proof by mathematical induction

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subject is PreCalulus.

2^(k+3) = and < (k+3)!

i know how to do proving using mathematical induction when its just an equal sign, but I don't understand what to do when its an inequality.

thank you!!!

Here is what I would do

Step 1: check if it is true for k=1

that is: let k=1
2^4<= 4! ?
16<=24 ? YES

Step 2: assume it is true for k=n

or 2^(n+3) <= (n+3)! is true

Step 3: test to see if it is then true for k= n+1

or, is 2^(n+4) <= (n+4)! ??

subtract the two inequations (step 4 minus step3)

is 2^(n+4) - 2^(n+3) <= (n+4)! - (n+3)! ???

the left side factors to
2^(n+3)(2-1)
or 2^(n+3)

the right side is (n+4)*(n+3)! .... {just like 8! - 7! = 8*7!}

so we are looking at:

is 2^(n+3) <= (n+4)*(n+3)! ??

Well, the left side is the original left side of step 2, and the right side is the original right side of step 2 multiplied by a positive number > 1.

Since the right side was already >= the left side, muliplying it by a positive integer would cerainly make it even larger.



thank you lots!!

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