# mathematical induction methods

Mathematical induction I know there is one specific way of proving it which is say for instance the example:

Method 1

Prove using mathematical induction that: $$2^n>n+4, n\ge 3$$

I will skip straight to the induction step: We assume $$P(k)$$ is true and hence we have: $$2^k>k+4, k\ge 3$$ Now $$P(k+1)$$ states that: $$2^{k+1}=2\times2^{k}, k\ge 3$$ $$2^{k+1}=2\times2^{k}> 2(k+4)=2k+8>k+8>k+5=(k+1)+4 , k\ge 3$$ $$2^{k+1}>(k+1)+4$$

Alternative Method:

Going straight to the induction step:

$$2^k>k+4, k\ge 3$$ $$2^{k+1}>(k+1)+4$$ $$2\times 2^k-k-5>0$$ $$2\times 2^k-k-5>2(k+4)-k-5>0$$ $$2\times 2^k-k-5>k+3>0$$

But this is true as $$k\ge 3$$.

My question is, are both methods valid and is this valid for any mathematical induction inequality problems?

• Your alternative method breaks down when you go from $2\times 2^k-k-5\gt 0$ to $2\times 2^k-k-5\gt 2(k+4)-k-5\gt 0$. Note that $x\gt 0$ and $x\gt y$ doesn't imply $x\gt y\gt 0$ – learner Mar 20 at 16:06
• Why on earth would the alternative method be valid? Why would you assume $2^{k+1} > 2^k + 1$... if that is what you were assuming. Also it is very unclear what your "Methods" are. These are specific to this question about an exponential value's inequality to a linear value. How what you apply this to any other question, say a question about the number of divisors, or number of combinations, or paths? – fleablood Mar 20 at 16:11
• Learner But we are saying $y>0$ and therefore $x>y$ implies $x>y>0.$ – Aurora Borealis Mar 20 at 17:06

Basically you did the same thing twice with a different form. Anyway I don't like that you write $$2^{k+1}>(k+1)+4$$ since this is to be proved. It is better to write $$2^{k+1}-(k+1)-4=...$$
$$2^{k+1}\stackrel{?}{>}(k+1)+4$$