Classes - Page 4

davieddy · 2012-09-16, 12:21

Quote:

Originally Posted by gollin

You could always come see me to ask about an approach to the problem-- GG

Welcome to the Pleasuredome.

I'm sure Paul will correct me, but a good model of diatomic molecules involves a potential that goes as a/x¹² - b/x⁶.

The first term would best be explained by Fermi or Pauli.
The second is Van der Waals.

Tell your star pupil that he will be thrilled with Lagrange

David

Dubslow · 2012-09-18, 06:06

(I'll continue to post my hw questions here until someone says otherwise.)

Let a,b,m,n be positive integers, let p be prime. Then p^a exactly divides n (p^a||n) if p^a|n and p^a+1 -|- n.
Let p^a||m and p^b||n.

What power of p exactly divides m+n?

The answer is clearly min(a,b), except that the book says that a=b => no general answer.
I fail to see why: If p^a||m and p^a||n, then p^a|m+n (we have a linear combination lemma called Proposition 1.2). Also, p^a+1 -|- m => p^a+1 -|- m+n WLOG. So what gives?

axn · 2012-09-18, 06:18

Quote:

Originally Posted by Dubslow

Also, p^a+1 -|- m => p^a+1 -|- m+n WLOG.

False. Consider m=5, n=20.

EDIT:- This is very relevant for aliquot sequences!

Dubslow · 2012-09-18, 06:27

Quote:

Originally Posted by axn

False. Consider m=5, n=20.

EDIT:- This is very relevant for aliquot sequences!

Okay, but then is that true for a!=b?

axn · 2012-09-18, 06:36

Quote:

Originally Posted by Dubslow

Okay, but then is that true for a!=b?

Is what true?

Quote:

Originally Posted by Dubslow

The answer is clearly min(a,b), except that the book says that a=b => no general answer.

The text book is correct. I've shown counterexample to your assertion. The only thing we can say when a=b is that the power of p is at least a.

Dubslow · 2012-09-18, 06:54

Quote:

Originally Posted by axn

Is what true?

Quote:

Originally Posted by Dubslow

Also, p^a+1 -|- m => p^a+1 -|- m+n WLOG.

Since this fails for a==b, is it still true for a != b (since the standard rule applies)? What about a==b makes it fail? (It's one thing to see a counter-example, it's another to understand why.)
Actually, that statement doesn't make any reference to b, so then my proof for a != b is wrong...

axn · 2012-09-18, 07:25

Quote:

Originally Posted by Dubslow

Let a,b,m,n be positive integers, let p be prime.

One observation. The problem (and result/proof) remains valid if we extend it to a,b>=0 and m,n any integer.

Another observation. If p=2, a=b, then the power of p that divides m+n is at least (a+1) (instead of 'a' in the general case).

jyb · 2012-09-18, 18:59

Quote:

Originally Posted by Dubslow

Since this fails for a==b, is it still true for a != b (since the standard rule applies)? What about a==b makes it fail? (It's one thing to see a counter-example, it's another to understand why.)
Actually, that statement doesn't make any reference to b, so then my proof for a != b is wrong...

Yes, it is still true when a != b.

Remember that if x | y and x -|- z, then x -|- (y+z). Whereas if x -|- y and x -|- z then we don't really know whether x | (y+z).

In your case, WLOG let a > b (i.e. min(a,b) = b). Then p^b+1 | m and p^b+1 -|- n. So p^b+1 -|- (m + n). But of course p^b | (m+n). So p^b || (m+n).

But if a = b, then we just have p^b+1 -|- m and p^b+1 -|- n, so we don't know whether p^b+1 | (m+n).

Dubslow · 2012-09-18, 19:03

Quote:

Originally Posted by jyb

I guess I just need to convince myself of this. (I'm sure it's not that hard. It's just that we haven't explicitly covered these rules of non-divisibility, only the rule of divisibility.)

jyb · 2012-09-18, 20:18

Quote:

Originally Posted by Dubslow

I guess I just need to convince myself of this. (I'm sure it's not that hard. It's just that we haven't explicitly covered these rules of non-divisibility, only the rule of divisibility.)

Ah, well then good! Try proving this, just using the definition of divisibility (and a little algebra, of course):

Code:

If x | y and x -|- z, then x -|- (y+z).

Hints available if needed.

And find examples that go both ways for this one:

Code:

If x -|- y and x -|- z, then sometimes x | (y+z) and sometimes x -|- (y+z).

davieddy · 2012-09-18, 22:41

Quote:

Originally Posted by gollin

You could always come see me to ask about an approach to the problem-- GG

When they are totally familiar with keeping equations dimensionally correct, theoretical physicists are happy to let hbar = c = e = 1.

But in your problem, the distance scale has to be chosen in a problem-specific way to render (1/y² - 1/y) sensible.
Furthermore you call it "potential" which presumably means potential energy per unit mass, but then inform us that the mass is m.

David

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2012-09-18, 06:06	#35
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 3×29×83 Posts	(I'll continue to post my hw questions here until someone says otherwise.) Let a,b,m,n be positive integers, let p be prime. Then p^a exactly divides n (p^a\|\|n) if p^a\|n and p^a+1 -\|- n. Let p^a\|\|m and p^b\|\|n. What power of p exactly divides m+n? The answer is clearly min(a,b), except that the book says that a=b => no general answer. I fail to see why: If p^a\|\|m and p^a\|\|n, then p^a\|m+n (we have a linear combination lemma called Proposition 1.2). Also, p^a+1 -\|- m => p^a+1 -\|- m+n WLOG. So what gives?