Hi Everyone,
Still trying to grasp doing inductions, saw this question today but i got it completely wrong.
Any assistance would be most appreciative.
Using induction to pove that for all n≥4 the inequalities 2ⁿ ≥ n² and 2ⁿ < n! hold.
P(k)
......
P(k+1)
.......
Proof
......
by multiplying out
We are showing whether or not P(k) being true causes P(k+1) to be true.
This is how induction works.
The comments in the next post "correcting" this are incorrect
.......
If P(k) is true, then
therefore is
Therefore if
then also.
P(k)
P(k+1)
Proof
We are showing that P(k) being true causes P(k+1) to be true
The whole thing about basic mathematical induction is tedious algebra expansion and simplification. If you can expand one small step at a time, you should be able to do any form of induction in your first course of math in college. I am going to show you how to take the small steps. If you observe the example, you should not have trouble with simple induction.
Remark: You can break down you proof into two parts: One for , and the other one,
You are to prove for all , where
and
Note: Keep this in mind that (Let's call it the precondition.) We will return to this later.
Here we prove statement :
Basis step:
For . Since this is true, we move on.
Inductive Step:
Suppose for a positive integer that is true. Then multiplying through by 2, we obtain
. Observe each step:
Next, we substitute the precondition for one of the
Next, we substitute the precondition for the in the last term.
Therefore,
Now, we will prove the statement
Basis Step:
For all positive integer , . This one checked.
Inductive Step:
Suppose for all positive integer . Then multiplying through by 2, we obtain
since the smallest in the parenthesis is greater than 4. Therefore,
Consequently, for all positive integer