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Helpless dumbfuck calling for help by Sidney Passlenotch - Mon, 04 May 2015 16:43:13 EST ID:3dz0uQ7J No.14724 Ignore Report Reply Quick Reply
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Is a horizontal or vertical line that expands to infinity it's own asymptote? Or is this in this case not applicable and a really dumb question?
>>
Phineas Hecklesot - Tue, 05 May 2015 10:18:46 EST ID:xOOOxXVr No.14725 Ignore Report Quick Reply
An asymptote is defined as a value that is approached but never reached. Y=3 reaches Y = 3, so it's not its own asymptote

nice question though
>>
Reuben Chacklefield - Sun, 17 May 2015 06:24:17 EST ID:WtAxPZi7 No.14734 Ignore Report Quick Reply
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tl;dr not really but I can prove otherwise

So consider the horizontal line y (x) = 3 i.e. y is independent of x, so for all values of x, y is always 3.
If x/x = 1,
then y (x) = 3 = 3*1 = 3x/x
It would still give you a horizontal line, but at x = 0 shit fucks ass.
You can then repeat this idea with other values like (x-1)/(x-1) = 1 so you get a "hole" at x=1, so on and so forth. Repeat this for all values of x and insert it into the equation then your line will be asymptotic to itself.
>>
Shit Bottingbire - Sun, 17 May 2015 13:24:26 EST ID:xOOOxXVr No.14737 Ignore Report Quick Reply
>>14734

An asymptote is a discontinuity, but not all discontinuities are asymptotes. What you just described is a function that is not continuous at all integer values of x, but not one with an asymptote.

I would also tend to argue that y = 3(x/x) is a different function from y = 3 simply because it returns different results


How can I relearn math? by Nicholas Cluzzleshaw - Sun, 26 Apr 2015 16:13:04 EST ID:L+X0k1Ap No.14703 Ignore Report Reply Quick Reply
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Hello /math/. I have retrograde amnesia.
I've forgotten math essentially, so much so that my abilities have regressed to that of a high school freshman.
Where and how do I relearn what I've forgotten? I've lost my job because of this.
2 posts and 1 images omitted. Click Reply to view.
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Hamilton Siddlenare - Wed, 29 Apr 2015 14:00:24 EST ID:L/U2K+oV No.14712 Ignore Report Quick Reply
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This reason is why i was so pissed about school. I was in honors during middle and high school. About a year ago I realized it had been years since i did any math. I found that i couldnt remember how to do SIMPLE SUBTRACTION. Fuck grade school.
>>
Henry Murdbury - Fri, 01 May 2015 06:40:52 EST ID:gp8hje0w No.14715 Ignore Report Quick Reply
I would say to check out your local community college and do a college mathmatics placement test to see where you are at. I would say you should be able to enroll in pre calculus if you have some memory of simple algebra and geometry
>>
Barnaby Fizzletetch - Sat, 02 May 2015 19:18:23 EST ID:Hj/F401x No.14716 Ignore Report Quick Reply
>>14712
I've been in the exact same situation. Didn't like math in HS, didn't do it for years afterward. When I started taking classes again, I got owned hard.
Fast forward to the day before yesterday and I feel like I did pretty good on the back-to-back midterms I took on Fourier analysis + linear ODEs and algorithms (which is proof-heavy).

The same boiling water that softens the potato hardens the egg.
>>
Martin Dribberdark - Sat, 09 May 2015 01:52:04 EST ID:YlDX0MWs No.14727 Ignore Report Quick Reply
My memory has always been shit, but it turned out to work in my favor for maths because I couldn't just memorize everything, I ended up having to derive everything, and re-derive it and re-derive it sometimes, until it made sense in an I guess "intuitive" level. Turns out this is a pretty good way of learning math, if a bit "slow".

My advice is to start MORE basic than you think. If I say "learn about fractions" and your reaction is "no that's too easy I want to start higher", I would suggest not to skip it. Spend a little bit actually working problems so you definitely have a very solid foundation. When I tutored calculus, there were students who got all frazzled at when where and how (not to mention why) they could "cancel out" a numerator and denominator, so their learning got held up because somewhere along the way they were like "yeah yeah ok whatever got it"
>>
Basil Giffingford - Mon, 11 May 2015 10:33:10 EST ID:bG7/Mgyv No.14731 Ignore Report Quick Reply
Buy/torrent the book "Basic Mathematics" by Serge Lang. It's highschool and precalc but written from the perspective of a mathematician so you get a rigorous logic and analysis course out of highschool math instead of just being a calculator.


Soo by Nell Gabblewell - Sun, 03 May 2015 14:03:43 EST ID:zDln8d4D No.14718 Ignore Report Reply Quick Reply
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denominators dont cancel eachother out ,right? Tf do i do here?
>>
Fuck Funkinpit - Sun, 03 May 2015 14:43:22 EST ID:qz3c7Bt+ No.14719 Ignore Report Quick Reply
>>14718
>>
Fuck Funkinpit - Sun, 03 May 2015 14:44:27 EST ID:qz3c7Bt+ No.14720 Ignore Report Quick Reply
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You don't cancel a+2 like that.
>>
Sophie Perringchure - Sun, 03 May 2015 20:11:51 EST ID:hudlvJsh No.14721 Ignore Report Quick Reply
>>14718
Looks right to me
Is the equation supposed to equal something? are you trying to solve for a?
>>
Hamilton Fundlebick - Sun, 03 May 2015 21:18:36 EST ID:zDln8d4D No.14722 Ignore Report Quick Reply
>>14721
Nah I fired it out
I was supposed to multiply the 3s
>>
Hamilton Fundlebick - Sun, 03 May 2015 21:21:48 EST ID:zDln8d4D No.14723 Ignore Report Quick Reply
>>14720
Yes I believe you do
I got it right


Discrete Maths: Video Courses by Betsy Pubberchadge - Thu, 23 Apr 2015 20:57:44 EST ID:Y+QSKjuy No.14700 Ignore Report Reply Quick Reply
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Hey guys,

I'd really like to learn discrete maths, preferably starting with some video series. Ones that would cover the subjects well in depth, rather than giving brief overviews. I will turn to books after that, it's just that having an actual teacher on such media is for me way more stimulating and makes me feel more involved.

Could anyone be kind enough to recommend such video serie(s) that go well in depth for each subject?

That would be greatly appreciated, thanks!
>>
Nigger Sungerforth - Thu, 30 Apr 2015 05:00:01 EST ID:i84x+n57 No.14713 Ignore Report Quick Reply
A while back, I torrented a lecture series by Arthur T. Benjamin in order to teach myself discrete math. I found them pretty helpful. Give 'em a try; they shouldn't be too hard to find.
>>
Nigger Sungerforth - Thu, 30 Apr 2015 06:55:06 EST ID:i84x+n57 No.14714 Ignore Report Quick Reply
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>>14713
Here:

http://www.thegreatcourses.com/courses/discrete-mathematics.html

magnet:?xt=urn:btih:fb04ab6c9e6072cbb26cc810a599c906672d3774&dn=TTC+Video+-+Discrete+Mathematics&tr=udp%3A%2F%2Fopen.demonii.com%3A1337&tr=udp%3A%2F%2Ftracker.coppersurfer.tk%3A6969&tr=udp%3A%2F%2Ftracker.leechers-paradise.org%3A6969&tr=udp%3A%2F%2Fexodus.desync.com%3A6969


finalss by Martin Sovingcocke - Tue, 28 Apr 2015 00:24:22 EST ID:eI+mFgPX No.14708 Ignore Report Reply Quick Reply
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I have my analysis final Wednesday. I'm redoing past tests and I cannot figure out number 4 part a! Some kind soul please help!
>>
Thomas Sommerville - Tue, 28 Apr 2015 00:39:31 EST ID:qz3c7Bt+ No.14709 Ignore Report Quick Reply
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DICKS EVERYWHERE


HALP. by Lydia Fancocke - Sun, 26 Apr 2015 03:31:01 EST ID:6wsrD9dK No.14701 Ignore Report Reply Quick Reply
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I'm allowed to ask for help here?

I haven't done maths since high school and I cannot work out why my final projection total is different if you work it out as a row (As in with the totals of the year) or if you work it out as a column (as in the totals revenues of each market).

This is obviously a pretty newby question but if its any community consolation I help out newbies on boards I know shit about. Thanks /math/
>>
Hamilton Fellerbuck - Sun, 26 Apr 2015 12:40:31 EST ID:qz3c7Bt+ No.14702 Ignore Report Quick Reply
I don't know what the hell you're talking about but the "Residential Services Total" for 2008 is not equal to the sum of the first three rows in "2008 Projection". This disagrees with the way 2005, 2006, and 2007 "Residential Services Total" are being computed. Are you doing this by hand?


Have mercy on my undeserving soul // Pass Calc Exam by Albert Blackstock - Mon, 02 Mar 2015 01:53:56 EST ID:oIF65CiW No.14625 Ignore Report Reply Quick Reply
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Look, I'll get right to the point.

I'm smart, but I put no effort into school. It's fucked up, and I'm sorry.

I need to pass my 2nd Calc 1 exam, which covers limits, differentiation, implicit differentiation... basically all of Chapter 2 in the standard Stewart Calculus book, and that exam is on Wednesday around 6pm.

How can I best prepare for this exam? I am able to grasp concepts, I understand math up until Calc, and limits/differentiation fairly intuitively, but I fail in the details. I really haven't kept up to speed with this class because, as previously stated, I'm a jackass.

Just... have mercy on my soul, /math/ friends. I am trying to clean my act up.
2 posts omitted. Click Reply to view.
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Lillian Gashbat - Thu, 12 Mar 2015 12:28:25 EST ID:Dk8yywxc No.14635 Ignore Report Quick Reply
>>14634

That's not necessarily true. Many schools begin introducing more rigorous proofs and abstract notions that can make Calc 3 quite difficult if you dont keep up with it.
>>
Betsy Niggerlock - Thu, 12 Mar 2015 13:43:46 EST ID:nxCnniYW No.14636 Ignore Report Quick Reply
As a side note, does anyone else think the calc1,2,3 progression is interesting? It almost seems badly organized, like it should go single var, multi var, then series/etc. Most people seem to think calc2 is a departure from what they were learning, and calc3 is a return to it. That's not exactly how I felt, but I can see their point.

This is assuming you live in a place where this is how it's done. And if you experienced a different progression through calculus, I would be very interested in how the material was presented, chronologically.
>>
Sidney Suzzleson - Fri, 27 Mar 2015 21:47:51 EST ID:FF4db3LF No.14671 Ignore Report Quick Reply
>>14634
Seconding this. I'm in Cal 3 right now, and it's nothing to worry about. Cal 2 can be rough the first time around, though.
>>
Beatrice Claywater - Sat, 28 Mar 2015 19:32:07 EST ID:uBbxSpMx No.14674 Ignore Report Quick Reply
>>14626
I'd say look over the past papers right away so that you have an idea of what kind of questions you may be asked. Practice those questions over an over again, do them looking at your notes so you get how to do the questions. Then do all that other stuff.
>>
Phineas Suffinglock - Thu, 16 Apr 2015 15:25:59 EST ID:xwcXgACR No.14698 Ignore Report Quick Reply
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>>14625
Myself and my group of friends consider ourselves, "smart" but my friend and I went through an enormous shift during our first encounter with calculus. Part of it was realizing that our algebra skills, as developed by the education system, were sub-par and we had to take the reigns to make sure our mathematical life was in order. The second thing was that we went through our first mathematical "growing pains", alluding to the idea of mathematical maturity.

Besides doing practice problems, you might want to soak up and wonder about what you're actually seeing. The feeling you get when you realize another level of generalization can be pretty dramatic. For example, realizing that you can construct mathematical objects from thin air and apply calculus tools to them, as long as you do this properly. It's the first time you're asked to be creative mathematically instead of just remembering things. This leads me to my next point.

When you're studying math the best thing you can do is collect and understand the hard facts relating to what you're studying. Theorems, differentiation rules, etc. The facts you know will determine what moves you can confidently make when you don't have any references around you. The depth of your imagination, as I referred to earlier, will determine what moves will occur to you. It doesn't matter if you have all the facts if you can't use them. My impression is people generally feel insecure when they do math, because it seems like a very authoritative subject. You might try something, "your way" only to be told that's not right. The only way to get fluent in math is by making creative decisions motivated by facts and then being reassured when you find out you're right.

Do the homework, but don't be afraid to sling dick and do every problem you can get your hands on. The problems that take an entire day to solve can be the most rewarding because it's likely not computationally difficult, it's conceptually difficult. If you have all the facts and still can't do the problem quickly, then you're probably going to experience the kind of mind expansion required to get to the next l…
Comment too long. Click here to view the full text.


what is the two consecutive number? by Phineas Punningsedging - Wed, 15 Apr 2015 10:35:58 EST ID:iHlPHeD4 No.14692 Ignore Report Reply Quick Reply
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found this puzzle on the internet, and it drives me crazy. say, you got a number that we call "A", "A" is divisible with all the numbers between 1 and 10 001. excluding two consecutive numbers in that range. We dont care about "A", we want to know the two consecutive number. How should i start this?
>>
Phyllis Drocklewore - Thu, 16 Apr 2015 05:02:00 EST ID:qz3c7Bt+ No.14693 Ignore Report Quick Reply
Use fundamental theorem of arithmetic. Find the least common multiple 'a' of all numbers between 1 and 10001. Then, choose a prime factor p of 'a' such that y divides p+1 and p+1 does not divide x/y.

In general p doesn't need to be a prime factor, but it's been a while and I'm too tired to work out the precise number theory for this problem.

ex: p=8191, p+1=8192, y=2
>>
Phyllis Drocklewore - Thu, 16 Apr 2015 05:03:44 EST ID:qz3c7Bt+ No.14694 Ignore Report Quick Reply
>>14693
Oops. Let the least common multiple be 'x'. Your 'a' is then x /((p)(p+1)).
>>
Doris Clupperstock - Thu, 16 Apr 2015 05:11:36 EST ID:TG1mst7r No.14695 Ignore Report Quick Reply
>>14692
You know it has to be the largest power of some prime factor of lcm(2,3,...,10001) and ±1 of it being a prime. If ±1 is even then you already have the factors in A so the prime^n had to be 2^n. Log_2(10001)≈13 so we need 2^13±1 to be prime. Since 2^m≠13 for any integer m by some number theory theorem 2^13+1 can't be prime. Thus if it's possible at all, 2^13-1 must be prime and wolfram alpha says it is.
>>
Phyllis Drocklewore - Thu, 16 Apr 2015 14:20:25 EST ID:qz3c7Bt+ No.14697 Ignore Report Quick Reply
>>14695
The theorem you're describing is Mersenne's conjecture-- it turned out to be false.


Might by Fanny Crorryman - Wed, 18 Dec 2013 07:58:21 EST ID:CGJx9sbH No.13496 Ignore Report Reply Quick Reply
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I did some math right now overly blazed. I hope I did good, mathgods.
I really hope I did well.
8 posts and 3 images omitted. Click Reply to view.
>>
Charlotte Bittingwater - Sat, 21 Dec 2013 16:01:26 EST ID:Cid0x3u0 No.13512 Ignore Report Quick Reply
>>13511
By this I mean that psychedelics are probably the only drug that you can do math on without it impairing you, but even then you're probably better of doing it sober.
If you NEED to get math done, sober or stims. If you want to have fun, take whatever is fun.
>>
Doris Sanderhit - Tue, 13 Jan 2015 10:24:15 EST ID:Kpw3fDyk No.14562 Ignore Report Quick Reply
>>13496

Try smoking some high quality dog excrement.
>>
Nigger Dubberdet - Fri, 23 Jan 2015 15:12:27 EST ID:OlFjx/Q0 No.14573 Ignore Report Quick Reply
Copypasta from wiki on Paul Erdős:

>After 1971 he also took amphetamines, despite the concern of his friends, one of whom (Ron Graham) bet him $500 that he could not stop taking the drug for a month.[17] Erdős won the bet, but complained that during his abstinence, mathematics had been set back by a month: "Before, when I looked at a piece of blank paper my mind was filled with ideas. Now all I see is a blank piece of paper." After he won the bet, he promptly resumed his amphetamine use.
>>
Priscilla Crallerfidging - Sat, 24 Jan 2015 16:35:06 EST ID:jEbtLayo No.14575 Ignore Report Quick Reply
>>13512
That's bullshit. When I'm tripping I can't do math worth shit. The major problem is short term memory impairment and lack of focus. I'm sure you could do a very low dose of lsd and force yourself to do some work but there really isn't any reason to.
>>
hybridgrace - Thu, 09 Apr 2015 19:26:45 EST ID:A1LnfKBL No.14688 Ignore Report Quick Reply
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LSD is ideal for finding novel solutions and perspectives to problems that you're already familiar with, I assume this applies to mathematics as well


Cryptography where to start by Nigger Ganningfoot - Tue, 24 Mar 2015 06:01:37 EST ID:ZkV6uFpl No.14663 Ignore Report Reply Quick Reply
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I need recommendations for books that teach a solid theoretical base of cryptography, and also what previous knowledge i need?
>>
Reuben Bardfield - Mon, 30 Mar 2015 03:06:15 EST ID:UqThtfYr No.14676 Ignore Report Quick Reply
Lot's of linear algebra and number theory. Here's a sample paper this cryptographer uses Lattice reduction to do analysis https://www1.lip6.fr/~joux/pages/papers/ToolBox.pdf

You need "mathematical maturity" which means you can read and write proofs, and you can figure out mathematical notation if given a key for what the symbols mean. Spivak's Calclulus is a good book to get said maturity, so is "How to Solve it" or "How to prove it" (look them up on Amazon).

There's some intro cryptography courses around http://bryanpendleton.blogspot.ca/2012/05/comparing-coursera-and-udacity.html
they usually use this book: https://books.google.ca/books?id=1YwIcpDtQPEC (Schneier's first book is too outdated, this version with Ferguson is good).

Or just go through various university calendars and see what kind of information you can find, like lecture notes, recommended reading to look up yourself, ect http://cseweb.ucsd.edu/classes/wi10/cse206a/

Universities consider advanced crypto courses their specialist shit so almost always make them expensive and not open to the public. You should look up whoever was a finalist at the SHA-3 competition and read through the papers (or public comments to NIST) of their analysis http://www.groestl.info/analysis.html where you'll learn how modern cryptographers break each other's shit and thus learn a solid theoretical base.
>>
Doris Clupperstock - Thu, 16 Apr 2015 05:22:05 EST ID:TG1mst7r No.14696 Ignore Report Quick Reply
>>14663
>I need recommendations for books that teach a solid theoretical base of cryptography

What do you mean by theoretical base? Provable security or the mathematical basis of underlying operations (elliptic curves, algebra, etc)...

"Introduction to Modern Cryptography" by Katz and Lindell for the former
"An Introduction to Mathematical Cryptography" by Hoffstein, Pipher, and Silverman for the latter


Mathematical induction by Mr.TickleDicks - Sat, 28 Mar 2015 19:11:07 EST ID:fghQwkMe No.14672 Ignore Report Reply Quick Reply
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Hello /math/ can you guys help me?
I need to prove that these propositions are true for every whole positive N.
does any one know how to solve it ? pls explain how did he arrive at the solution.

A) 2 + 4+ 6 + ...+(4n-2)=2n^2

C)1 + 3 + 6 + .... ( n(n+1) )/2 = ( n(n+1)(n+2) )/6

D) 1/1.2 + 1/2.3 + 1/3,4+ . . . + 1/n(n+1) = n/(n+1)
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Beatrice Claywater - Sat, 28 Mar 2015 19:28:32 EST ID:uBbxSpMx No.14673 Ignore Report Quick Reply
>>14672
A) Test the case where n =1: LHS: 2, RHS 2*(1)^2=2
It works!
Assume it works for k.
Want: works for k+1.
sum to k+1: 2 + 4 +6 + ... + (4k-2) + (4(k+1)-2)
Write the sum up to k as 2k^2. (we're assuming it works)
we get: 2k^2 +4k-2
what does 2(k+1)^2 equal?
...
It works for k+1!
Now by principle of induction ... ... ...V-allz n in the o'naturelle we just proved blah blah blah

I'm not sure what the homework rules are on this bored. If it says "no homework threads" please delete this thread and take it to http://mathoverflow.net/, I'm sure they'll be more than happy to help
>>
Caroline Hubblechetch - Sun, 29 Mar 2015 02:27:13 EST ID:qz3c7Bt+ No.14675 Ignore Report Quick Reply
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I like these


Quadratic Reciprocity by Wesley Middlewudging - Fri, 13 Feb 2015 08:52:14 EST ID:iRDOjfyp No.14607 Ignore Report Reply Quick Reply
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Hey /math/, I've been reading a lot of maths in the last year because I always enjoyed it at school. I'm comfortable with all the maths one might do in high school and bits and pieces of other areas.
I'm told that some proofs for quadratic reciprocity are very beautiful, although I just can't get my head around what it actually IS
Can anyone enlighten me?
Thanks
>>
Fanny Sacklested - Sun, 08 Mar 2015 07:56:03 EST ID:nxCnniYW No.14631 Ignore Report Quick Reply
I haven't come across it until just now, and I hate learning things from wikipedia. I hope someone comes along and enlightens us, Wes.

https://www.youtube.com/watch?v=HhWqQQI5HJU
This guy talks about Legendre symbols which I guess helps (introduced by Legendre while trying to prove quadratic reciprocity). I'm at about 4min, it's okay so far, a bit stumbly but still better than trying to learn from wikipedia imo.
>>
Phineas Dreshwell - Tue, 24 Mar 2015 17:59:19 EST ID:PeQJ0tWt No.14665 Ignore Report Quick Reply
>>14607
I've just finish a Number theory course which covered it.

First you should be pretty comfortable with modular arithmetic. If you're not then I'm kinda wasting my time here but I'll try and give a summary:
If we have a 12 hour clock, then the hour 1 is thought of as the same as the hour 13. We want a way of relating any integer to one of the numbers 1 to 12. We do this by saying that two numbers, x and y represent the same hour if (x-y) is divisible by 12 which is the same as saying x=y+12n for some integer n, this is kind of obvious as we deal with clocks on a day to day basis.

Now we can obviously generalize this to a "clock" of any size, where each hour is represented by a number from 0 to m. Then we say that x is "congruent to" y "modulo m" if (x-y) is divisible by m. So, 0 and m represent the same "class" as does 1 and m+1, -1 and 241m-1 etc.

If you're familiar with that, then we can also set up a quadratic congruence: so recall how we'd normally want to find an x such that ax^2+bx+c=0, well now we want to find a solution x (not a number but a class of numbers) such that ax^2+bx+c is /congruent/ to 0 modulo p, where p is an odd prime. We can complete the square to get y=2ax+b and the problem becomes: find a class, y such that y^2 is congruent to d (modulo p).

Now to skim over some details, refer to the video Fanny posted to go over Legendre Symbols, but basically if we stick to the notation from before then we just define (d/p) to be 1 if p doesn't divide a and there's a solution, y to " find a class, y such that y^2 is congruent to d (modulo p)", -1 if it doesn't have a solution and 0 if p divides a.

Now it turns out that there are a bunch of algorithms which make (d/p) easy to calculate and so we can easily tell if the quadratic congruence has a solution or not. For example:

Euler's criterion: (d/p) is congruent to d^((p-1)/2) (modulo p)
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Phoebe Bonnerbetch - Wed, 25 Mar 2015 10:22:53 EST ID:HZ1lfOQk No.14666 Ignore Report Quick Reply
>>14665
(a/b) is real bad notation, if I'm wrong I'd love to hear an argument in favor of it.

Fucking sloppy shit imo.

Way too much gets forgiven by "I'm familiar with it so it's okay"
>>
Whitey Mangerbare - Wed, 25 Mar 2015 16:10:23 EST ID:RzF3HHjt No.14667 Ignore Report Quick Reply
>>14666
I'm sorry, I was pretty tired last night when I typed it up.

Replace the (d/p) with [d/p], (p/q) with [p/q] and anything else involving (x/p) where x is anything by [x/p].

If you're talking about just in general then I personally don't see it as that much of a problem.
>>
Oliver Sobberchetch - Wed, 25 Mar 2015 21:10:05 EST ID:IaC61I0h No.14668 Ignore Report Quick Reply
>>14666
i wrote this in another thread on /math/ like a year ago but yeah (a/b) sucks ass and i prefer to do L_b (a) for legendre symbol


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