Yahoo Answers: Answers and Comments for Maths question 10 points =)? [Mathematics]
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From Anonymous
enUS
Sat, 07 Nov 2009 14:59:48 +0000
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Yahoo Answers: Answers and Comments for Maths question 10 points =)? [Mathematics]
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From Anonymous: at the end of the question
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https://answers.yahoo.com/question/index?qid=20091107145948AAwO3YN
Sat, 07 Nov 2009 15:03:42 +0000
at the end of the question

From edgelets: Pf: Let k = 8. Let m = (8 + n), where n is a n...
https://answers.yahoo.com/question/index?qid=20091107145948AAwO3YN
https://answers.yahoo.com/question/index?qid=20091107145948AAwO3YN
Sat, 07 Nov 2009 15:23:39 +0000
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