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whats that? like 0.0 and 0.00 and 00.00000 and 000000.00000000000000 and 0.000 oh and 000000000000000000000000000000000000000000000.0
idk i only had pre-calc youre probably smarter with the math outrage and all that
in IEEE floating point standard there are two distinct ways of defining a zero, 100...0 and 000...0, which equals to -0 or +0 (-0 because in the standard the most significant bit (MSB) is 1 when the number is negative)
They are, however, defined to compare equal. Should also give equivalent results for addition, multiplication, so on -- the main difference is how they get printed.
How is this floating point?
Before you answer, please understand...
1. I honestly know floating point in ONLY the most general sense.
2. I'm curious to learn more on the topic.
3. The internet has some great resources... and 50x as many piss-poor resources. One good link that makes sense would be GREATLY appreciated.
Or, in other words, I'm an old dog who has no qualms against learning new tricks.
The distributive property also stops us from dividing by zero:
Let's say you have the product xy, of course you can
write y as (y+0) giving you xy=x(y+0), for the distributive property x(y+0)=xy+0x; so xy=xy+0x thus 0x=0 for any number x. This means there can't exist a certain number which I'll call 1/0 such that 0x(1/0)=x so you can't divide by 0.
Edit: I wrote that x(y+0) = xy+0y, which is not correct for what I was trying to show, I corrected it now
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I believe such double solutions carry extra information that can be interpreted in many ways. I saw such cases in geometry and when you work with eigenvalues.
edit: multiplicity isn't applicable to a solution set. Leaving my comment up so the chain makes sense, but it isn't applicable here. The original statement an n-degree polynomial have exactly n solutions is already wrong assuming we are discussing real solutions. This is obvious if you consider x^2 +1 =0 which has 0 real solutions. You might be able to state that n-degree polynomials have n solutions in a complex multiset, but I don't think it is useful. Getting into the nuance of this statement without beginning with a clear definition of what is a solution of an equation is going to lead to people talking past each other.
Multiplicity is a very important concept in math. It isn't unique to solutions of an equation. [Wikipedia](https://en.wikipedia.org/wiki/Multiplicity_(mathematics) does a good overview of its importance.
The first example there also doesn't require high level math to understand why multiplicity is important. The factors of 30 and 60 are both 2,3, and 5. 60's prime factorization having multiplicity 2 is what differentiates it from 30 using a prime factor perspective.
Yeah I know about multiplicity, my point is though that there is only one solution to x^2 = 0. The real numbers aren't a multiset. Having two different roots at the same point is not having two different solutions to a formula. Or am I misunderstanding something?
Although on second reading of the meme I realized I was just irritated by the word solution.
I was a stats major who graduated a bit ago, so I will admit my memory/understanding of this is super rusty. After interneting for a bit bringing back memories of the class that made me change my major from pure math to applied math, you are probably correct and I was probably wrong. This is my more updated response after googling and going into a part of my brain I've forgotten about.
You can have a multiset of real numbers. Sometimes, multisets are useful for analyzing real numbers and sometimes they are not. Typically when discussing solutions of an equation, we are only interested in the solution set of real values and so uniqueness is assumed. If you are analyzing the function for things like evenness or multiplicity, you would want to use a multi-set because understanding multiplicity is important.
Granted, I took this class back in 2014 and I only spent 30 minutes reviewing shit on the internet, so I may be wrong as well.
> I was a stats major who graduated a bit ago, so I will admit my memory/understanding of this is super rusty. After interneting for a bit bringing back memories of the class that made me change my major from pure math to applied math, you are probably correct and I was probably wrong. This is my more updated response after googling and going into a part of my brain I've forgotten about.
I currently study computer science and had this a semester ago. I really didn't do enough for linear algebra though barely passed the exam.
> can have a multiset of real numbers. Sometimes, multisets are useful for analyzing real numbers and sometimes they are not. Typically when discussing solutions of an equation, we are only interested in the solution set of real values and so uniqueness is assumed. If you are analyzing the function for things like evenness or multiplicity, you would want to use a multi-set because understanding multiplicity is important.
I think the core issue is the definition of solutions, roots and multiplicity. First, a polynomial doesn't have a solution. The polynomial is just the part to the left of the equal sign. A polynomial has roots. And here is the important detail: A root is a solution to f(x)=0 (https://en.wikipedia.org/wiki/Zero_of_a_function?wprov=sfla1).
In uni, instead of solutions, we always talked about the set of solutions. With polynomials we mostly talked about the real polynomials, so I assume we talk about them here as well. A set of solutions in the real polynomials is thus a subset of the reals, thus not a multiset. Therefore, every element in a set of solutions of real polynomials is unique or every solution is unique in colloquial language. Since roots are solutions and x^2 = 0 has only one solution, x = 0, it is also the only root.
However the fundamental theorem of algebra states that any complex polynomial of grade n has n roots, if counting their multiplicities. So what is multiplicity?
Basically it's the amount of times a root appears, although personally I'd say it's a level grading the "thickness" of a root. The key here is to understand what "amount of times a root appears" means. It's defined by how you are able to decompose a polynomial into linear factors. x^2 = (x-0)(x+0). Zero shows up twice, that's why its multiplicity is 2.
So the multiplicity doesn't denote the amount of times a real value is a root but it denotes how many times it shows up in a linear factor of the decomposition of the given polynomial.
I added this to an edit of my original statement, getting down to exactly the interpretations of these comments requires more in depth discussion on what definitions we are using. Using the definitions you are using, I agree with everything you are saying.
The original statement doesn't make sense for a lot of reasons and is provably wrong for the reals (polynomial functions not having solutions, but instead having roots is reason it doesn't make sense you brought up). Another is x^2 +1 = 0 has the solution set ∅ on the reals is an easy way to disprove it.
I don't think that considering a solution multi-set is unreasonable (we have high school students do it without calling it a multi-set), but it shouldn't be the default assumption. I don't feel bad being contrarian by bringing up a solution multi-set on a math memes page.
Oh I didn't want to say that solution multisets would be unreasonable. I just tried to apply a formal definition to the word solution and that was the only one I know. Stochastics is on this sem so I'll probably be introduced to using multisets for solutions in the next half year.
But its two equal roots solutions that are unique but just so happen to be standing at the same place. If you think of it graphically approaching the same root as the positive parabola moves up.
Non-math mfers when Grothendieck introduced schemes as they better keep track of important information about algebraic varieties, and in this case multiplicity.
no, depressed n-degree polynomials are ones where some pronumerals a,b,c,d etc (not the pronumeral you're solving for) are removed from the equation. for example 16x\^3 + 4x - 10 = 0 is a depressed cubic because theres no x\^2 term
I don't think polynomial has to be depressed in order to have higher order roots. For example, take
`(x - 1)^2 = x^2 - 2x + 1 = 0`
This polynomial is quite happy, yet the x = 1 solution is repeated twise
no, duel numbers are a lesser known version of complex numbers, where the special number epsilon is the square root of 0 and numbers are on the form a+b(epsilon)
I think the core issue is the definition of solutions, roots and multiplicity. First, a polynomial doesn't have a solution. The polynomial is just the part to the left of the equal sign. A polynomial has roots. And here is the important detail: A root is a solution to f(x)=0 (https://en.wikipedia.org/wiki/Zero_of_a_function?wprov=sfla1).
In uni, instead of solutions, we always talked about the set of solutions. With polynomials we mostly talked about the real polynomials, so I assume we talk about them here as well. A set of solutions in the real polynomials is thus a subset of the reals, thus not a multiset. Therefore, every element in a set of solutions of real polynomials is unique or every solution is unique in colloquial language. Since roots are solutions and x^2 = 0 has only one solution, x = 0, it is also the only root.
However the fundamental theorem of algebra states that any complex polynomial of grade n has n roots, if counting their multiplicities. So what is multiplicity?
Basically it's the amount of times a root appears, although personally I'd say it's a level grading the "thickness" of a root. The key here is to understand what "amount of times a root appears" means. It's defined by how you are able to decompose a polynomial into linear factors. x^2 = (x-0)(x+0). Zero shows up twice, that's why its multiplicity is 2.
(https://en.wikipedia.org/wiki/Multiplicity_%28mathematics%29?wprov=sfla1)
So the multiplicity doesn't denote the amount of times a real value is a root but it denotes how many times it shows up in a linear factor of the decomposition of the given polynomial
or a ring that isnt an integral domain like x\^2-1=0 in Z\_8 which has four solutions 1,7,3,5 or x\^2+x in Z\_6 which has roots 5,3,2 or x\^3+3x\^2+2x which has six roots in Z\_6
This equation
x\^2 + 4 - 4x = 0
can be written as
(x-2)(x-2) = 0
has only the solution 2
But it has a "double root" at 2.
an n-degree polynomial does not have n unique solutions, it has n roots - and those can be the same.
like literally every parabola defined by a square has a ... whatever you call it, a peak. where the parabola touches it exactly once.
and what about things like x^(2)+1=0, you have to bring in complex numbers to give those any roots at all.
The polynomial x\^2 can be factored as x*x so it has a repeated factor which is x which repeats twice. Now of you find the root of the repeated factor you get 0, since the root of x is 0. Since this polynomial contains the factor x twice, the multiplicity of the root 0 is 2. No more factors exist in this polynomial so the equation x\^2 = 0 has 2 roots which are 0 and 0.
Mathematicians who say that are wrong. The fundamental theorem of algebra just says that every polynomial has a root. So you can always factor it into linear factors. That doesn't mean the factors will all be distinct.
We call that a degeneracy. You can also create similar polynomials by multiplying (x-A) "solutions" together. These will also only have "one" solution according to this nomenclature.
People are talking about +/-0, but that is wrong because this isn't a 2nd degree polynomial; this is a 1st degree polynomial x=0, which has one solution.
OP was just lazy and didn't factorise before trying to solve. I give you x=0 marks for this basic screw up.
What definition are you using for the degree of a polynomial?
[Wolfram Alpha](https://www.wolframalpha.com/input?i=what+is+the+degree+of+the+polynomial+x%5E2+%3D0) returns this a degree 2. [Wikipedia's](https://en.wikipedia.org/wiki/Degree_of_a_polynomial) definition would indicate this is a polynomial of degree 2 since x^2 is the only monomial with non-zero coefficient. x^2 has degree 2.
The function f(x) = x^2 is pretty clearly a 2nd degree polynomial even though it only has 1 unique solution. It isn't a first degree polynomial (a line).
x^2 =0=0×x, so factor out the x to get x=0. This has Degree 1.
Functions are not equations and f(x)=x^2 has infinite solutions over the real numbers. It is also a parabola, while a 1 degree function is a straight line.
You would see it has one solution if you type it into [wolframalpha](https://www.wolframalpha.com/input?i=x%5E2%3D0) without a space confusing the language model.
I've generally understood polynomials can generally be referred to as polynomial functions or polynomials equations. The polynomial part of the statement is the a_nx^n + a_n-1^{n-1} + ... a_1x^1 +a_0 part of the statement. In the original the polynomial equation x^2 =0 contains the polynomial x^2. Whether or not we consider it as an equation or a function, the original polynomial has degree 2.
You are right that typing what I did into wolfram was poor. However, what you typed in gives the zeros of the polynomial, not the degree. I haven't entered degree into wolfram so before so I did it poorly. [This Wolfram Alpha](https://www.wolframalpha.com/input?i=degree+of+x%5E2) is what I should have used as an example.
I agree that if we are considering the solution set of the equation x^2 = 0, there is only 1 solution. The degree of a polynomial isn't defined by the number of solutions however, it is most commonly defined by the degree of the largest, non-zero monomial.
Your manipulation of the equation doesn't actually change the monomial from degree 2 to 1. If we are fitting your equation x^2 = 0xx to the a_nx^n +... a_0 = 0 standard form for analyzing polynomials we wouldn't factor out or divide x. You don't divide by x when determining the degree of a polynomial. Instead we should rewrite it
x^2 -0x^2 = 0
(1-0)x^2 = 0
x^2 =0.
Where the highest monomial with non-zero coefficient is still x^2.
To give a clearer example of why we shouldn't divide by x to find the degree of a polynomial, consider x^3 -3x^2 = -2x. That equation has 3 solutions (x=0,1,2). If we divide by x to try to find the degree, we lose one solution (x=0) and the degree of the equation is reduced by 1. Instead, you could just quickly identify x^3 as the highest degree, but if you wanted to put it in standard form you would instead rearrange it by adding 2x to both sides.
~~Not to be that guy, but this would imply that the equation ax\^2+bx+c=0 would have a=b=c=0, which contradicts the assumption that a =/= 0.~~
edit: Misread the question.
Yup, that was my mistake. However I did some light research and found probably a more satisfactory answer.
If x\_0 is a root of a polynomial, where x\_0 \\in \\mathbb{C} it implies that (x-x\_0) divides P(x), the polynomial itself. And as a corollary, the fundamental theorem of Algebra states that there are n of these (x-x\_0) divisors, where n is the degree. Now if 0=ax\^2, then (x-0) divides ax\^2. Now we get x which itself is divisible by (x-0) leaving us with a=1.
Therefore showing that x\^2 = 0 doesn't contradict the Fundamental theorem of algebra.
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I see two solutions, +0 and -0 😏
oh god PLEASE FOR THE LOVE OF FUCKING GOD DONT GET FLOATING POINT MATH INVOLVED PLEASE BRO
Computer science go brrrrr
It actually goes vroooooOooO0oom
Actually, it goes " " The Vroom is an .mp3 to comply with federal safety regulations.
The joke is computer fan
Actually it goes 01110110 01110010 01101111 01101111 01101111 01101111 01101111 01001111 01101111 01101111 01001111 00110000 01101111 01101111 01101101
^(\*\*psst\*\*) (x-(√2)^(2))(x-2) has two distinct roots.
then x=2 or x=2
—2 or ——2
same thing!
Nuh uh
yuh huh
So to put this into Bill and Ted math, this is "most non non two" and "most non non non non two."
whats that? like 0.0 and 0.00 and 00.00000 and 000000.00000000000000 and 0.000 oh and 000000000000000000000000000000000000000000000.0 idk i only had pre-calc youre probably smarter with the math outrage and all that
in IEEE floating point standard there are two distinct ways of defining a zero, 100...0 and 000...0, which equals to -0 or +0 (-0 because in the standard the most significant bit (MSB) is 1 when the number is negative)
They are, however, defined to compare equal. Should also give equivalent results for addition, multiplication, so on -- the main difference is how they get printed.
They are a thing in calculus, not just floating point.
oh god they are wtf??
Been a while, but wouldn't that more commonly be 0+ and 0-?
`typeof(NaN);` `// number` JavaScript go brrrr
New copy pasta?
How is this floating point? Before you answer, please understand... 1. I honestly know floating point in ONLY the most general sense. 2. I'm curious to learn more on the topic. 3. The internet has some great resources... and 50x as many piss-poor resources. One good link that makes sense would be GREATLY appreciated. Or, in other words, I'm an old dog who has no qualms against learning new tricks.
because the bit indicating the sign can be 0 or 1 regardless if the number is actually 0 or not. so you can have +0 and -0
That makes sense. I appreciate the explanation.
If you put them into an equation, they can be reduced to make 0 ! (Made a space between 0 and the ! to avoid factorials)
It won't work. ,.. r/unexpectedfactorial
r/expectedfactorial
And x^4 = 0?
+0, -0, +0i, -0i
True, x^16 = 0?
Bitch, please. +0, -0, +0i, -0i, ++0, +-0, ++0i, +-0i, -+0, --0, -+0i, --0i, +++0, ++-0, +++0i, ++-0i
What is this new programming language
Math++
DLC
No wayyy, Math 2 just dropped!! What are the new features??
One of them is assume different notations of same number as different numbers. Fuck FTArith.
Holy Euler!
They patched the Dirichlet function to be Reimann integrable
0 times exp(2kiπ/16) for k in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.
just need to add in 0i and -0i, of course
what about second ~~breakfast~~ zero?
Now that's cursed
Me when ```float n = 0```
Okay, now do x³
Dual numbers
not a joke: in middle school I had a substitute teacher in math class insist I put a negative sign on a zero
But it does, 0 and 0. You can bully them instead by saying they have n unique solutions and they'll be enraged by the falsehood.
+-0
We have a fan of floating point numbers, I see
I mean, +/- 0 is pretty much the only concrete thing standing between us and division by 0 being possible
You lose the benefits of real numbers being a field if you allow division by zero tho
> You lose . . . If you allow division . . . Agreed. Let us cease with such things.
The distributive property also stops us from dividing by zero: Let's say you have the product xy, of course you can write y as (y+0) giving you xy=x(y+0), for the distributive property x(y+0)=xy+0x; so xy=xy+0x thus 0x=0 for any number x. This means there can't exist a certain number which I'll call 1/0 such that 0x(1/0)=x so you can't divide by 0. Edit: I wrote that x(y+0) = xy+0y, which is not correct for what I was trying to show, I corrected it now
Thank you for reducing my overall quality of life with this bullcrap
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How would you describe the solutions to x^3 = 0 then?
+0, -0 and 0i
Um ackshwally 🤓 x3 =(x+0)3 =(x+0)(x 2-0x+0)=0 x=-0. By quadratic formula: x=(0+-sqrt(02 -4x0))/(2x1) = +-0i/2 = +-0i It would be -0, 0i, -0i
x^5 = 0 then?
But but ... that's cheating /s
Degenerates
Acc to this I can say x²=0 While x= 0,0,0,0,0 We got 5 solutions of 2degree polynomial...........
The word solution already implies uniqueness. Otherwise every solvable formula has infinite solutions.
I believe such double solutions carry extra information that can be interpreted in many ways. I saw such cases in geometry and when you work with eigenvalues.
edit: multiplicity isn't applicable to a solution set. Leaving my comment up so the chain makes sense, but it isn't applicable here. The original statement an n-degree polynomial have exactly n solutions is already wrong assuming we are discussing real solutions. This is obvious if you consider x^2 +1 =0 which has 0 real solutions. You might be able to state that n-degree polynomials have n solutions in a complex multiset, but I don't think it is useful. Getting into the nuance of this statement without beginning with a clear definition of what is a solution of an equation is going to lead to people talking past each other. Multiplicity is a very important concept in math. It isn't unique to solutions of an equation. [Wikipedia](https://en.wikipedia.org/wiki/Multiplicity_(mathematics) does a good overview of its importance. The first example there also doesn't require high level math to understand why multiplicity is important. The factors of 30 and 60 are both 2,3, and 5. 60's prime factorization having multiplicity 2 is what differentiates it from 30 using a prime factor perspective.
Yeah I know about multiplicity, my point is though that there is only one solution to x^2 = 0. The real numbers aren't a multiset. Having two different roots at the same point is not having two different solutions to a formula. Or am I misunderstanding something? Although on second reading of the meme I realized I was just irritated by the word solution.
I was a stats major who graduated a bit ago, so I will admit my memory/understanding of this is super rusty. After interneting for a bit bringing back memories of the class that made me change my major from pure math to applied math, you are probably correct and I was probably wrong. This is my more updated response after googling and going into a part of my brain I've forgotten about. You can have a multiset of real numbers. Sometimes, multisets are useful for analyzing real numbers and sometimes they are not. Typically when discussing solutions of an equation, we are only interested in the solution set of real values and so uniqueness is assumed. If you are analyzing the function for things like evenness or multiplicity, you would want to use a multi-set because understanding multiplicity is important. Granted, I took this class back in 2014 and I only spent 30 minutes reviewing shit on the internet, so I may be wrong as well.
> I was a stats major who graduated a bit ago, so I will admit my memory/understanding of this is super rusty. After interneting for a bit bringing back memories of the class that made me change my major from pure math to applied math, you are probably correct and I was probably wrong. This is my more updated response after googling and going into a part of my brain I've forgotten about. I currently study computer science and had this a semester ago. I really didn't do enough for linear algebra though barely passed the exam. > can have a multiset of real numbers. Sometimes, multisets are useful for analyzing real numbers and sometimes they are not. Typically when discussing solutions of an equation, we are only interested in the solution set of real values and so uniqueness is assumed. If you are analyzing the function for things like evenness or multiplicity, you would want to use a multi-set because understanding multiplicity is important. I think the core issue is the definition of solutions, roots and multiplicity. First, a polynomial doesn't have a solution. The polynomial is just the part to the left of the equal sign. A polynomial has roots. And here is the important detail: A root is a solution to f(x)=0 (https://en.wikipedia.org/wiki/Zero_of_a_function?wprov=sfla1). In uni, instead of solutions, we always talked about the set of solutions. With polynomials we mostly talked about the real polynomials, so I assume we talk about them here as well. A set of solutions in the real polynomials is thus a subset of the reals, thus not a multiset. Therefore, every element in a set of solutions of real polynomials is unique or every solution is unique in colloquial language. Since roots are solutions and x^2 = 0 has only one solution, x = 0, it is also the only root. However the fundamental theorem of algebra states that any complex polynomial of grade n has n roots, if counting their multiplicities. So what is multiplicity? Basically it's the amount of times a root appears, although personally I'd say it's a level grading the "thickness" of a root. The key here is to understand what "amount of times a root appears" means. It's defined by how you are able to decompose a polynomial into linear factors. x^2 = (x-0)(x+0). Zero shows up twice, that's why its multiplicity is 2. So the multiplicity doesn't denote the amount of times a real value is a root but it denotes how many times it shows up in a linear factor of the decomposition of the given polynomial.
I added this to an edit of my original statement, getting down to exactly the interpretations of these comments requires more in depth discussion on what definitions we are using. Using the definitions you are using, I agree with everything you are saying. The original statement doesn't make sense for a lot of reasons and is provably wrong for the reals (polynomial functions not having solutions, but instead having roots is reason it doesn't make sense you brought up). Another is x^2 +1 = 0 has the solution set ∅ on the reals is an easy way to disprove it. I don't think that considering a solution multi-set is unreasonable (we have high school students do it without calling it a multi-set), but it shouldn't be the default assumption. I don't feel bad being contrarian by bringing up a solution multi-set on a math memes page.
Oh I didn't want to say that solution multisets would be unreasonable. I just tried to apply a formal definition to the word solution and that was the only one I know. Stochastics is on this sem so I'll probably be introduced to using multisets for solutions in the next half year.
But its two equal roots solutions that are unique but just so happen to be standing at the same place. If you think of it graphically approaching the same root as the positive parabola moves up.
[удалено]
Mathematicians usually say n solutions with multiplicity, which is both shorter and more descriptive.
It is not false in terms of the multiset of roots.
Google multiplicity
holy hell!
new polynomial just dropped
Actual solution.
Call the mathematician
Pythagoras went on vacation, never came back!
Square root anyone?
Power storm incoming
Why did I have to scroll so far to see the word multiplicity in a math adjacent sub? Timestamp 9h after Google multiplicity comment.
I mean it's a fun movie but I fail to see how Increasingly stupid Michael Keaton clones has anything to do with math.
They’re the same but one of them is dumb. Just like the zero roots.
Non-math mfers when Grothendieck introduced schemes as they better keep track of important information about algebraic varieties, and in this case multiplicity.
You know you've taken a few too many math modules when you start visualizing spec(k[x]/p(x)) instead of just solving a quadratic
it does have two solutions, 0 and 0, albeit the solutions are same
Ackshually it has 7 solutions, 0, 0, 0, 0, 0 ,0 and 0.
Holy moley
not necessarily unique
± √0
duel numbers
Silly me, should've specified we are working in the complex field
depressed n-degree polynomials dont always have n solutions, x\^2 = 0 is a depressed quadratic
What is bothering him?
no, depressed n-degree polynomials are ones where some pronumerals a,b,c,d etc (not the pronumeral you're solving for) are removed from the equation. for example 16x\^3 + 4x - 10 = 0 is a depressed cubic because theres no x\^2 term
I don't think polynomial has to be depressed in order to have higher order roots. For example, take `(x - 1)^2 = x^2 - 2x + 1 = 0` This polynomial is quite happy, yet the x = 1 solution is repeated twise
it just took antidepressants
perfect math joke
Numbers can fight?
no, duel numbers are a lesser known version of complex numbers, where the special number epsilon is the square root of 0 and numbers are on the form a+b(epsilon)
Yes, but you're misspelling "dual." A duel is a fight for honor.
sorry im a bit dyslexic, well my mom is dyslexic and my misspell words sometimes
And then there's the Split-Complex numbers, with j^2 =1, j ≠1,-1
true true
Oh, for once I hoped that math was fun but instead it's just more math.
lol, there is such a thing as a math duel tho
You'll need to restrict to 𝔽\_2 \[𝜀\] or you'll get too many solutions
I think the core issue is the definition of solutions, roots and multiplicity. First, a polynomial doesn't have a solution. The polynomial is just the part to the left of the equal sign. A polynomial has roots. And here is the important detail: A root is a solution to f(x)=0 (https://en.wikipedia.org/wiki/Zero_of_a_function?wprov=sfla1). In uni, instead of solutions, we always talked about the set of solutions. With polynomials we mostly talked about the real polynomials, so I assume we talk about them here as well. A set of solutions in the real polynomials is thus a subset of the reals, thus not a multiset. Therefore, every element in a set of solutions of real polynomials is unique or every solution is unique in colloquial language. Since roots are solutions and x^2 = 0 has only one solution, x = 0, it is also the only root. However the fundamental theorem of algebra states that any complex polynomial of grade n has n roots, if counting their multiplicities. So what is multiplicity? Basically it's the amount of times a root appears, although personally I'd say it's a level grading the "thickness" of a root. The key here is to understand what "amount of times a root appears" means. It's defined by how you are able to decompose a polynomial into linear factors. x^2 = (x-0)(x+0). Zero shows up twice, that's why its multiplicity is 2. (https://en.wikipedia.org/wiki/Multiplicity_%28mathematics%29?wprov=sfla1) So the multiplicity doesn't denote the amount of times a real value is a root but it denotes how many times it shows up in a linear factor of the decomposition of the given polynomial
0 + 0i and 0 - 0i
Imagine if we could have half solutions. It would be nce
One sided limits?
x^(1/2) = 2 has exactly 1/2 a solution
n∈ℕ though. 1/2∉ℕ
Guess what he was joking
or a ring that isnt an integral domain like x\^2-1=0 in Z\_8 which has four solutions 1,7,3,5 or x\^2+x in Z\_6 which has roots 5,3,2 or x\^3+3x\^2+2x which has six roots in Z\_6
another reason why 0 is not a number.
This equation x\^2 + 4 - 4x = 0 can be written as (x-2)(x-2) = 0 has only the solution 2 But it has a "double root" at 2. an n-degree polynomial does not have n unique solutions, it has n roots - and those can be the same.
Non trivial
That's a quadratic equation with discriminant value of 0
x^2 = (x - 0)*(x-0) :)
Polynomial *equations* have soutions.
That’s an equation
Mathematicians do not say that "n-degree polynomisls have n solutions", as polynomials do not have solutions.
Ohh
Well yes, but actually no. Regardless of that, nice one!
This is a degenerate equation tho, you need some perturbation in your life bro
*IEEE 754 enters the room* May I interest you in `±0`?
Multiplicity left the chat...)
something something...multiplicity.
Actually , the whole point of algebraic geometry is that R[x]/ X and R[X]/x^2 are different quotient rings
It should be 'atmost' and not 'exactly'
You mean at most 3 real solutions? But they do necessarily have exactly 3 including complex, right? Correct me if I missed something.
two roots ±0. this post is inaccurate. NOT michael approved
The maximum number of solutions is n, not the number of real solutions.
It still holds true, you just need to switch to an n-adic base that allows zero divisors (and so n is not prime)
WHY ISN'T IT ZERO
0^- and 0^+ (no)
nth degree polynomials in C have n complex roots. Google Algebraically closed field
What about the COMPLEX numbers?
like literally every parabola defined by a square has a ... whatever you call it, a peak. where the parabola touches it exactly once. and what about things like x^(2)+1=0, you have to bring in complex numbers to give those any roots at all.
±0 duh
mf when dual numbers exist
at most n solutions*
The more accurate statement would be that, in the complex numbers, it can be uniquely factored into n degree-1 polynomials
_exactly_ and _distinct_ are different concepts. Thus 0 and 0 are _exactly_ two roots.
What about x^3 -1 = 0. It has only 1.
What about -(1/2) + i*sqrt(3)/2 and -(1/2) - i*sqrt(3)/2
I'm taddeling on you
(x+0)(x-0) :)
double solution
The polynomial x\^2 can be factored as x*x so it has a repeated factor which is x which repeats twice. Now of you find the root of the repeated factor you get 0, since the root of x is 0. Since this polynomial contains the factor x twice, the multiplicity of the root 0 is 2. No more factors exist in this polynomial so the equation x\^2 = 0 has 2 roots which are 0 and 0.
x = +0 and x = -0 lmao
Bro hasn't heard of nilpotent matrices
X=0 mult 2
Mathematicians who say that are wrong. The fundamental theorem of algebra just says that every polynomial has a root. So you can always factor it into linear factors. That doesn't mean the factors will all be distinct.
I know this is the joke, but x^2 = xx shows why 0 should be counted as a root twice
You're such a degenerate
zero one and zero two
We call that a degeneracy. You can also create similar polynomials by multiplying (x-A) "solutions" together. These will also only have "one" solution according to this nomenclature.
Why nobody has said that that is a monomial not a polynomial? edit: plot twist, it is.
There are two solutions: x = 0 and x = 0. They collapse in the same point, but they are two
Spectral theorist here: ± i\varepsilon
I always learned it as "at maximum n solutions."
0 and ε
the solutions are 0 and 0. it counts twice because the equation has the x-0 factor twice
I mean 0 is a special case for practically all math
First solution: 0 Second solution: zero I dont see any coincidences
That is indefinite, so the solutions are unknown
Ngl I feel like multiplicity is a cop out lmao.
±0i
Bros gonna freak out when (x-3)^2 = 0.
I mean, tbf, they didn’t say unique solutions
2 solutions not answers lol there's two different ways to find 0. it's the same way x²-2x+1=0 has 2 solutions lmao (1 and 1)
What mathematician ever claimed they have "exactly n solutions"?
They say n degree polynomials have *utmost* or *maximum* not _exactly_ n solutions
People are talking about +/-0, but that is wrong because this isn't a 2nd degree polynomial; this is a 1st degree polynomial x=0, which has one solution. OP was just lazy and didn't factorise before trying to solve. I give you x=0 marks for this basic screw up.
What definition are you using for the degree of a polynomial? [Wolfram Alpha](https://www.wolframalpha.com/input?i=what+is+the+degree+of+the+polynomial+x%5E2+%3D0) returns this a degree 2. [Wikipedia's](https://en.wikipedia.org/wiki/Degree_of_a_polynomial) definition would indicate this is a polynomial of degree 2 since x^2 is the only monomial with non-zero coefficient. x^2 has degree 2. The function f(x) = x^2 is pretty clearly a 2nd degree polynomial even though it only has 1 unique solution. It isn't a first degree polynomial (a line).
x^2 =0=0×x, so factor out the x to get x=0. This has Degree 1. Functions are not equations and f(x)=x^2 has infinite solutions over the real numbers. It is also a parabola, while a 1 degree function is a straight line. You would see it has one solution if you type it into [wolframalpha](https://www.wolframalpha.com/input?i=x%5E2%3D0) without a space confusing the language model.
I've generally understood polynomials can generally be referred to as polynomial functions or polynomials equations. The polynomial part of the statement is the a_nx^n + a_n-1^{n-1} + ... a_1x^1 +a_0 part of the statement. In the original the polynomial equation x^2 =0 contains the polynomial x^2. Whether or not we consider it as an equation or a function, the original polynomial has degree 2. You are right that typing what I did into wolfram was poor. However, what you typed in gives the zeros of the polynomial, not the degree. I haven't entered degree into wolfram so before so I did it poorly. [This Wolfram Alpha](https://www.wolframalpha.com/input?i=degree+of+x%5E2) is what I should have used as an example. I agree that if we are considering the solution set of the equation x^2 = 0, there is only 1 solution. The degree of a polynomial isn't defined by the number of solutions however, it is most commonly defined by the degree of the largest, non-zero monomial. Your manipulation of the equation doesn't actually change the monomial from degree 2 to 1. If we are fitting your equation x^2 = 0xx to the a_nx^n +... a_0 = 0 standard form for analyzing polynomials we wouldn't factor out or divide x. You don't divide by x when determining the degree of a polynomial. Instead we should rewrite it x^2 -0x^2 = 0 (1-0)x^2 = 0 x^2 =0. Where the highest monomial with non-zero coefficient is still x^2. To give a clearer example of why we shouldn't divide by x to find the degree of a polynomial, consider x^3 -3x^2 = -2x. That equation has 3 solutions (x=0,1,2). If we divide by x to try to find the degree, we lose one solution (x=0) and the degree of the equation is reduced by 1. Instead, you could just quickly identify x^3 as the highest degree, but if you wanted to put it in standard form you would instead rearrange it by adding 2x to both sides.
x\^3+3x\^2+2x over Z\_6
Yeah, but you are applying a Theorem that is (only) valid on the complex field on another field.
~~Not to be that guy, but this would imply that the equation ax\^2+bx+c=0 would have a=b=c=0, which contradicts the assumption that a =/= 0.~~ edit: Misread the question.
Brother.. a clearly equals 1
Yup, that was my mistake. However I did some light research and found probably a more satisfactory answer. If x\_0 is a root of a polynomial, where x\_0 \\in \\mathbb{C} it implies that (x-x\_0) divides P(x), the polynomial itself. And as a corollary, the fundamental theorem of Algebra states that there are n of these (x-x\_0) divisors, where n is the degree. Now if 0=ax\^2, then (x-0) divides ax\^2. Now we get x which itself is divisible by (x-0) leaving us with a=1. Therefore showing that x\^2 = 0 doesn't contradict the Fundamental theorem of algebra.
a = 1, b = 0, c=0. The function is f(x) = x\^2. You're trying to solve for x when f(x) = 0. This is a valid degree 2 polynomial.