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# maths question 10 points =)?

prove the following by induction:

m! > 4^m where m>k, where k can be found

Relevance

Pf: Let k = 8. Let m = (8 + n), where n is a natural number.

Claim: m! > 4^m.

Part 1: True for n = 1.

If n = 1, m = 9.

9! = 362,880 > 262,144 = 4^9.

Thus, m!>m^4 for n=1 is true.

Part 2: If true for n, true for n+1.

Assume true for n a natural number.

m! = (8+n)! > 4^(8+n) = 4^m

Multiply both sides by (n+9), a positive number:

(9+n) (8+n)! = (9+n)! > (9+n) 4^(8+n)

We have (9+n) > 4 because n is positive. Multiply both sides by 4^(8+n), a positive number:

(9+n) 4^(8+n) > 4 ( 4^(8+n) ) = 4^(9+n)

"Greater than" is an equivalence relation, so we can combine the two in equalities to find:

(9+n)! > 4^(9+n)

To further drive the point home, we can even write it like this:

(8 + (n+1) )! > 4^(8 + (n+1) )

Thus, m! > 4^m is true for n+1.

By the induction hypothesis, the statement is true for all n natural numbers (positive integers).

So it's true for all m = (8+n) where n is a positive integer. Thus, it is true for all m > k = 8.

qed

Source(s): math major
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