This next post will focus on using mathematical induction to prove an inequality. Our statement to be proven is as follows:

for all integers n

Solution:

The first step as always in theses types of proofs is to prove the basis step. This is accomplished by just inputting values.

Show that P(3) is true:

As was shown above, the basis step checks out in our proof and we are now ready to move onto our inductive step.

Show that the inductive hypothesis is true:

Suppose that k is some arbitrarily chosen but particular value such that:

for all integers k

We must show that

for all integers n

We are allowed to use part of our inductive hypothesis because that’s what we are assuming to be true. In order to take advantage of our inductive hypothesis we must re-arrange out current statement P(k+1). The left side of this inequality can be written as follows:

With the left side of the inequality in this form we can make the following claim:

This result was achieved by simply substituting the 2k+1 term from our inductive hypothesis. Now if we can prove that is strictly less than we will have finished the proof.

This next step takes a little finesse because we must remember how inequalities work. We must remember that by adding/subtracting/multiplying/dividing the same POSITIVE term for both sides we are maintaining the inequalities validity (try a few cases if your not satisfied). This means that we can go through the following set of manipulations:

Our final inequality is clearly valid for all terms thus by default it will be valid for all

for all integers

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