Classes

[Dubslow]

While I'm waiting for that poly-select to finish, I thought I'd post my provisional (read: desired) schedule for next semester (FA12). There's probs a less than 50% chance it will actually be this, but anyways:

MATH 453: Elementary Theory of Numbers

Basic introduction to the theory of numbers. Core topics include divisibility, primes and factorization, congruences, arithmetic functions, quadratic residues and quadratic reciprocity, primitive roots and orders. Additional topics covered at the discretion of the instructor include sums of squares, Diophantine equations, continued fractions, Farey fractions, recurrences, and applications to primality testing and cryptopgraphy.

MATH 423: Differential Geometry

Applications of the calculus to the study of the shape and curvature of curves and surfaces; introduction to vector fields, differential forms on Euclidean spaces, and the method of moving frames for low- dimensional differential geometry.

PHYS 325: Classical Mechanics I

Kinematics and dynamics of classical systems, including a review of Newtonian kinematics and dynamics. Three dimensional motion, variable mass, and conservation laws; damped and periodically driven oscillations; gravitational potential of extended objects and motion in rotating frames of reference; Lagrangian and Hamiltonian mechanics.

PHYS 435: Electromagnetic Fields I

Static electric and magnetic fields, their interactions with electric charge and current, and their transformation properties; the effect of special relativity is incorporated. Macroscopic fields in material media are described.

MUS 133: Introduction to World Music

A survey of various musical traditions from different regions and peoples of the world.
For music and non-music majors.

Edit: Okay, so I only posted for MATH 453 :P

[Dubslow]

Heh, here's question one of my number theory hw due tomorrow:
26. Prove or disprove the following statements.
a) If p is a prime number, then 2^p-1 is a prime number.
b) If 2^p-1 is prime, then p is prime.

Last week I had to prove that if a^n-1 is prime, n is prime and a==2.

[LaurV]

Haha, that is nice!
But from your former posts I understood (and I was totally convinced) that you go to some college/university, not to kindergarten...

[Dubslow]

Originally Posted by LaurV

Haha, that is nice!
But from your former posts I understood (and I was totally convinced) that you go to some college/university, not to kindergarten...

Hey, that's only two weeks of class, give us time... (I'm half-hoping we'll be able to get into the workings of NFS, but I'm rather doubtful, especially that our scope is broad/deep enough for NFS.)

[Raman]

Originally Posted by Dubslow

Heh, here's question one of my number theory hw due tomorrow:
26. Prove or disprove the following statements.
a) If p is a prime number, then 2^p-1 is a prime number.
b) If 2^p-1 is prime, then p is prime.

Last week I had to prove that if a^n-1 is prime, n is prime and a==2.

Come on, guess what following statements implies the other thing with, given p is being a prime ≥ 3.

Γ: 2 is a primitive root (mod p)
Λ: 2 is a quadratic non-residue (mod p)
Σ: p ≡ 3, 5 (mod 8)
Ω: p is NOT a prime factor of some Mersenne number 2^q-1, with q < p-1

For these four things, you can be able to be, being include the NOT clause for these following four statements, given p is being a prime ≥ 2.
Π: p ≡ 3 (mod 4)
Ψ: p is the sum of two squares
Θ: 2p+1 is prime, with 2p+1 | 2^p-1
Φ: 2p+1 is prime, with 2p+1 | 2^p+1

By the way, this was just simply being a fun problem I gave, don't mind it taking it, solving it very seriously, during the time periods.

[Dubslow]

This really belongs in the Puzzles forum, but here goes...

Another of the problems from the homework I turned in today was the following:

Prove that every integer greater than 6 can be expressed as the sum of two relatively prime integers greater than 1.

We have a comparatively small basis of material we've covered with which to derive a proof; such includes that
-- a|b and a|b --> a|mb+nc for some m,n in Z;
-- if a=qb+r for a given a,bq,r in Z, then 0<=r<b, and q and r are unique;
-- we've defined GCD(a,b), and
-- proven that GCD(a,b) = min{ma+nb | m,n in Z, ma+nb>0}, i.e. that GCD(a,b) is the smallest positive integer that's a linear integral combination of a and b (and in particular, such a combination exists).

The (semi-constructive) proof I came up with at 3am only depended on what we've covered, except that I needed Chebyshev's theorem. (I can reproduce it if anyone desires.)

Is there a proof of the statement without relying on that theorem?

[science_man_88]

Originally Posted by Dubslow

This really belongs in the Puzzles forum, but here goes...

Another of the problems from the homework I turned in today was the following:

We have a comparatively small basis of material we've covered with which to derive a proof; such includes that
-- a|b and a|b --> a|mb+nc for some m,n in Z;
-- if a=qb+r for a given a,bq,r in Z, then 0<=r<b, and q and r are unique;
-- we've defined GCD(a,b), and
-- proven that GCD(a,b) = min{ma+nb | m,n in Z, ma+nb>0}, i.e. that GCD(a,b) is the smallest positive integer that's a linear integral combination of a and b (and in particular, such a combination exists).

The (semi-constructive) proof I came up with at 3am only depended on what we've covered, except that I needed Chebyshev's theorem. (I can reproduce it if anyone desires.)

Is there a proof of the statement without relying on that theorem?

For me personally using a basic arithmetic property and a basic property of GCD I can prove for odd numbers, it's the even numbers that have me messed up. never mind I'm used to greatest common divisor but I don't think that's what you mean.

[ewmayer]

Originally Posted by Dubslow

Prove that every integer greater than 6 can be expressed as the sum of two relatively prime integers greater than 1.

How much "material" do you need for that?

1. If n odd and n > 3, n = 2 + k with k odd and k > 1, hence 2,k coprime;

2. If n even and n > 4 with n == 0 (mod 4), we have n = (n/2-1) + (n/2+1) with each summand odd and > 1. Since the 2 summands differ by 2, they cannot share any odd factors;

3. If n even and n > 6 with n == 2 (mod 4), we have n = (n/2-2) + (n/2+2) with each summand odd and > 1. Since the 2 summands differ by 4, they cannot share any odd factors.

[science_man_88]

Originally Posted by ewmayer

How much "material" do you need for that?

1. If n odd and n > 3, n = 2 + k with k odd and k > 1, hence 2,k coprime;

2. If n even and n > 4 with n == 0 (mod 4), we have n = (n/2-1) + (n/2+1) with each summand odd and > 1. Since the 2 summands differ by 2, they cannot share any odd factors;

3. If n even and n > 6 with n == 2 (mod 4), we have n = (n/2-2) + (n/2+2) with each summand odd and > 1. Since the 2 summands differ by 4, they cannot share any odd factors.

interesting I was going towards:

odd1=even+odd2=floor(odd1/2)+ceiling(odd1/2) these 2 values have a difference of one. since for a>b, GCD(a,b) must divide a-b and a-b = 1 GCD(a,b)==1 hence they are coprime.

[Dubslow]

Originally Posted by ewmayer

2. If n even and n > 4 with n == 0 (mod 4), we have n = (n/2-1) + (n/2+1) with each summand odd and > 1.

3. If n even and n > 6 with n == 2 (mod 4), we have n = (n/2-2) + (n/2+2) with each summand odd and > 1.

See, I thought it couldn't possibly be that hard. That's why I asked. (The formulas for n are not yet apparent to me... :razz:) Nevermind, I see how they work now. (The odd case was clear to me from the beginning, but I dislike "case-based" proofs. The proof I turned in doesn't depend on the factors of n )

[science_man_88]

Originally Posted by Dubslow

See, I thought it couldn't possibly be that hard. That's why I asked. (The formulas for n are not yet apparent to me... :razz:) Nevermind, I see how they work now. (The odd case was clear to me from the beginning, but I dislike "case-based" proofs. The proof I turned in doesn't depend on the factors of n )

so you hate Mathematical induction I assume, also aren't phrasings like "suppose..." kinda case based if so it looks like you hate proof by contradiction

also:

In proof by exhaustion, the conclusion is established by dividing it into a finite number of cases and proving each one separately.

quite a few types of mathematical proof you seem not to like.