Mathematics - 2013-03-26 (page 3 of 3)

« first day (965 days earlier) ← previous day next day → last day (4353 days later) »

Peter Tamaroff

23:00

@caveman Can't you do some integration swapping using some theorems?

caveman

how

Peter Tamaroff

@caveman this?

caveman

how does that help

Peter Tamaroff

No, no that.

pourjour

Q: Proving $3n^3-11n+48 | n+3$

I'm really stuck while I'm trying to prove this statement: $\forall n \in$ $\mathbb{N}$ $3n^3-11n+48 | n+3$. I couldn't even how to start.

Gustavo Bandeira

23:08

@pourjour What's the meaning of the |?

pourjour

@GustavoBandeira it means $3n^3-11n+48 divise n+3$

caveman

@mixedmath, not at all pleased with your comment on my answer

Gustavo Bandeira

What's the meaning of $\stackrel{?}{=}$ in $P\stackrel{?}{=}NP$?

@pourjour Divided by n+3?

pourjour

@GustavoBandeira the opposite

Gustavo Bandeira

@pourjour ?

pourjour

23:12

@GustavoBandeira n+3 divised by 3n^3-11n+48

it seems weird

ok so could someone please show how to prove that

caveman

@pourjour, dont you know long division

pourjour

@caveman I mean n+3 divised by 3n^3..............

caveman

@pourjour, you have ti backwards

d|n means there exists k with dk = n

pourjour

@caveman yeah

I know what it means and I know what I've written

caveman

@pourjour, then why can you see that it's obviously false?

pourjour

23:17

@caveman contre example

caveman

@pourjour, 48 doesn't divide 3 for example

ugh

pourjour

sorry

I'm really stupid

caveman

@pourjour, yuo can accept my answer?

pourjour

I couldn't notice that because I tough it's obvious

caveman

I really dislike mixedmaths comment on my answer

he was here and ignored me when I told him that

thanks

pourjour

23:21

@caveman u deserve it

caveman

@mixedmath, it's called long division - most learn in school.. Maybe we can let the OP decide for himself what helps him. — caveman 1 min ago

pourjour

$\rm (ac,db) \color{#C00}{\stackrel{(1)}{=}} (ac,\color{#0A0}{(d,ac)}\color{blue}{(b,ac)})$ can you explain to me this

math.stackexchange.com/questions/342092/…

caveman

@pourjour, its just using (X,YZ) = (X,(X,Y)(X,Z))

which is deduced from (X,YZ) = (X,(X,YZ)) and (X,Y)(X,Z)|(X,YZ)

pourjour

@caveman but this line mean Y=(X,Y) and Z=(X,Z)

caveman

pourjour

23:32

can u please translate it to $(A|B)$

I'm not familiar with GCD

caveman

did you see my answer to your question?

A: number theory $\gcd(a,bc)=\gcd(a,c)$

Definition The gcd $g = (x,y)$ is the universal $g$ satisfying $g|x$ and $g|y$. Universal here means that if $g'$ satisfies $g'|x$ and $g'|y$ then $g'|g$. Note that universals are unique. Theorem $(a,b)=1$ implies $(a,bc) = (a,c)$. We need to show that $g=(a,c)$ is the universal $g$ such that $g...

pourjour

@caveman it seems a little bit complicated

caveman

I think you should understand it before saying that

@pourjour, here is a different way math.stackexchange.com/a/342228/58512

if you imagine a,b,c written in prime factorization

b doesn't h ave any primes that a does

so (a,bc) = (a,c)

you just ignore b

since it doesn't share any primes with a

pourjour

@caveman that's the problem becasue I'm trying to show that $d/bc ==> d/c$ using gauss theorem but I need (d,b)=1

caveman

what

pourjour

23:45

to show this

so (a,bc) = (a,c)

we suppose that d1=(a,bc)

and d2 = (a,c)

caveman

(d,b) = 1 and d|bc implies d|c?

pourjour

@caveman yes Gauss theorem am I wrong

caveman

I prove that here math.stackexchange.com/a/342123/58512

what is Gauss theorem?

(Gauss proved many things)

Peter Tamaroff

@caveman Many? A gazillion things!

caveman

yeah :)

pourjour

23:48

hhh yeah exactly the one which says a|bc and (a,b)=1 implies a/c

ok is there any lemma such (a,b)=d an d(a,c)=d implies (a,bc)=d

caveman

@pourjour,

> let $g\mid a$ and $g\mid bc$, we will show that $g|(a,c)$. By $(a,b)=1$ if $g\mid b$ then $g=1$, so suppose $g \not \mid b$, then $g \mid c$ and we are done.

if $a|bc$ and $(a,b)=1$ then we have some $k$ with $ak = bc$ and wish to prove there is some $h$ such that $ah = c$

since $ak=bc$ we can divide by $b$ to get $a(k/b)=c$ and $k/b$ is an integer

so take $h=k/b$

pourjour

but a can't divise b and c at the same time this is an exception

caveman