1. ## Inequalities

Can anyone help confirm if I have solved this correctly. Specifically the part preceded by A, where I have defined the coefficient as 2k. I had originally thought that it might to be 2k-1, but subbing 3 into that version yielded negative answers.

Many thanks.

Q.
n! > 2n-1 for n > 3, n $\in \mathbb{N}$

Attempt: Step 1: For n = 3...
3! = 1 x 2 x 3 = 6 & 23-1 = 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! > 2k-1
We must now show that the statement is true for n = k + 1,
i.e. (k + 1)! > 2(k+1)-1 = (k + 1)! > 2k
(k + 1)! = (k + 1)k! > (k + 1)2k-1...(k! > 2k-1...assumed)
If it can be shown that (k + 1)2k-1 > 2(k+1)-1, then (k + 1)! > 2k
If (k + 1)2k-1 > 2k
If (k + 1)2k-1 - 2k > 0
A: If 2k[(k + 1)2k-2 - 1] > 0...true when > 3
(k + 1)! > 2k
Therefore, the statement is true for n = k + 1, if true for n = k. Thus the statement is true for all n > 3, n $\in \mathbb{N}$

2. ## Re: Inequalities

Your argument preceding step A is correct but the following step is incorrect
Originally Posted by GrigOrig99
A: If 2k[(k + 1)2k-2 - 1] > 0...true when > 3
it has to be $2^k[\frac{(k+1)}{2} -1] > 0$.

~Kalyan.

3. ## Re: Inequalities

Originally Posted by GrigOrig99
i.e. (k + 1)! > 2(k+1)-1 = (k + 1)! > 2k
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.

Originally Posted by GrigOrig99
If (k + 1)2k-1 - 2k > 0
A: If 2k[(k + 1)2k-2 - 1] > 0...true when > 3
(k + 1)2k-1 - 2k ≠ 2k[(k + 1)2k-2 - 1] because 2k[(k + 1)2k-2 - 1] = (k + 1)22k-2 - 2k. Rather, (k + 1)2k-1 - 2k = 2k-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)2k-1 - 2k > 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)2k-1 > (since k + 1 > 2 for k ≥ 3)
2 * 2k-1 =
2k

4. ## Re: Inequalities

Great. Thank you.