Im not good at induction involving inequalities; can someone help me prove by mathematical induction:
1) 2^n - n^2 > 0 where n>4
2) n^2 - 3n + 2 ≥ 0
3) 4^n - 1 - 7n ≥ 0 where n ≥2
P(n) :2^n - n^2 >0 When n>4
Now 7 >0 hence its correct for n = 5
Assumption : P(k) is correct
Proof for P(k+1) using assumption:
But since (k+1)^2 is always a positive number (greater than 16 atleast ) thus proved that its correct for k+1 hence through the principle of mathematical induction this is proved
I know the rules state create a new thread for each problem, but this isn't really a new problem. Rather, it's a question regarding the solution posted, as I'm unsure as well about the application of induction to inequalities.
Reading the Assumption step, I'm not sure how that would work if p(n) = 2^n > n^2, for n>4. Would it essentially be the opposite to what was posted, becoming "hence: 2^n - n^2 > 0" ?
It is more convincing to begin with and end with . By beginning with P(k+1) is less desirable though it is not always the case for Strong Principle of Mathematical Induction.
As much as we could, we would begin with
Now subtract both sides by , and we arrive at
, which is precisely as desired; namely, P(k+1).
Notice that the first line of proof was
and the last line of proof is .
That is to say we have proved