Can anyone help confirm if I have solved this correctly. Specifically the part preceded by A, where I have defined the coefficient as 2^{k}. I had originally thought that it might to be 2^{k-1}, but subbing 3 into that version yielded negative answers.
Many thanks.
Q. n! > 2^{n-1} for n > 3, n
Attempt: Step 1: For n = 3...
3! = 1 x 2 x 3 = 6 & 2^{3-1 }= 4
Since 6 > 4, the statement is true for n = 3.
Step 2: Assume the statement is true for n = k, i.e. assume k! > 2^{k-1}
We must now show that the statement is true for n = k + 1,
i.e. (k + 1)! > 2^{(k+1)}^{-1} = (k + 1)! > 2^{k}
(k + 1)! = (k + 1)k! > (k + 1)2^{k-1}...(k! > 2^{k-1}...assumed)
If it can be shown that (k + 1)2^{k-1} > 2^{(k+1)-1}, then (k + 1)! > 2^{k}
If (k + 1)2^{k-1} > 2^{k}
If (k + 1)2^{k-1} - 2^{k} > 0
A: If 2^{k}[(k + 1)2^{k-2} - 1] > 0...true when > 3
(k + 1)! > 2^{k}
Therefore, the statement is true for n = k + 1, if true for n = k. Thus the statement is true for all n > 3, n
It's OK to write = between numbers, but between propositions (which can be true or false). The notation "x > y = y' > z" is confusing and ambiguous. Use "iff" or "⇔" to claim that two propositions are equivalent.
(k + 1)2^{k-1} - 2^{k} ≠ 2^{k}[(k + 1)2^{k-2} - 1] because 2^{k}[(k + 1)2^{k-2} - 1] = (k + 1)2^{2k-2} - 2^{k}. Rather, (k + 1)2^{k-1} - 2^{k} = 2^{k-1}(k + 1 - 2).
Finally, an (informal) proof should be a correct English text. In particular, all sentences should be complete. You use phrases like "If (k + 1)2^{k-1} - 2^{k} > 0" that are not complete sentences. I believe you showed an example from the book that said "A if B if C ..." meaning that C implies B, which implies A. This at least can be considered as a complete sentence. In your case, "If" with a capital "I" starts a new sentence with no "then." I prefer a chain of equalities and inequalities like this:
(k + 1)! = (k +1)k! > (by IH)
(k + 1)2^{k-1} > (since k + 1 > 2 for k ≥ 3)
2 * 2^{k-1} =
2^{k}