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?
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 $$2^{k+1}-(k+1)-4=...$$
and estimate that expression. And after all, if you write first one it is better to write it like this:
$$2^{k+1}\stackrel{?}{>}(k+1)+4$$