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#1 |
I Floop the Pig
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Math geeks only
- OR -
Yes, I am geeky enough to be kept awake at night by number theory I need some formal proof help. Never my strong suit. I grasp the general concept, but when it comes to being properly formal and pedantic, I get a bit fuzzy. What I'm struggling with now is whether a certain kind of deductive conclusion is permissable, or if I need more rigor. What I'm trying to prove is that for any integer value of n > 2, n*7 < 10^(n-1). It's certainly a true statement, but I'm trying to formalize it. The best I've come up with is to know that for the value n=3 it is true and that the expression on the left grows linearly while the expression on the right grown exponentially. Is that sufficient proof? That given any 2 functions f and f` such that f(n) < f`(n) AND f grows linearly while f` grows exponentially, f(x) < f`(x) for all values > n?
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#2 |
I throw stones at houses
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Not a math geek, but is 2 the absolute top of the non-qualifying values for n in this? I mean, short of testing it myself, is there any value of n=2.x that might also be disqualified?
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#3 |
I Floop the Pig
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I specified integer values only.
But to answer your question, yes. There is a non-integer value (that I'm ill equipped to work out at the moment) of n, between 2 and 3, for which n*7 = 10^(n-1). Below that value, the result is > 10^(n-1), above it, it's > 10^(n-1).
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'He who receives an idea from me, receives instruction himself without lessening mine; as he who lights his taper at mine, receives light without darkening me.' -TJ Last edited by Ghoulish Delight : 06-18-2008 at 12:29 AM. |
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#4 |
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Ah, that would be me forgetting that integer meant a whole number
(Proof that I'm not a math geek... now ask me the integer value that represents how many times I had to take algebra II in order to pass it. Hint: n>2)
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#5 |
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Doesn't it also work for zero?
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#6 | |
I Floop the Pig
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Quote:
But in this particular case, we do happen to be dealing with only whole numbers.
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#7 |
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That wouldn't nullify the equation though. He's stating for all integer values greater than 2, but not addressing values 2 and under. Right?
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#8 |
I Floop the Pig
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Yes it does work for zero (7*0 = 0, 10^-1 = .1). However, I'm not concerned with any of the infinite other values of n may or may not work. All I'm interested in proving is that for intergers > 3, it's true. Anything else is irrelevant to the overall problem.
ETA: Thank you Mori, precisely.
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#9 |
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How about proving it with a curve? Does that do anything for you? I mean, demonstrating that one increases linearly while the other increases exponentially... a curve would demonstrate that, yes?
(ETA: I can't believe I have even this many posts in a math thread... did I mention n>2?)
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Then how much of a number theory problem is it?
Is "For all C > -40: 9/5 * C + 32 > C" a number theory problem?
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