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March 17

is mathematics subjective?

So my understanding is that math isn't falsifiable, because it doesn't make physical predictions. Is it then subjective?

For example, whether "something has been proved", is that a question of viewpoint, a subjective thing, or is it objective? —Preceding unsigned comment added by 79.122.42.52 (talk) 01:41, 17 March 2008 (UTC)[reply]

I can't speak for all of mathematics but probability is subjective.

The probability of an event depends on what the observer/questioner knows. So two different observers can assign two different probability to the same event if they have different knowledge. 202.168.50.40 (talk) 02:26, 17 March 2008 (UTC)[reply]

Mathematics is objective in the sense that all proofs are derived from a set of axioms, so unlike the physical sciences where different experiments can lead to different results, whenever you start from the same axioms you'll get the same result for your proof. That said, the choice of axioms is somewhat subjective, since there is nothing to say that axiom set A is any "better" than axiom set B, although set A may result in something more familiar such as the natural numbers. Confusing Manifestation(Say hi!) 03:26, 17 March 2008 (UTC)[reply]

Sets of axioms are necessarily completely different, according to Gödel's incompleteness theorems. Because there are an infinite number of sets, subjectivity can not exist. Mac Davis (talk) 05:06, 17 March 2008 (UTC)[reply]

Mathematics is falsifiable, in that guesses can be checked. That's what falsifiability, as a doctrine, is for in the sciences - make sure your clever guesses aren't dead wrong. In mathematics, what's called proof serves roughly the same purpose. We can check our guesses. Whether we can check the system we use to check our guesses is open to philosophical debate, as the system of empirical validation used in the sciences is. Maybe it works, but maybe we live in god's big toe and he's playing games with us. That's a different level of fact-checking, and one that even falsifiable theories can't pass. So, to answer your question, mathematics is subjective to roughly the same extent that science is. Black Carrot (talk) 06:14, 17 March 2008 (UTC)[reply]

Mathematics differs in subjectivity from other sciences in a quantitative, not qualitative, way. Mathematics certainly has a higher standard of derivation - while in science, repeating an experiment several times passes as "proof" for its result, in mathematics we prefer to formulate our statements in an abstract way and derive them in a way we consider "logical". But don't be fooled into thinking that mathematics is completely rigorous, objective and devoid of human weaknesses. Not so much because of the choice of axioms (since it is understood that establishing a theorem doesn't mean that it's absolute, but rather that it follows from the specified axioms), but because of the choice of phrasing and logical inference. We have rules like "if we know A and we know A->B then we can deduce B", but there is nothing objectively "true" about this rule, other than that humanity's experience over the eons seems to support it. To this extent, mathematics relies on empirical evidence the same way as other sciences, the only difference being the amount of expected evidence. -- Meni Rosenfeld (talk) 06:49, 17 March 2008 (UTC)[reply]

The axioms of the logic in use are, to me, just as much part of the "choice of axioms" as the more mathematical axioms. It turns out that these are deeply intertwined anyway; a lovely result I saw once was that DeMorgan's Laws as logical axioms are equivalent to axiomatically stating that maximal ideals of commutative rings are necessarily prime -- one statement appears purely logical, the other most definitely in the realm of math not logic, yet in practice they are equivalent axioms. What I'm getting at with this little digression is that I don't really see the distinction you make between "choice of axioms" and axioms of logic that we assume. -- Leland McInnes (talk) 14:19, 17 March 2008 (UTC)[reply]

As I understand it - something that is 'subjective' depends on who is examining it. So a good maths proof should not be subjective.

Really though 'subjective' and 'objective' have no meaning here ie all mathematical knowledge should be 'objective'. Subjective examples of mathematical things include: imperfect models of phyical behaviour, best estimates of statisical behaviour/probabilities - eg things that are (educated) guesse.

I suggest reading about 'subjective and objective' and forming your own opinion. Picked these definitions 'at random' from the net:

Subjective: characteristic of or belonging to reality as perceived rather than as independent of mind : phenomenal

Objective: of, relating to, or being an object, phenomenon, or condition in the realm of sensible experience independent of individual thought and perceptible by all observers : having reality independent of the minds objective reality

Emphasis mine.87.102.13.144 (talk) 11:59, 17 March 2008 (UTC)[reply]

Answer - when something has been proved mathematically that should be objective, and not subjective knowledge. (It's work noting that peoples perception of what the words 'subjective' and 'objective' mean can be a bit subjective...)87.102.13.144 (talk) 14:07, 17 March 2008 (UTC)[reply]

Yes, something that has been proved should be objective, in the sense that we would want it to be. Unfortunately, it is not, per the responses above. -- Meni Rosenfeld (talk) 15:06, 17 March 2008 (UTC)[reply]

Our choice of what to try and prove is very subjective, but the actual proof is objective. We don't prove XYZ, we prove that our choice of axioms implies XYZ. That choice of axioms is very subjective, but the implication is entirely objective (give or take Godel, anyway - even then, I don't think it's subjective, it's just not wholly reliable). So, I guess the answer to the question "Is maths subjective?" is that maths is a subjective collection of objective facts (there is more to maths than a collection of facts, of course, but for the sake of this discussion I think that definition will do). --Tango (talk) 15:59, 17 March 2008 (UTC)[reply]

I disagree. First, nobody proves theorems in a completely formal language. People use natural language to describe their proofs, and hope that others will interpret it as they meant and agree that each of their implications is indeed valid. Sometimes it works, but there is still a lot variance in people's interpretation of arguments in a language such as used in articles. Second, even if the proof was written in a completely formal language, it still relies on the person's ability to validate each step. This validation is also composed of steps that can only be described in a natural language ("look at this line, see what's written here, compare it to what you see there..."). Especially for a long and complicated proof (like most proofs will be when written in a formal language), the possibility of human error is huge. It's also not unthinkable for the same person to make the same error repeatedly because of a certain mindset he is in. Thus different people will report different views on the correctness of a proof. In fact, when given a certain proof, there is no way for us to tell if it is truly valid (whatever that means) or is it just that every person who has ever examined it happened to mistakenly accept it. Thus I reiterate my claim that everything in mathematics is subjective, but to a much lesser extent than some other sciences. -- Meni Rosenfeld (talk) 16:15, 17 March 2008 (UTC)[reply]

What you say is certainly true, but I'm not sure if that's really a form of subjectivity - if it is, then nothing is truly objective and the question is moot. --Tango (talk) 17:16, 17 March 2008 (UTC)[reply]

Personally, I do believe there is an "objective truth", but that anything a human can ever discover is only a crude approximation of it, and thus subjective. The original question is not moot, as a priori one might think that modern mathematics has, at least in some parts, achieved the holy grail of true objectiveness - my argument is that this is not the case. It's probably as close as we'll be for a while, though. -- Meni Rosenfeld (talk) 18:38, 17 March 2008 (UTC)[reply]

Simultaneous Equation

how to solve this simultaneous equation 4x+y+8=x²+x-y=2 —Preceding unsigned comment added by 60.48.198.225 (talk) 07:32, 17 March 2008 (UTC)[reply]

You have the two equations

4x+y+8=2\;\!

and

x^{2}+x-y=2\;\!

. There is a simple operation you can take to get rid of the y's and have an equation with only x. Can you see it? -- Meni Rosenfeld (talk) 07:35, 17 March 2008 (UTC)[reply]

Infinitely Many Factors

Consider the ring of formal power series over something like the integers. Is there any power series that can be split into infinitely many (non-unit) factors? That is, is there one where the process of pulling off factors will never stop with some number of irreducible factors? Black Carrot (talk) 10:02, 17 March 2008 (UTC)[reply]

Maybe I'm missing some subtlety in the question, but isn't

\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =x\prod _{n=-\infty }^{\infty }\left(1-{\frac {x}{\pi n}}\right)

an example of a power series with an infinite number of irreducible factors ? Gandalf61 (talk) 10:44, 17 March 2008 (UTC)[reply]

Another example (unless I, too, misunderstand the question) is

\prod _{n=0}^{\infty }\left(1+2^{-n}x\right)=\sum _{n=0}^{\infty }{\frac {2^{n}}{\prod _{k=1}^{n}(2^{k}-1)}}x^{n}

. -- Meni Rosenfeld (talk) 10:50, 17 March 2008 (UTC)[reply]

Those are units: (1-ax)(1+ax+a^2*x^2+a^3*x^3...) = 1. You can always pull off infinitely many unit factors, even from something irreducible. Black Carrot (talk) 11:12, 17 March 2008 (UTC)[reply]

Oh, I see. No idea then. -- Meni Rosenfeld (talk) 11:22, 17 March 2008 (UTC)[reply]

Let's see if I have got this straight. You want an infinite number of factors each of which has, say, integer coefficients, is irreducible and is not a unit when considered as a formal power series over the integers ? Let's take out the largest power of x that we can, so we have an initial factor of x^k for some k (possibly 0). Doesn't this mean that the constant term of every other factor cannot be 0 (otherwise it is reducible) or +/-1 (otherwise it is a unit when considered as a formal power series) ? So you have x^k times an infinite number of factors each of which has constant term with absolute value greater than or equal to 2 - which seems to imply that the coefficient of the lowest power of x in the product is not finite, so the product is not a power series. Gandalf61 (talk) 11:29, 17 March 2008 (UTC)[reply]

What is meant by "non unit factor"?87.102.13.144 (talk) 11:47, 17 March 2008 (UTC)[reply]

In this context, a unit is something you can divide by without leaving the number system. For instance, in the whole numbers 1 is a unit, but anything bigger isn't. That's why 1 isn't considered prime - every number would have infinitely many prime factorizations, eg 5 = 5*1 = 5*1*1 = 5*1*1*1... You're right Gandalf, I hadn't thought of looking at it like that. So, does that mean that power series over the rationals also have finitely many factors? Since everything with a constant term would be a unit, and everything without could be divided by x. Black Carrot (talk) 12:04, 17 March 2008 (UTC)[reply]

Meni, should that be 2^n at the top of your last formula? Black Carrot (talk) 12:09, 17 March 2008 (UTC)[reply]

Yes, indeed. Fixed. -- Meni Rosenfeld (talk) 14:55, 17 March 2008 (UTC)[reply]

Was the first example given not an example? eg a power series over rational fractions expressed, but made of factors which are irrational (contain 1/pi)?87.102.13.144 (talk) 14:12, 17 March 2008 (UTC)[reply]

The original question needs to be made more precise. What do you mean by "something like the integers"? For example, are algebraic integers something like the integers? And wat about the complex numbers? Should the factors be over the same ring as the power series? Can the factors again be formal power series, or should they be polynomials? --Lambiam 14:49, 17 March 2008 (UTC)[reply]

Gandalf has solved one form of the question: I believe his proof shows that if R is a Noetherian integral domain, then any element of R[[X]] can be expressed as a product of finitely many irreducibles in R[[X]]. Algebraist 16:09, 17 March 2008 (UTC)[reply]

math: polynomial as the difference of two squares

How do you determine if a polynomial is the difference of two squares? —Preceding unsigned comment added by 72.94.230.129 (talk) 14:08, 17 March 2008 (UTC)[reply]

Assuming that by "square" you mean the square of a polynomial, and that the coefficients are from a number system that is closed under division by 2, such as the rationals and the reals, any polynomial P can be rewritten as:

\left({\frac {P+1}{2}}\right)^{2}-\left({\frac {P-1}{2}}\right)^{2}.

--Lambiam 14:33, 17 March 2008 (UTC)[reply]

I'm not sure that's what the OP means. We studied polynomials last year in algebra I. We learned the difference of two squares as

(x+S)(x-S)\,\!

, in which S is any number. This then takes the form

x^{2}-S^{2}\,\!

. So, to determine if a polynomial is the difference of two squares, just look for two square terms (such as

x^{2}-64

). The solution to a polynomial of the form

(x+S)(x-S)

is

x=\pm {S}

. So the solution to

x^{2}-64

would be

x=\pm {8}

.

Disclaimer: I learned these things last year so they are subject to possible error.

Zrs 12 (talk) 19:48, 17 March 2008 (UTC)[reply]

That's right, but the principle can be used for more complicated polynomials, as well. For example

x^{2}-y^{2}+8x+4y+12=x^{2}+8x+16-y^{2}+4y-4=(x+4)^{2}-(y-2)^{2}=(x+4+y-2)(x+4-y+2)=(x+y+2)(x-y+6)

. If you want everything to have integer coefficients, then I'm not sure there is a general algorithm to find out if you can write a polynomial as the difference of two squares, you just have to fiddle around with it and see what happens (you can try completing the square as a starting point). --Tango (talk) 20:08, 17 March 2008 (UTC)[reply]

We also have an article on this. See difference of two squares. Zrs 12 (talk) 20:06, 17 March 2008 (UTC)[reply]

Tango: A polynomial (over Z, say) is the difference of two squares iff it can be factorised with the sum of the (two) factors being divisible by 2 (i.e. corresponding coefficients in the factors have to be of the same parity). ~~Thus a poly is a difference of two squares iff its reduction mod 2 is a square.~~ Algebraist 23:25, 17 March 2008 (UTC)[reply]

Or rather not. We get an algorithm, anyway. There's an algorithm that factors a polynomial over Z into its irreducible factors, so we can just test all possible factorizations. I've no idea if this can be improved on. Algebraist 23:28, 17 March 2008 (UTC)[reply]

need a more indepth determination

Bold textHow do you determine if a polynomial is the difference of two roots? Thats how the question was brought to me.I don't quite get it either. —Preceding unsigned comment added by 72.94.241.142 (talk) 19:36, 17 March 2008 (UTC)[reply]

The difference of two roots? Well, assuming you mean square root, any number is a square root because it is able to be squared. Just go down a number line and mulitiply each term by itself. For example

S

is the square root of

S^{2}

; 4 is the square root of 16. However, every number is not a square because not every number has an integer square root. I think the answer to your previous question above may be what you seek. Zrs 12 (talk) 19:55, 17 March 2008 (UTC)[reply]

Solve for y: x = y - 1/y

How can you solve the equation $x=y-{\frac {1}{y}}$ for y in terms of x? I know that the solution is $y={\frac {x\pm {\sqrt {x^{2}+4}}}{2}}$ , and this works when you substitute it back into the original equation, but how can this be found analytically? (I used a calculator). --BrainInAVat (talk) 20:59, 17 March 2008 (UTC)[reply]

Try multiplying both sides by $y$ $y$ ... you should then have a quadratic in terms of $y$ $y$ which can be solved using the "usual method". --Kinu ^t/_c 21:14, 17 March 2008 (UTC)[reply]
- I'm at a loss, sorry. You get $xy=y^{2}-1$ , but then what?--BrainInAVat (talk) 01:29, 18 March 2008 (UTC)[reply]

Take it all over to one side and you have a quadratic in y (just treat x like a number), you can then solve it using the quadratic formula and you'll get the answer you gave above. --Tango (talk) 01:53, 18 March 2008 (UTC)[reply]

If you are not familiar with it, you can take a look at out Quadratic equation article. -- Meni Rosenfeld (talk) 07:19, 18 March 2008 (UTC)[reply]

This can be a little confusing the first time you see it - here:

x=y - 1/y

xy=y² - 1 (equation 1)

y² - xy - 1 = 0

Now complete the square

(y-x/2)² - x²/4 - 1 = 0

As you should be able to see multilpling by y gives a quadratic in y (equation 1)

The solve the quadratic by completing the square - you can see that x is a parameter of the quadratic. Hopefully from the last equation you should be able to get the result.

Once you've got it, keep an eye out for equations of this type as it's a useful thing to remember.87.102.74.53 (talk) 11:35, 18 March 2008 (UTC)[reply]