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Proposition:
for all n ≥ 2.Proof: If n=2, we have
or , which is true.I can't get k+1 in the base of the RHS! Help plz
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Estimates are a mathematician's best friend, especially if you are studying analysis. Don't be trying to show equality. Rather, show inequality:
(k+1)k^k < (k+1)(k+1)^k
This should be fairly straightforward.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Thanks heaps, could you please help me on another proof?
Exercise 2.9. Prove that if n ≥ 12 then n can be written as a sum of 4s and 5s. For example, 23 = 5 + 5 + 5 + 4 + 4 = 3 · 5 + 2 · 4. [Hint. In this case it will help to do the cases n = 12, 13, 14, and 15 separately. Then use induction to handle n ≥ 16.]
If n = 5a + 4b for some integers a and b, what will n+1 be equal to? n+1 = 5c+4d? Then 5c+4d-1=5a+4b... I don't think I'm getting anywhere.
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So the trick here is that 5 - 4 = 1. That is, if you take out one 4, and you add in one 5, you will get +1.
3*4 + 2*5 = 22
2*4 + 3*5 = 23
Now that is not the entire problem, but it's part of it. What if you have 5*x, i.e. only 5's? The above trick won't work. But what will?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Hmm, if n = 5x, then n+1 = 5(x-3)+4(4)
n = 4x, then n+1 = 4(x-1)+5
if n = 5x+4x, then n+1 = 5(x-1)+4(x+1)
So do I have to make cases for n= 5x, 4x, 5x+4x and evaluate them separately? Do they cover all possible cases?
Last edited by Identity (2007-09-17 03:36:42)
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Nah, just think about it this way:
case 1: n contains at least 1 4.
case 2: n contains no 4's.
Now get n+1 in each case.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Thanks for your help, I've finished the induction chapter
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