
LHS $\geq 1+2+2n+4n$ [algebra]
LHS $\geq 1+2(1+n)$ [since $n>0$]

How can the $4n$ be omitted by $n>0$? This really boggles me, appreciate any insights and help.

Thanks!

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If n>0 then 4n>0 and omitting it from the RHS reduces the value. So >= still applies - or applies more strongly, and the = could be dropped.

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Ah I see, thank you! Dropping the = makes sense, but does doing so change the proof? –  xlm Mar 1 '12 at 8:14
@xlm, it forces you to add the condition $n > 0$ and prove a new base case. Edit: hang on... is the base case not $n=0$? If not, why not? Equality holds then... I would have thought it should be $n \ge 0$, in which case you can't drop the $=$. –  Peter Taylor Mar 1 '12 at 9:34
Apologies for not adding it earlier, base is n=1 since the proof is for n>=1 –  xlm Mar 1 '12 at 13:18
Apologies for ambiguity there - I was not referring to the base case in dropping the equality, but to the inductive step. It is exploring such things which leads to extensions and generalisations. –  Mark Bennet Mar 3 '12 at 9:08

Since $n>0$, we have $4n>0$ so $1+2+2n+4n>1+2+2n=1+2(1+n)$, thus $\text{LHS}\geq 1+2(1+n)$.

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