Mathematics - 2011-11-17 (page 1 of 4)

t.b.

12:00 AM

(and yes, add an @ in front of my name so that I get notified).

Potato

@t.b. I'm not concerned about easiness now as about whether or not that line of argumentation leads to a proof.

@t.b. an answer of "no, there's nothing there" is perfectly fine, I'm just curious

Rob

I also added another proof that uses the multiplicative property of zero

I believe the reason for people not understanding this topic is that they have not seen the proof of the mutiplicative property of zero at the beginning

t.b.

@Potato To be honest, I don't see how that can lead to a proof, sorry. But maybe I'm just dense.

Potato

@t.b., no, it is just a bad idea that I was clinging too. I just wanted to make sure there was nothing obvious I was missing. Thanks

t.b.

@potato After having given "my" argument. You can read it backwards and say that if there's a limit point then the two omega_1 and omega_2 were not linearly independent in the first place. But that's not really a different proof, just an "after the fact".

Potato

12:09 AM

oh, I see

Jonas Teuwen

@HenningMakholm Haha!

Potato

So consider the polynomial f(z,w)=z^2-w. This is singular at 0, correct? (trying to get some practice with definitions...)

t.b.

Looks pretty singular at zero to me, yes.

robjohn

@potato: there is no singularity at 0

Potato

But f(z,w)=zw+1 (over the complex numbers) is nonsingular and degree 2?

@robjohn but aren't both partials 0 at (0,0)?

robjohn

12:16 AM

@potato: a singularity means it blows up. Unless you are using some other definition of singular.

Do you mean critical?

Potato

"A polynomial is nonsingular at a root p is either partial derivative is not zero at p."

t.b.

@robjohn: I understood regular point versus singular point.

(what Potato said)

@Potato And yes, this looks correct to me, too.

robjohn

Okay, that is a different notion of singularity.

J. M.

Quite an alliterative title...

Srivatsan

@JM I don't get it. An undergrad majoring in math would focus almost exclusively on math, won't she?

Or does he want research-level math in undergrad?

J. M.

12:22 AM

Dunno. Probably they want a course where it's all math, math, math...

robjohn

@Srivatsan: there are breadth requirements

J. M.

(which sounds like a great recipe for making math stale...)

Potato

There are a few places without breadth requirements, but they aren't top math programs. I think the answer to the question is a simple no.

robjohn

At a lot of schools.

Potato

Really? The loosest ones I know of require only that you complete a major and do one freshman writing course. So that's a minimum of one required course.

robjohn

12:24 AM

I'm a bit slow, typing on my iPhone.

t.b.

There are some extremely prodigious young people here. Let's see what they have to say if they feel like it.

J. M.

Yes, I think I saw a few high-rep users in their teens.

Potato

Well, regardless of what they have to say, the answer is just no, I believe.

t.b.

So, how did Wolfram and some others manage to complete their Ph.D. at the age of 20?

Potato

I think they just went to college really early.

But I don't think even they got to bypass breadth requirements at top universities. Like I said, if you want to go to a lower tier school you have a lot more leeway.

t.b.

12:30 AM

Well, it's the prodigy's and his parents' call whether this is a good idea. If I had such a prodigious kid I would tell him to try and accomplish the mandatory things in as little time as possible and try to convince him that he might be missing out on a lot by trying to bypass that.

Potato

Oh, I don't doubt that getting through the core requirements as fast as possible is a good idea, merely that if he wants to go to Harvard or Princeton or wherever (and if he's really that good, he should accept no less) then he's going to have to do the core.

J. M.

@tb "A lot" indeed; "life isn't entirely math", and all that...

Srivatsan

@tb What do you mean by mandatory things?

J. M.

Stuff you have to go through to graduate. At places I know, physical education is one such thing, to give an example.

Srivatsan

@JM Ok, thanks.

t.b.

12:37 AM

Among all the really good mathematicians I've met in my life there's not a single one with a narrow focus, mathematically speaking and generally speaking.

Srivatsan

Apparently the OP is an associate prof, and not some "gen guy"... =)

Potato

Can I get on why a singular affine plane curve of degree 2 degenerates into the product of two factors (two intersecting lines)? I'm trying to just bare-hands it by writing out the derivatives but maybe I messed up the algebra.

*a hint

J. M.

@Potato: note that you can edit your comments (for a limited timeframe) by pressing the up arrow key. :)

Potato

Duly noted.

Srivatsan

=)

Jonas Teuwen

12:40 AM

Good night guys. I'll be thrown in front of the lions in about 12 hours =).

robjohn

the mobile interface has no corrective ability I found out.

Potato

Wait I think I got it. Nevermind.

robjohn

@JonasTeuwen good night and good luck!

t.b.

@Jonas: Good luck! All the best.

Jonas Teuwen

Thanks!

J. M.

12:41 AM

@JonasTeuwen Then bring a chair and a whip. Seriously, good luck!

Srivatsan

Wish you luck, @Jonas.

t.b.

So I'll cross my fingers at 2 :)

Jonas Teuwen

:D. Thanks. Bye.

Srivatsan

I guess I'll leave too. Bye all.

robjohn

@Srivatsan later :-)

J. M.

12:43 AM

See you, @Sri.

robjohn

Well, it appears my prediction of a 0-rep day for me was correct.

Work took over today.

Henning Makholm

Good luck, @Jonas. I'm off to bed too; ttfn.

robjohn

@tb: when I texted about the accident clearing, I was in my car and not moving at all.

on the way home from the exam.

Potato

So if we have a conic that degenerates into two parallel lines and it factors, normalizing the coefficient, it must factor in the form (1+px+qy)(1-px-qy)=0, yes?

robjohn

@HenningMakholm good night.

Potato

12:50 AM

Oh wait!

The conic isn't necessarily homogenous. Ugh.

robjohn

@Potato those are parallel lines.

That would happen under very specific circumstances.

Potato

oh wait maybe I'm right and that form is correct in general, for singular conics?

I need to prove that it factors into two lines when it has a root with both partials equal to zero at that root.

robjohn

The general form would be two lines, intersecting or parallel.

J. M.

or coincident

Potato

Ah, I mean intersecting lines.

The problem is: show an affine plane curve of degree two, defined by the zero locus of a polynomial f(z,w), factors into two linear polynomials (and thus two intersecting lines) if it has a singular point (a root where both partials are zero).

Could I get a hint? Not the full solution please, just a nudge.

Can you just rotate to eliminate the xy term, translate to center it at 0, and which point you know the constant must vanish and then it's trivial to write down the factorization?

oh wait, the rotating might screw up the derivatives...

J. M.

1:05 AM

"oh wait, the rotating might screw up the derivatives..." - it shouldn't; rotation is a rigid motion.

If anything, it might make for neater expressions for manipulation.

Potato

Right, but you used to be taking the derivative "in a different direction", because you rotated the axis

if that makes sense

J. M.

"different direction"?

Potato

oh wait, I don't think rotating screws up your derivatives after all

I take that back

For example, if you take a line of slope 1 and rotate, obviously its derivative changes

J. M.

If it helps: rotate, take derivatives, and undo rotation.

Potato

I don't think so. You're still taking derivatives with respect to the wrong coordinate system.

J. M.

1:11 AM

Well... I'd expect x^2-y^2=0 and xy=0 to have the same properties...

robjohn

1:36 AM

they are rotated by \pi/4

J. M.

Indeed. The conic being degenerate (factorable) shouldn't have to depend on orientation.

Potato

Well rotation doesn't turn out so great, but you can translate the root the 0,0 and everything works out great

robjohn

@Potato What kind of properties are you looking for?

Potato

I'm not looking for any properties. Anyway I have solved the problem.

robjohn

@Potato Oh, okay. I dropped into the middle of this, so I'm not sure what was really being asked.

I just poke my nose in where it's not wanted and comment on things ;-)

J. M.

1:50 AM

Well, I've to leave for work. Later, y'all.

robjohn

@JM Nice to see you. Have a good day.

Pete Clark's answer about the density of primes has sure proved popular.

Srivatsan

2:12 AM

Damn, this is why I shouldn't downvote. :(

t.b.

@Srivatsan: what happened?

Srivatsan

I downvoted an answer that was subsequently deleted. So do I lost the 1 rep for nothing?

(I did not get back the 1 rep, I believe.)

t.b.

@Srivatsan: no you haven't lost it but it isn't re-awarded automatically. You can get it back by triggering a rep recalc: go here and scroll to the bottom of the page and click the button (you can do this once a day and I would do this from time to time to balance fluctuations from deleted users, etc.)

See here for some more explanations.

Srivatsan

Oh cool =)

I got it back (along with 2 more points as bonus).

Hurrah =)

t.b.

So the sun is shining down on you :)

Srivatsan

2:17 AM

Thanks, tb.

robjohn

@tb Great. I just lost 22 rep! ;-)

Srivatsan

@robjohn How come?

robjohn

@Srivatsan When I did the rep recalc.

Srivatsan

How do you lose rep?

robjohn

I had 10222 before, and now I have 10200.

t.b.

2:19 AM

@Srivatsan: if you delete an upvoted answer your rep isn't recalculated, for instance. Also, when a user is nuked out of the system the the votes are deleted, too. So you lose rep when that happens but it isn't displayed. Triggering a recalc results in losses mostly (unless you downvote often enough :))

Srivatsan

Ok, thanks.

robjohn

I don't think I have ever downvoted.

I usually comment instead.

I think it is more useful to the OP

and I really hate being downvoted with no comments.

Srivatsan

@robjohn I comment or upvote an existing comment. // And I give time for the poster to correct the mistake.

t.b.

I usually comment long before I downvote but if the user fails to correct his answer or reacts like the limit guy recently; or somebody happens to get upvotes for nonsense, I do downvote.

robjohn

@tb sounds reasonable. I may change my habits.

Actually, when I comment on a mistake, it is usually corrected, so I don't need to downvote.

Srivatsan

2:27 AM

In my case today, the answer I downvoted was clearly wrong. I had pointed it out as such and waited for a few hours before downvoting. The answerer deleted it subsequently.

I wonder why Dilip is explaining my answer to me. =)

I think it goes to show that the answer is not clear.

And I find Mariano's comment hilarious for some reason: math.stackexchange.com/questions/82865.

t.b.

@Srivatsan: perfectly clear answer, I think.

Srivatsan

Thanks tb. [Now, I am embarrassed. =)]

t.b.

On the other hand, I don't understand the comments, though

Srivatsan

My comment or his comment? The contents of the comments or why the comments are there?

t.b.

I don't see why you should do a cardinality estimate.

Srivatsan

2:36 AM

Yes, I myself oscillate between the two views. Sometimes, I find it convenient to say |A| <= |B| and vice versa, so |A| = |B|.

t.b.

Yes, but you already have a bijection swapping the two, so why estimate?

Srivatsan

[contd] I used to find it more of a mental burden to exhibit an invertible map from A to B.

In the cardinality way, one can take each direction separately; potentially easier.

But in many cases, bijection can be much more direct. Like in this case.

My viewpoint favors the bijective proofs more these days.

t.b.

I'm rarely in position to use them, but usually a bijective proof is very neat. Like for instance Mariano's celebrated picture for 1+2+3+...+(n-1) = (n choose 2)

robjohn

A case of exponent tracking.

t.b.

Nomen est omen

Benjamin Lim

2:44 AM

@t.b. Hello

robjohn

@tb :-D

t.b.

@BenjaminLim Hi!

Benjamin Lim

I just realised I have a problem in the way I proved some things in rings

Like for example, I tried to prove that Z[x]/(x^2 -3, 2x + 4) is isomorphic to the zero ring

Srivatsan

@tb Nice thread. I was going through the answers again, and Asaf's made me crack up for some reason. // Of course, that picture is celebrated...

t.b.

So, what did you do?

Benjamin Lim

2:46 AM

I used the fact the kernel of the evaluation map x -> -2 is generated by the principal ideal (2x + 4)

which is not true

So in other words to do my approach I first need to reduce (x^2 -3 , 2x + 4) to an ideal generated by two monic polynomials

2x + 4 is not monic

t.b.

Very true, that 2 is somewhat annoying :)

So, what did you try instead?

Benjamin Lim

how can I get around that?

I guess you just set stuff equal to zero

I get from x^2 -3 = 0 and 2x + 4 = 0 that 1 =0

t.b.

Can you combine the two polynomials in some way? Like: what do you get if you take their sum?

Benjamin Lim

you get

x^2 + 2x +1 = 0

robjohn

Might this question better be calculus or inequality?

t.b.

2:54 AM

So (x^3 - 3, 2x+4) = (x^3 -3, (x+1)^2), right? now you have two monic polynomials. Does that help?

Benjamin Lim

ahhhhhhhh

t.b.

@robjohn I'd go for inequality and analysis or real-analysis

Always unsure how to classify basic (p,q)-stuff.

robjohn

@tb yeah, me too.

Should I remove the current tags, or just add inequality and real-analysis?

Benjamin Lim

@t.b. Wait my calculation now shows that Z[x]/(x^2 -3 ,2x + 4) is not the zero ring

robjohn

@BenjaminLim It isn't the 0 ring, is it?

t.b.

2:59 AM

@robjohn: as you please. I don't think this is measure theory, though. functional-analysis is a bit over the top as well, but fine

Benjamin Lim

@robjohn But if you do x^2 -3 = 0, 2x + 4 = 0

t.b.

I just looked for "Hölder" and the questions are tagged a bit arbitrarily.

Benjamin Lim

you get x = -2

robjohn

If it were Jensen, then I could see measure theory, perhaps, but not Hölder.

Benjamin Lim

(-2)^2 - 3 = 0

or 1 = 0

t.b.

3:02 AM

@Benjamin: so before worrying about that what did your new calculations give you?

Benjamin Lim

the zero ring

t.b.

You said

6 mins ago, by Benjamin Lim

@t.b. Wait my calculation now shows that Z[x]/(x^2 -3 ,2x + 4) is not the zero ring

Benjamin Lim

That calculation I don't think is right

t.b.

(sorry about this spastic attack above).

Benjamin Lim

I don't think that Z[x]/ (x^2 + 2x + 1) is isomorphic to Z

robjohn

3:06 AM

(2x^2-6)-(x-2)(2x+4)=2
I don't think you can get 1, so the ring is isomorphic to Z/2

Potato

I have a question about problem J here: books.google.com/…

Does it just follow from the fact that every holomorphic function with nonzero derivative at a point is a local analytic isomorphism at that point? This is in Ahlfors.

I'm not seeing how the implicit function theorem comes into play.

t.b.

@Potato: can you give the page?

Potato

sorry, page 13

Benjamin Lim

@robjohn Is it true that Z[x]/(x^2 + 2x + 1) is isomorphic to Z[-1] which is isomorphic to Z??

t.b.

@Potato You can use what we discussed a few days ago (what you outlined above). Else try applying the implicit function theorem to f(z,w) = (z,\phi(w))

well, the way it is phrased there to f(z,w) = (z-p, \phi(w) - \phi(p))

robjohn

3:13 AM

@BenjaminLim How would you reduce x ?

Potato

but if phi isn't a polynomial, then doesn't that cause f(z,w) to not be a polynomial?

Also, why is your f mapping to C^2?

t.b.

I didn't read the version given there properly. I was thinking of the implicit function theorem for analytic functions.

Let me skim what's in the book.

Potato

His statement is weird.

t.b.

(and mapping to C^2 was a goof)

Potato

So my issue is that the statement he gives requires a polynomial as part of the hypothesis, and phi(z) does not necessarily need to be a polynomial.

Benjamin Lim

3:20 AM

@robjohn What do you mean?

But I am confused from the relations x^2 -3 = 0, 2x + 4 = 0 I got that 1 = 0........

t.b.

@Potato: Yes, but I think he means to say that this is a particular instance of the implicit function theorem. Since he doesn't prove it, I guess you're free to use whatever formulation you like.

robjohn

@BenjaminLim You cannot divide by 2

there is no 1/2 in Z[x]

gotta go for a bit... bbl

Benjamin Lim

yeah you're right

I am just wondering

Which ring homomorphism from Z[x] into some other ring has (2x + 4) has its kernel?

t.b.

@Ben: Are you asking what Z[x]/(2x+4) is?

Benjamin Lim

yeah

Also, I am trying to figure out now what is Z[x]/(x^2 + 2x + 1)

t.b.

3:33 AM

which one first?

Benjamin Lim

Z[x]/(2x + 4) first

t.b.

I'm waiting for you to say something :)

Benjamin Lim

I don't even know how to start

t.b.

You can do Z[x]/(x+2), right?

Benjamin Lim

yeah

that's the thing

then I know that Z[x]/(x+2) is isomorphic to Z[-2] which I think is isomorphic to Z

But then the problem with Z[x]/(2x + 4) is that I can't say that the ideal (2x+4) is the kernel of some evaluation map

Also, I am confused about the fact that we know if you have an element x in a ring say R and adjoin R[x] we get exactly the same ring back

but if you adjoin and element x in a ring R such that x^2 = 1, then it is possible not to get the ring back but instead a product ring or something like we talked about

Is it because in the first case you are adjoining just one element already in the ring to itself

whereas in the second case you are sort of adjoining two things?

@t.b.

t.b.

3:52 AM

I'm listening.

Benjamin Lim

that's all I've got

t.b.

I need to fix myself some coffee, just a sec.

Okay, while the coffee boils up...

Benjamin Lim

I am out of ideas theo

t.b.

Let's sort this out slowly: Once again, what do you mean by saying "adjoin an element"

Benjamin Lim

Given a ring R, to adjoin an element say x is to form the smallest ring containing R and that x

t.b.

3:57 AM

but then the x is what?

Benjamin Lim

the root of a polynomial with coefficients in the ring

so elements in R[x] are of the form a1 + a2 x^2 + a3 x^3 + ... a_n x^n

t.b.

yes. Now by factoring out an ideal generated by a polynomial you introduce a relation, right?

Benjamin Lim

yeah

t.b.

So what does this mean, exactly?

Benjamin Lim

so for example, when we do Z[x]/(2x + 4)

we are imposing the relation that 2x + 4 = 0

in the quotient

t.b.

4:04 AM

Yes, so you identify any two polynomials that differ by integral multiples of 2x+4

Benjamin Lim

yeah

How does this tell us what Z[x]/(2x + 4) is isomorphic to?

t.b.

you're so impatient :)

Benjamin Lim

where do we go next?

t.b.

Now you said that Z[x]/(x+2) was Z. how did you see that?

Benjamin Lim

Z[x]/(x+2) is isomorphic to the ring Z[a], a the element satisfying a + 2 = 0

t.b.

4:06 AM

why is that?

Benjamin Lim

So in Z[a], we have linear combinations x + ay, x,y in Z

first isomorphism theorem

t.b.

Now how are (2x+4) and (x+2) related?

Benjamin Lim

(2x + 4) is contained in (x+2)?

t.b.

and what does this tell you?

Benjamin Lim

I don't understand

what do you mean?

You mean what does it tell me about Z[x]/(2x + 4)

Maybe that somehow by comparing size, Z[x]/(2x + 4) is bigger than Z?

It does not make sense, forget what I said

t.b.

4:13 AM

Well, anything that gets identified when factoring out (2x+4) also gets identified when factoring out (x+2) but not the other way around.

Benjamin Lim

so Z[x]/(2x + 4) is isomorphic to something smaller than Z, a subring of it?

t.b.

That's not what I'm saying. You have this ideal (x+2) in Z[x]. It contains the ideal (2x+4), so...

Benjamin Lim

I don't get

t.b.

what happens to (x+2) when passing to Z[x]/(2x+4)?

Benjamin Lim

it is zero?

t.b.

4:16 AM

why?

That's not true, why do you think so?

Benjamin Lim

if we declare 2x + 4 to be zero

then 2x = -4

t.b.

yes. that's true.

Benjamin Lim

so that 2x + 6 = 2

hence x + 2 = x + (2x + 6)

which is 3x + 6

x+2 becomes 3x + 6

is that what you're trying to mean?

t.b.

Well, sort of. But that doesn't lead too far...

Benjamin Lim

ok, then that is not it

t.b.

4:21 AM

What do you know about the correspondence between ideals in Z[x]/(2x+4) and the ideals of Z[x]?

Benjamin Lim

there is a one to one correspondence between them, by the correspondence theorem

t.b.

what exactly does the correspondence theorem say?

Benjamin Lim

Let f be the map from Z[x] to Z[x]/(2x + 4)

then if I is an ideal in Z[x]

J = f(I) an ideal in Z[x]/ (2x + 4)

then Z[x]/ I is isomorphic to (Z[x]/(2x + 4) )/J

t.b.

what is the relation between I and J?

Benjamin Lim

I is the image of I

t.b.

4:23 AM

Is I just any ideal?

Benjamin Lim

Sorry, it contains the kernel

t.b.

the kernel being (2x+4)

Benjamin Lim

yeah

OH wai

(x + 2) contains the kernel (2x + 4)

so that

is isomorphic to Z[x]/(2x + 4) / ( f(x + 2) )

t.b.

Z[x]/(x+2) is isomorphic to ...

(what you said)

Benjamin Lim

yeah but what is it

hmmm

t.b.

4:27 AM

Now this x+2 is mapped to something in Z[x]/(2x+4)

Benjamin Lim

yeah

t.b.

that something, let's call it y, do you see a relation it satisfies?

Benjamin Lim

no

t.b.

well, how are x+2 and 2x+4 related?

Benjamin Lim

2(x+2) = 2x + 4?

t.b.

4:31 AM

yes, so you've got this y in Z[x]/(2x+4) satisfying 2y = 0

Benjamin Lim

yeah

so

t.b.

y generates an ideal and when you factor it out you get Z[x]/(x+2)

Benjamin Lim

yes

t.b.

and Z[x]/(x+2) was what again?

Benjamin Lim

Z

t.b.

4:35 AM

so, make a conjecture what Z[x]/(2x+4) could be.

Benjamin Lim

hah

Z/(2)

t.b.

well, not quite, you want Z to be a quotient of it.

Benjamin Lim

I don't understand

t.b.

well, you want a ring R containing an element y such that 2y = 0 and R/(y) = Z

Benjamin Lim

Ok

t.b.

4:39 AM

So if you're asked to produce such a ring. How would you proceed to construct such a thing?

Benjamin Lim

Via some evaluation map?

e.g. like how we say that R[i] is C

t.b.

What are the very first examples of rings you knew: Z, Z/(n), right?

Using these how can you produce such a ring R?

Benjamin Lim

I don't even know

t.b.

products, for example?

Benjamin Lim

product rings?

t.b.

4:44 AM

yes

Benjamin Lim

I am really lost now

t.b.

How about Z x Z/(2) ?

Benjamin Lim

I am looking for idempotents now in Z[x]/(2x + 4)

How did you come up with Z x Z/(2) ??

t.b.

Oh, I just produced an example of such a ring R. The generator y of Z/(2) satisfies the relation I want.

Benjamin Lim

worst part is I have an algebra exam tomorrow.....

t.b.

4:52 AM

oh, good luck then!

Benjamin Lim

I don't think anyone in my course gets these things on quotient rings and stuff

anyway, I still don't get how you plucked Z x Z/(2)

t.b.

It is somewhat subtle.

I was mischievously a bit misleading...

Benjamin Lim

Anyway

I think if given to identify what Z[x]/(2x + 4, x^2 -3 ) is

I could do:

2x = -4

4x^2 = 16

but then x^2 = 3

so that 12 =16

4 = 0

t.b.

How did you get from 4x^2 = 16 to x^2 = 3?

Benjamin Lim

x^2 - 3 = 0

t.b.

4:56 AM

ah. sorry, missed that.

Benjamin Lim

so in fact we should have Z[x]/ (2x + 4, x^2 -3) is isomorphic to Z/(4)

Rob

What level of Algebra is this?

Benjamin Lim

commutative rings

Theo I have to go now, I will have to revise on other topics in algebra

I think if I am stuck in an exam

I will try and reduce the ideals to some generated by monics

and then apply the usual strategy that i've got

Rob

Is that second year university?

Benjamin Lim

i'm first year, but yes it's a second year course

t.b.

5:00 AM

@BenjaminLim Yes that reduction is definitely a good thing to practice. See you and good luck at your exam!

Benjamin Lim

Well if I am out of luck I will use it

see you man! à tout l'heure!

t.b.

Ciao!

Benjamin Lim

J'utiliserai ce forum comme un cours!!

Rob

5:14 AM

Hello? is there anybody out there...

does anybody really care?

What do you think this Cantor quote means:

In mathematics the art of asking questions is more valuable than solving problems.

Einstein also stressed the importance of asking the right question

but is it more valuable than solving problems?

t.b.

6:26 AM

@Srivatsan: I found the first comment here (as well as the downvote) completely unnecessary. What do you think?

Alexei Averchenko

7:48 AM

why do we have christianity.SE?

they're a good trolling material, sure, but come on, REALLY?

is Zoroastrianism.SE next?

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$Mathematics$ Mathematics