anybody here a math major?

[1800bigk]

any body? I have a question

[stclark816]

Talk to Brooke, I think she used to tutor at her college.

[1800bigk]

is brooke her hq name or is that her real name?

[SICK BOY]

Brooke a tutor?Now that's hot.

[chris642]

is brooke her hq name or is that her real name?

322877[/snapback]

hq name

[frocashmoney24]

im a master at flunking math. im sure theres a few hidden smart ones on the ol hq

[racer]

im pretty handy with a calc, what do you need to know?

[NugShee]

you have to first ask us the question before we can give you an answer.

[1800bigk]

its an analysis question

 

let a,b,c be integers,

prove that if a divides (bc) then a divides b or a divides c.

 

i cant get past the first step

[SICK BOY]

[racer]

proofs suck. sorry bro, i cant help you with that. if you have any trig shit lemme know.

[1800bigk]

yeah they blow, way too much thinking involved, this is the only one I cant get on my take home test. I searched everywhere

[cb67rs1]

Let a and b be nonzero integers. Prove that there is a

natural number m such that

(i) ajm and bjm, and

(ii) if c is an integer such that ajc and bjc, then mjc.

Proof. Let S = fn 2 N : ajn and bjng. Since ab 2 S, S 6= ;. Thus,

by the Least-Natural-Number-Principle, there is a smallest element of

S. Let m be this smallest element. Now suppose ajc and bjc. Then

c 2 S and so c m. So by the division algorithm for integers, there

exist integers p; r such that 0 r < m such that c = mq+r. Since ajm

and bjm, then ajr and bjr. If c 6= 0, then r is an element of S smaller

than m. Thus, r = 0 and mjc.

[y2kbanshee9187]

is this a math course, if so, what math are you in? I am in PDM(Precalculous and Discrete Mathematics). My teacher always said with proofs in PDM, you start the first step putting the opposite of what your trying to prove. I can ask my teacher tommorow, she will have to know.

[1800bigk]

yeah its a math course its called Real Analysis, its a real pain in the ass.

putting the opposite of what I want to prove first is called proof by contrapositve ,and with this question since what I have to prove is compound, it makes it messy to do it that way.

Edited February 14, 2005 by 1800bigk