Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

maths question 10 points =)?

prove the following by induction:

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

2 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    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
    • Login to reply the answers
  • Anonymous
    1 decade ago

    at the end of the question

    • Login to reply the answers
Still have questions? Get your answers by asking now.