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
thanks
All the questions are almost same try them after seeing this
P(n) :2^n - n^2 >0 When n>4
First step:
Now 7 >0 hence its correct for n = 5
Assumption : P(k) is correct
ie;
Hence
Proof for P(k+1) using assumption:
since
Hence
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" ?
When we prove by Principle of Mathematical Induction, as much as it is possible, we want it be more convincing. Since the theorem says
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
for