Classes - Page 5

wblipp · 2012-09-19, 00:11

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.)

When confused, try a simple case. let n=4x and and m=4y, x and y odd. You know that 4 divides n+m, but so does 8 because 2 divides x+y, and 16 might, depending on whether or not 4 divides x+y. Generalize.

Dubslow · 2012-09-19, 08:22

Quote:

Originally Posted by jyb

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).

Quote:

Originally Posted by wblipp

When confused, try a simple case. let n=4x and and m=4y, x and y odd. You know that 4 divides n+m, but so does 8 because 2 divides x+y, and 16 might, depending on whether or not 4 divides x+y. Generalize.

I never meant to imply that it was hard or that I required hints. Though it took a good few hours to get around to it, even the initial half-second of thought was more than enough to make the solution clear:

y=0 (mod x) => y+z=z (mod x), but z!=0 (mod x) by hypothesis. If x -|- y, then y+z!=z (mod x), so no general conclusion can be drawn.

Dubslow · 2012-10-10, 07:42

Alright, here's another problem, dealing with a simple generalization of the Chinese Remainder Theorem.

Let x==b1 (mod m1) and x==b2 (mod m2).

It's easy to prove (read: I managed to prove) that a solution x exists iff (m1, m2) | b1 - b2. How do you prove the solution is unique mod [m1,m2]=m1m2/(m1,m2)?

(I'm half expecting the solution to appear in my sleep our something

)

only_human · 2012-10-11, 18:04

Quote:

Originally Posted by Dubslow

So you all, go look at the PHYS 325 description; this is my professor.

Cool article. I googled him a bit. The article should have mentioned the smear campaign against him that lingers on the web. It is so transparent. A good litimus test in my opinion is to be very skeptical of sites that lean heavily on the words truth or facts. His courage, conviction and persistence make him a true mensch; it is awesome to meet people like that. Taking a class from him could be a real treat.

Dubslow · 2012-10-11, 18:12

Quote:

Originally Posted by only_human

Cool article. I googled him a bit. The article should have mentioned the smear campaign against him that lingers on the web. It is so transparent. A good litimus test in my opinion is to be very skeptical of sites that lean heavily on the words truth or facts. His courage, conviction and persistence make him a true mensch; it is awesome to meet people like that. Taking a class from him could be a real treat.

I'm taking the first of his midterms tonight, and he said that while he'll try to make the test about an hour and a half, there will be no time limit: the exam will start at 7, but we can stay until we're kicked out of the building to work on the test.

only_human · 2012-10-11, 23:41

Quote:

Originally Posted by Dubslow

I'm taking the first of his midterms tonight, and he said that while he'll try to make the test about an hour and a half, there will be no time limit: the exam will start at 7, but we can stay until we're kicked out of the building to work on the test.

I'd wish you luck but am sure you will do well. Nice of him to remove time pressure.

Dubslow · 2012-10-11, 23:46

Quote:

Originally Posted by only_human

I'd wish you luck but am sure you will do well. Nice of him to remove time pressure.

No offense to him, but so far it's been more or less review of linear ODEs, admittedly with some fairly complex driving functions. All the homework up to what was handed out today is what my friends and I would call "easy". Today's hw, OTOH, looked significantly harder. (I'm looking forward to it!) So, I'm not worried at all

only_human · 2012-10-11, 23:52

Quote:

Originally Posted by Dubslow

No offense to him, but so far it's been more or less review of linear ODEs, admittedly with some fairly complex driving functions. All the homework up to what was handed out today is what my friends and I would call "easy". Today's hw, OTOH, looked significantly harder. (I'm looking forward to it!) So, I'm not worried at all

And a function of a driven, confiscated Jaguar XK8.

Dubslow · 2012-10-16, 03:50

I got a 92, the average was in the 70s somewhere. (The interested reader could easily use google to find the course website, where he stats are posted.)

More importantly, I need some help understanding the main consequence of the Chinese Remainder Theorem. I understand the 'canonical ' form and statement, but I don't quite get how that implies x==y mod ni iff x==y mod N. I get that x is unique mod N, but I don't see why that solution has to be y.

wblipp · 2012-10-19, 13:56

I'm not sure that I understand the question. But if you are asking what I think you are asking, it's sufficient to observe that y is a solution and the solution is unique. But perhaps you aren't clear about why y is a solution.

If we need further discussion, it would help to have a reference statement of the theorem so that everybody is sure what all the variables mean. Does Wikipedia work, or would you rather post your own statement?

Dubslow · 2012-10-19, 20:39

Quote:

Originally Posted by wblipp

But perhaps you aren't clear about why y is a solution.

Bingo.

Quote:

Originally Posted by wblipp

If we need further discussion, it would help to have a reference statement of the theorem so that everybody is sure what all the variables mean. Does Wikipedia work, or would you rather post your own statement?

Wiki is fine, and is identical to me book, except for my book uses b,m instead of a,n , which doesn't matter.

I guess I need to see that $\sum_i \frac{N}{n_i}\left[\left(\frac{N}{n_i} \right)^{-1}\right]_{n_i} \equiv 1 mod N$ , using Wikipedia's notation. I also don't see why the converse is true, i.e., x==y mod N => x==y mod n_i.

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2012-10-10, 07:42	#47
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 1110000110101₂ Posts	Alright, here's another problem, dealing with a simple generalization of the Chinese Remainder Theorem. Let x==b1 (mod m1) and x==b2 (mod m2). It's easy to prove (read: I managed to prove) that a solution x exists iff (m1, m2) \| b1 - b2. How do you prove the solution is unique mod [m1,m2]=m1m2/(m1,m2)? (I'm half expecting the solution to appear in my sleep our something )

2012-10-16, 03:50	#53
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 3×29×83 Posts	I got a 92, the average was in the 70s somewhere. (The interested reader could easily use google to find the course website, where he stats are posted.) More importantly, I need some help understanding the main consequence of the Chinese Remainder Theorem. I understand the 'canonical ' form and statement, but I don't quite get how that implies x==y mod ni iff x==y mod N. I get that x is unique mod N, but I don't see why that solution has to be y.

2012-10-19, 13:56	#54
wblipp "William" May 2003 New Haven 2·7·13² Posts	I'm not sure that I understand the question. But if you are asking what I think you are asking, it's sufficient to observe that y is a solution and the solution is unique. But perhaps you aren't clear about why y is a solution. If we need further discussion, it would help to have a reference statement of the theorem so that everybody is sure what all the variables mean. Does Wikipedia work, or would you rather post your own statement?