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bah! Induction

Alkatran

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I didn't realise induction was so tricky to get the hang of. Now I've just finished my homework (for linear algebra and discrete structures) (due tomorrow) over the past 4 hours.

Sigh... oh well. At least it lets me prove things about numbers...

I've already solved this problem (the trick is to use the 3<=n inequality) but here it is for fun: Is n^3 <= 3^n for all n contained in the natural numbers? It gets really close to breaking it at n=3...
 
Now prove that n^p <= p^n for all n contained in the natural numbers is only true for p=3. :)
 
Art Vandelay said:
Yes, you did. Think about it a bit more.
Click to expand...

No, I didn't. I can tell you right now I can't conclude for all p, n^p <= p^n for all n, implies n = 3 from the proof I have.

I ONLY proved the case where p = 3. I didn't prove p = 3 was the only case, I only showed it was A case.
 
What I said isn't precisely correct. If we reformulate the statement as "n^3 <= 3^n for all n, with a strict inequality for n != 0", then it is equivalent to saying that it is the only such case.

"I ONLY proved the case where p = 3. I didn't prove p = 3 was the only case, I only showed it was A case."
But there can only be one case. Suppose that there is someone who is taller than everyone else. There can only be one such person, no?

Suppose I were to say that n is "greater" than p if n^p > p^n. Under this definition, you've proved that 3 is "greater" than every other number, right? So you can't have another number that is also "greater" than every other number.
 
Alkatran said:
Is n^3 <= 3^n for all n contained in the natural numbers? It gets really close to breaking it at n=3...
Click to expand...
If you let n be real instead of just integral, you can break it. Let n = 2.9, for example.

Exercise: What number should replace 3, if you don't want it to break for any real n > 0?
 
Art Vandelay said:
What I said isn't precisely correct. If we reformulate the statement as "n^3 <= 3^n for all n, with a strict inequality for n != 0", then it is equivalent to saying that it is the only such case.

"I ONLY proved the case where p = 3. I didn't prove p = 3 was the only case, I only showed it was A case."
But there can only be one case. Suppose that there is someone who is taller than everyone else. There can only be one such person, no?

Suppose I were to say that n is "greater" than p if n^p > p^n. Under this definition, you've proved that 3 is "greater" than every other number, right? So you can't have another number that is also "greater" than every other number.
Click to expand...

So it's a logical consequence. I still need to extend the proof to show it. (I mean, technicly, all things in math are equivalent to the axioms being true...)

I see your point now. For all p, the case where n = 3 is an exception.
 
Art Vandelay said:
What I said isn't precisely correct. If we reformulate the statement as "n^3 <= 3^n for all n, with a strict inequality for n != 0", then it is equivalent to saying that it is the only such case.

"I ONLY proved the case where p = 3. I didn't prove p = 3 was the only case, I only showed it was A case."
But there can only be one case. Suppose that there is someone who is taller than everyone else. There can only be one such person, no?

Suppose I were to say that n is "greater" than p if n^p > p^n. Under this definition, you've proved that 3 is "greater" than every other number, right? So you can't have another number that is also "greater" than every other number.
Click to expand...
Maybe slow down a bit for me. I'm getting confused.

Where n=p, you will always have n^p = p^n. But why it is the case that for n^p <= p^n, where n and p are integers, that this is only true for all n when p=3? Or better yet, look at real numbers for p and n. It seems to come out at p=2.5... Any idea why?
 
DevilsAdvocate said:
Maybe slow down a bit for me. I'm getting confused.

Where n=p, you will always have n^p = p^n. But why it is the case that for n^p <= p^n, where n and p are integers, that this is only true for all n when p=3? Or better yet, look at real numbers for p and n. It seems to come out at p=2.5... Any idea why?
Click to expand...

As the other poster pointer out, IF it is true for ALL natural numbers n than n^3 <= 3^n then there is counter example for all natural numbers p if p != 3. (set n to 3 and you know the inequality goes the other way).

Come to think of it, that's not right. If n^p = p^n (which we haven't disproven) then this won't hold.
 
What kind of equations do you call these things, anyway? And what good do they do a person? Just curious.
 
Iamme said:
What kind of equations do you call these things, anyway? And what good do they do a person? Just curious.
Click to expand...

They're called mathematical or logical equations. And ALL of math is proven to be true (given that the axioms are true) using them. How do you know 2+2=4? Because you can prove it based on the definition of numbers.

In fact, math is really one big tautology in the form of implications.
 
Alkatran said:
As the other poster pointer out, IF it is true for ALL natural numbers n than n^3 <= 3^n then there is counter example for all natural numbers p if p != 3. (set n to 3 and you know the inequality goes the other way).

Come to think of it, that's not right. If n^p = p^n (which we haven't disproven) then this won't hold.
Click to expand...
Right. As Art Vandelay said, you need to show that n3 < 3n for all natural numbers n except 3. But you can do this using an induction proof that's very similar to the one you already used.

By the way, 24 = 42.
 
69dodge said:
Exercise: What number should replace 3, if you don't want it to break for any real n > 0?
Click to expand...

I'm guesing e but I really can't be arsed to check right now but I don't see it being anything else.
 
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