Mathematics

I agreed, @Null. So now understand why you're done with your problem. :P

Hi @Faker

And for (e), I essentially need to find a divergent series that satisfies $\lim na_n=0$…

Mike Miller

are we doing anything fun

DogAteMy is :)

@MikeMiller Lots and lots of beer

12:02 AM

LOL

That requires you consider analysis problems as fun.

We ran out, though; sorry

@MikeMiller MATH! :DDDDDDDDD

Young whippersnapper, DogAteMy.

Sorry, that was a little too saccharine, even for me.

Mike Miller

12:03 AM

I'm fond of analysis, but maybe not this

DogAteMy doesn't seem to be suffering unduly, @Semiclassic.

matter of taste

@Fargle XDDDD!!!1!1

@Brody holds up spork

OK, I vaguely remember $1/(n\ln n)$ being divergent, but I'm not sure why

12:03 AM

You're just angry cuz he came up with the counterexample in a millisecond, @Semiclassic.

mutter

There are different ways to do that, DogAteMy.

@Fargle Wuz dat mean?

It's an age-old Internet thing. Google it, you won't be disappointed.

"Penguin of Doom" copypasta.

12:05 AM

@TedShifrin bear with me. how do I show that $G$ is an element of this space? vectoraddition and scalarmultiplication is pointwise defined for this excercise.

Because $G$ is a coset. Go back and think about my $\Bbb Z/3\Bbb Z$ analogy.

You get to add any vector in $\Bbb R^2$ to things in your $U$ ...

@TedShifrin oh, i thought G should be an element of this space

@Fargle Nice. It's before my time but very satisfying

misconception!

$G$ is an element of $\Bbb R^2/U$, yes.

That quotient space consists of cosets (equivalence classes).

DogAteMy: Are you done? If so, I have one final mumble. If not, I'll wait.

12:07 AM

hehe, charges mah lazr

Fargle isn't much older than you, Brody.

@TedShifrin I need to go, I'll be back later

The title's familiar but not the pasta itself. Probably just missed it.

@TedShifrin Shhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.

No biggie. I'll just mumble Taylor's Theorem, DogAteMy. See ya.

@Fargle: From my perspective, you're both mere children.

12:09 AM

@TedShifrin I'd feel rude making any jokes here.

Good ...

@TedShifrin but this parallel analogy doesn't work anymore in R^3 and planes or? or R^4 and ???

lol

@Null: Sure it does.

more than one direction to be parallel now

12:09 AM

Like a whole one or two semesters ahead of me @Ted, what an old fart he is

It works any time you have a subspace and mod out by it. @Null

Mahmoud

@TedShifrin I'm still not going to give up :D Challenge accepted ! Bye.

Bye, @Mahmoud.

I'm outta here. I have other things to do ...

@TedShifrin thanks alot, and good day

Hey @Null, what does $\Bbb N$ usually denote for you?

the keyboard that lets me type 這个 interfered >.>

12:14 AM

@Brody some numberset

$\{0,1,2,3,\ldots\}$ in particular or no?

@Brody depending on the context, with 0, or without 0, but yes

or without some "starting" elements like 2,3,4...

@Null Gotchya. What about the nonnegative and strictly positive integers? using $\Bbb Z$ of course

@Brody depends really on what is defined as negative

Kk >.>

12:18 AM

@Brody why in particular you ask?

to start some 0 discussion? :D

Goodness, no. Just wondering what conventions you're used to @Null. lol

@Brody well, nonnegative: 0,1,2,3... positive: 1,2,3

I'm "used to" strictly positive integers, but for myself, I take the nonnegative ones.

I have to pull an all day er so I can finish my last abstract algebra homework of the semester ;p

@Null and the symbols for them respectfully?

using $\Bbb Z$ I mean

12:20 AM

I prefer it to be a monoid, rather than a semigroup.

No more of that talk @Fargle

What'd I do?

@Brody the easiest is always $\mathbb{Z}^{+}\cup \{0\}$ or $\mathbb{Z}^{+}\setminus\{0\}$

whatever fits the need

Steve

Here's a rather simple question, but I can't for the life of me remember it right now. How do I convert polar coordinates to rectangular form? (complex numbers)

@Null $\Bbb Z$ already contains 0...

12:21 AM

@Fargle i see

@Null Yes, with $\Bbb Z^+$ you mean?

yep

@Steve If $z = x + iy$, $x = r \cos \theta,\;y = r \sin \theta$.

@Brody i'd just notate it as N personally, to save work

Right, @Null

12:23 AM

not really any different than changing polar coordinates $(r,\theta)$ to cartesian coordinates $(x,y)$

(To go backwards, use that $x^2 + y^2 = r^2$ and that $\theta = \tan^{-1}\left(\frac{y}{x}\right)$, adjusting the angle as needed)

You did nothing @Fargle, just being cheeky :P

@Brody lol, just playing along.

I tend to have $\Bbb N$ as positive integers.

Steve

@Fargle thanks

12:24 AM

No problem @Steve!

possibly because I mostly think of them in the context of number theory, where $0$ isn't a sensible divisor.

On the other hand I typically have my sums start from n=0

I suppose the difference is whether you're thinking more about multiplication or addition. The latter is definitely what I have in mind for sums (e.g. shifting all indices by some amount)

@Brody if anyone thinks he's a genius because he relies on some wierdo definition of positive numbers, he probably know fully well that other people view it different. so fool that.

The positive numbers are precisely those I'd prefer to see in my bank balance @Null

Take this as a definition

@Brody a 1, with lots of zeroes? ;)

how do I find the ideals in mod 6?
ideal is closure under subtraction and closure under multiplication
mod 6 = 0,1,2,3,4,5
so maybe it's closed under subtraction because if I subtract 5 and 4 then 5-4=1, 5-2=3 and that's in mod 6.
and then for closed under multiplication then (5)(1) = 5 in mod 6 ??

12:31 AM

Is there a name for that: the angle between the spanning vectors is the same between all pairs of vectors.

in R^3 this would mean orthogonal i think

but is there a name for that in R^4?

12:41 AM

@usukidoll It can't just be in mod 6, it has to be in the ideal itself.

One way to find ideals: every ideal is also a subgroup of the ring's underlying abelian group.

so it's the ENTIRE ideal with the condition of mod 6

It just has some additional structure (it absorbs multiplication, much like $2\Bbb Z$ does in $\Bbb Z$: any number times an even number gives an even number).

@usukidoll Not sure what you mean by this.

all I can grab right now is that we need to find ideals mod 6
ideal is nonempty set, closure under subtraction and closure under multiplication
mod 6 = 0,1,2,3,4,5

so there's zero which is just the zero ideal

@usukidoll As I said above, try looking for subgroups under addition. These are necessarily going to be closed under subtraction.

then one
ideal for the ring with identity x.x.x

12:45 AM

Then just check and see if they're closed under multiplication by every element in the ring.

@usukidoll By my count, there are 3 more (including the entire ring).

2 would be
2 4

@usukidoll One more element you're missing.

Since every ideal has to be closed under multiplication, it must contain $0 \cdot x$ for some $x \in I$, but $0 \cdot x$ is just $0$.

like 0 2 4

Indeed!

and then 3 would just be 3?

0
1
0 2 4
3
4 - >2
5 -> 1
*scratches head*

12:52 AM

@usukidoll The ideal generated by 3 also contains 0.

so 0 3

Right.

but how come 4 goes back to 2 and 5 goes back to 1?

The line $G$ is defined by $y=1+\frac{x}{2}$ and is viewed as a mere subset of $\mathbb{R}^2$(and not as a function for example). Does $G+G$ make sense? I think not. because for that addition would have to be defined. And then the question might arise: what is $G$ + some circle?

@usukidoll Because 4 is already part of the ideal generated by 2.

12:53 AM

so 5 should be the part of the ideal generated by 1

@usukidoll Indeed. And this makes sense, as 5 = 1 - 1 - 1.

oh the negative -1 that's when mod goes backwards

Instead of thinking of it as closure under subtraction, think of it as closure under addition, and under inverses. The former is a good criterion for checking, but if you're trying to generate examples, the latter may be more useful.

like under additive inverses?"

Yes, sorry.

So, $(1)$ would contain $1$, $1+1 = 2$, $1 + 1 + 1 = 3$, and so on, until $1 + 1 + 1 + 1 + 1 + 1 = 0$ mod $6$.

So $(1) = \Bbb Z/6\Bbb Z$.

arctic tern

12:57 AM

@Null $\Bbb R^2$ has an addition operation. Given two subsets $X,Y\subseteq Z$ and a binary operation $\bullet$ on $Z$ sometimes people write $X\bullet Y$ for the subset $\{x\bullet y:x\in X,y\in Y\}\subseteq Z$.

@arctictern ok it makes sense when defined, but not immidiatly. Would you agree?

arctic tern

if by "makes sense" you mean "has an interpretation that people use"

never seen anybody use it for affine subspaces specifically though

@arctictern thanks for the tip, didn't know that (that it can have a meaning)

Can someone explain what contrapositive means? Im working on a proof for nonsingular iff eigenvalue is not equal to 0 and I have come across this term many times.

contrapositive is a double negative in reverse

1:08 AM

@Aksel'sRose Suppose you've got a statement like "if P then Q".

like
If P then Q is the original
if Not P then Not Q is the negation

the contrapositive is "If not Q, then not P."

Converse
If Q then P
Contrapositive
if NOT Q then NOT P

It's logically equivalent to the original statement, just stated in a different way.

is contrapositive a valid proof method? Ive never come across this before.

1:10 AM

It's valid, yes.

:O! Oh great how do I determine the R/I for my ideals in mod 6 which were
0
1
0,2,4
0,3
4 -> 2
5 -> 1
T_T

hmm, that may make things a bit easier then. Thanks!

yeah contrapositive is a valid proof method

It's just logic, really. "If I'm talking on this chat, then I've got an MSE account" is logically equivalent to "If i don't have an MSE account, then I can't talk on chat"

there's also proof by exhaustion and proof by cases

1:14 AM

@usukidoll which is equivalent? or not :s

the
0,2,4 and 4? maybe x.x *hides*

@usukidoll i mean by exhaustion and by cases

exhaustion is like everything possible you can do to the proof

cases is like case 1 case 2 case 3 yay the end... like it stops

so if my original statement is "nonsingular iff eigenvalue not zero" then my contrapositive would be "singular iff eigenvalue is zero"? (which is a well known proposition/definition)

Let's break it down a bit more.

1:22 AM

or do I use contrapositive within the proof? So for one direction of the iff say "eigenvalues are zero then singular"

The contrapositive of "nonsingular if eigenvalue not zero" would be "eigenvalue zero only if singular."

The contrapositive of "nonsingular only if eigenvalue not zero" would be "eigenvalue zero if singular."

@arctictern if $G$ defined by $y=1+x/2$ is viewed as an element of a quotientspace $\mathbb{R}^2/ U$, determine $G+G$. Can i use the distributive law: $G+G=2(G)$ and a line times 2 is still a line?

In that respect, you could say that the contrapositive of "nonsingular iff eigenvalue not zero" would be "eigenvalue zero iff singular."

The ordering doesn't actually make a difference, though. A iff B isn't different from B iff A.

@Semiclassical oh I see. I really should have taken a proof and logic course...

It's not such a big deal when it comes to iffs, though

Logical equivalence is simpler than logical implication.

1:26 AM

http://prntscr.com/dffdwj
if we have a commutative ring and $a \in I, b \in J$ won't we have something like well for $r \in R$
$r(ab)=ab(r) \in IJ$
for closure under multiplication??

Okay, one last question.When writing a contrapositve proof, do you claim contrapositve in the beginning or conclusion? ie. By contrapositive we can show , or therefore, by contrapositive_?

(no idea how I got bold)

principal ideal is generated by a for the ideal I and b for the ideal J

errr... I wouldn't write by contrapositive

I'd just say "We will prove the contrapositive, i.e. there is a zero eigenvalue iff the operator is singular."

but @Semiclassical has a good idea.

Contrapositive isn't a method, it's simply a restatement of whatever you wanted to prove.

1:29 AM

@Semiclassical Im going to have to do a bit more reading on it I think. (When my brain is a bit fresher) Thanks for all the help!

If the claim is fairly simple to state, then that's really all you'll need to say.

np