2005 Q8 (1 Viewer)

[Templar]

they have to firstly know what transcendental means. Be interesting if they have like a whole paragraph telling you just exactly what transcendental means.

A number that is not algebraic

or

A number that is not the root of any polynomial with integer coefficients.

The meaning isn't hard, just that it takes a lot of working and effort to prove it.

 

[withoutaface]

Does it also mean that no exponent of it will be rational?
EDIT: nm, that was covered in the 2nd point.

*stupid*

 

[who_loves_maths]

Originally Posted by Templar
[a transcendental number is] A number that is not the root of any polynomial with integer coefficients.

The meaning isn't hard, just that it takes a lot of working and effort to prove it.

if the meaning isn't hard, then how come a university student can't even get it right? and at the same time expect HSC students to know it fully?

A transcendental number is an algebraic number that is not the root of any polynomial with rational (not just integral) coefficients.
the set of integers is a small subset of the rationals...

 

[withoutaface]

if the meaning isn't hard, then how come a university student can't even get it right? and at the same time expect HSC students to know it fully?

A transcendental number is an algebraic number that is not the root of any polynomial with rational (not just integral) coefficients.
the set of integers is a small subset of the rationals...

Any polynomial with rational coeffecients can be expressed as one with integer coefficients by multiplying both sides by the product of the denominators.

 

[Templar]

if the meaning isn't hard, then how come a university student can't even get it right? and at the same time expect HSC students to know it fully?

I don't. As a matter of fact I don't even expect an honours student to get it right, as you say. It was thrown in purely as a joke.

Just a random point, the proof that e^(i*pi)+1=0 forms the basis of proving that e is transcendental.

 

[who_loves_maths]

Originally Posted by withoutaface
Any polynomial with rational coeffecients can be expressed as one with integer coefficients by multiplying both sides by the product of the denominators.

yes of course {although, you are not addressing the real problem which is the wording of the definition of a trascendental number}, but Templar ignored to mention that in his post - giving the impression that he did not actually know this fact...

but more importantly, mathematicians are very technical and pedantic about things such as notation, definition, and wording of questions, etc...
so in this case, as i said before, with the set of integers being a subset of the set of rationals, it is more accurate and correct that the definition of a trascendental number in terms of polynomials include the rational numbers - since they present a large set and range of numbers than the integers.

moreover, apart from wanting to be general, there is a good logical reason to say "polynomials with rational coefficients" rather than "polynomials with integral coefficients" because if you are presented with a polynomial P(x) with rational coefficients, then by Templar's definition, you can say that the roots to this polynomials CAN be transcendental since the polynomial does not have integral coefficients {and, again, this is because the set of integers is a subset of the rationals, meaning that many rational numbers are not part of or an element of the set of integers}! - and that deduction is obviously wrong!

[Note: the above logic shows the power of the idea of Sets in mathematics, that's why it is so important to get definitions and wording, etc, right in maths.]

so there is a logical flaw (albeit subtle) in Templar's definition, and it is NOT technically correct to simply say "polynomials with integral coefficients" - you must mention rational coefficients.


as to solving a polynomial: although the method by which you go about FINDING the actually roots might involve you multiplying to get integer coefficients and equating to 0, what you are actually doing is you are TRANSFORMING {and thus changing} the initial polynomial P(x) with rational coefficients into another polynomial that has the same roots but different shape - so although your job of finding the actual roots is easier, the transcedence of numbers, you'll notice, is NEVER defined in terms of the METHODS of finding roots to polynomials!

^ eg. P(x) = (1/2)x + 1, and, P(x) = x + 2, are NOT the same polynomials - they just share the same zeroes. But when we talk about transcendental numbers, the polynomial P(x) is considered BEFORE equating it to 0, beyond which you can multiply all you want.

so through pure definition, and this is the difference between definition and practice, it is not complete to not mention "rational", and it does not suffice to simply mention "integral" in defining a transcendental number.


i hope you can see the logic of this situation

 

[withoutaface]

Definitions have corrollaries, it doesn't make them any less valid. In the case of that definition it had the corrollary you mentioned, it's still accurate to say that because it is a trivial matter to get between the definition and its corrollary.

moreover, apart from wanting to be general, there is a good logical reason to say "polynomials with rational coefficients" rather than "polynomials with integral coefficients" because IF you are presented with a polynomial P(x) with rational coefficients, then by Templar's definition, you can say that the roots to this polynomials CAN be transcendental since the polynomial does not have integral coefficients {and, again, with the set of integers as subset of the rationals, meaning that many rational numbers are not part of or an element of the set of integers}! - and that deduction is obviously wrong!

The statement was not that integer coefficient polynomials do not have trancendental roots, although that is a converse of the statement. It was that transcendant numbers are not the root of any integer coefficient polynomials, and since it would be easier to prove transcendentalism through only considering integer polynomials, and the proof would be just as valid, I feel this definition is more apt for describing transcendental numbers. You're thinking of working from polynomials to roots, the definition of a transcendental number has to move from roots to polynomials.

 

[who_loves_maths]

^ NO, this is where you are wrong... you can't consider equating P(x) to 0 BEFORE considering trascendentalism, because you will be changing the initial polynomial.

like i editted into my last post: P(x) = (1/2)x +1 is not the same polynomial as P(x) = x + 2, even though they have the same zero.

this is why Templar is inaccurate (but not wrong) as well.

with the set of integers being a subset of the rationals, then according to Templar 's definition (and evidently yours):

[a transcendental number is] A number that is not the root of any polynomial with integer coefficients.

means that when you get a P(x) with rational coefficients, then Templar would be saying that the roots of this polynomial with rational coefficient CAN (but not necessarily are) very well be transcendental since they don't fit the his definition of "A number that is not the root of any polynomial with integer coefficients." - so that a number that IS the root of a polynomial with RATIONAL coefficients is then NOT included as a part of his definition.
but of course, you and i (and Templar) both know that is not the case, because roots of polynomials with rational coefficient simply ARE NOT (not even a maybe in "can") transcendental.


i don't think you are getting the logic of the situation at all! and in maths, contrary to what you might believe, it is not sufficient to be lazy with definitions - if you are, then that's where you miss the subtleties, such as this case.

eg. assume for a moment that ALL coloured bananas are coloured. and assume you don't know this "fact" already, but then someone comes to you and says "ALL yellow coloured bananas are very bitter tasting" and then hands you a green banana... would you bite into it (assuming that you are ravenous)?

^ of course, this example is different to the machinations of mathematics, but the principle is very similar - given only the information about yellow bananas being bitter, then you would bite into the green banana because there is a chance that it CAN be sweet ... but the fact is, ALL coloured bananas are in fact bitter, green included, and moreover, when you let a green banana sit for a while you can change its colour to yellow - but that doesn't mean when you originally saw it's green colour you would automatically know that it was bitter.

The parallel is that Templar is the person telling you that all yellow bananas are bitter while in fact all coloured bananas are bitter - so when he goes on to give you a green banana there is no reason why that green banana, to you, can't be sweet and so you might well take a bite - in which case you fall into the trap of tasting bitterness.


it's a simple, but subtle, logical process that Templar's definition does not fit into correctly that makes his definition insufficient (not wrong) to encompass the totality of transcendentalism.

 

[withoutaface]

One person defines 2 is a member of R.
Another defines 2 as a member of N.

Which, in your opinion, is closer to the correct definition?

 

[who_loves_maths]

^ what sort of exampleis that? you analogy does not depict Templar's problem of defining transcendental numbers.

clearly you don't fully grasp the concept of Sets:

both your statements are accurate, indeed 2 is a member of R (reals i assume), and also member of N (natural numbers i assume). but that is because N is a subset of R, which we know already... so there is no dilemma here.

the analogy, extending your example, that would be closer to the problem of defining transcendental numbers here would be something like:

If a person says that ALL member of N is = X, where 'X' is an unknown 'entity/quality/property', then when you meet a member '@' of R, but that which also belongs to N, then you can say that @ = X.

now, if, in fact, ALL members of R is also = X, but you don't know that (because the person telling you about N didn't mention it), then the next time you come across a member '#' of R, but that which is NOT a member of N, then would you say 100% of the times that # = X ? of course you wouldn't, cause it'd be silly since you don't know anything of R being = X, only that members of N = X.

so in another words, for you, you would think that '#' would not necessarily = X ; there will be an element of uncertainty and doubt in your mind - which translates into a probability that you will say that '#' is NOT = X.
in which case you'd be completely wrong! because all members of R are in fact = X.


P.S. i'm not going to argue with you further on this ... you can reply if you want, and if you don't agree with what i'm saying and think you are in the right then we'll just leave it at that, i have no problems with that It's just too bad that i don't have anyone around me other than my maths teacher who i can ask and confirm this, and i don't suspect my teacher will be too interested or even know... but since you're at uni, then i hope that maybe you can confirm your view, or otherwise, with someone who teaches there if you have the time to ask? if you are right on this, then it's worth confirming on i guess? and i wouldn't mind if you tell me afterwards what your teachers/tutors/lecturers has to say on it ?

 

[Meads]

me good at general math

 

[Templar]

Thank you for your concerns. Now please direct them to Wolfram, as I quoted from them on the definition instead of my own thinking that it would remove any possible errors or ambiguity.

 

[withoutaface]

Thank you for your concerns. Now please direct them to Wolfram, as I quoted from them on the definition instead of my own thinking that it would remove any possible errors or ambiguity.

 

[yrtherenonames]

I dunno i reckon who_loves_maths would bitch-slap wolfram

 

[ogmzergrush]

yes of course {although, you are not addressing the real problem which is the wording of the definition of a trascendental number}, but Templar ignored to mention that in his post - giving the impression that he did not actually know this fact...

but more importantly, mathematicians are very technical and pedantic about things such as notation, definition, and wording of questions, etc...
so in this case, as i said before, with the set of integers being a subset of the set of rationals, it is more accurate and correct that the definition of a trascendental number in terms of polynomials include the rational numbers - since they present a large set and range of numbers than the integers.

moreover, apart from wanting to be general, there is a good logical reason to say "polynomials with rational coefficients" rather than "polynomials with integral coefficients" because if you are presented with a polynomial P(x) with rational coefficients, then by Templar's definition, you can say that the roots to this polynomials CAN be transcendental since the polynomial does not have integral coefficients {and, again, this is because the set of integers is a subset of the rationals, meaning that many rational numbers are not part of or an element of the set of integers}! - and that deduction is obviously wrong!

[Note: the above logic shows the power of the idea of Sets in mathematics, that's why it is so important to get definitions and wording, etc, right in maths.]

so there is a logical flaw (albeit subtle) in Templar's definition, and it is NOT technically correct to simply say "polynomials with integral coefficients" - you must mention rational coefficients.


as to solving a polynomial: although the method by which you go about FINDING the actually roots might involve you multiplying to get integer coefficients and equating to 0, what you are actually doing is you are TRANSFORMING {and thus changing} the initial polynomial P(x) with rational coefficients into another polynomial that has the same roots but different shape - so although your job of finding the actual roots is easier, the transcedence of numbers, you'll notice, is NEVER defined in terms of the METHODS of finding roots to polynomials!

^ eg. P(x) = (1/2)x + 1, and, P(x) = x + 2, are NOT the same polynomials - they just share the same zeroes. But when we talk about transcendental numbers, the polynomial P(x) is considered BEFORE equating it to 0, beyond which you can multiply all you want.

so through pure definition, and this is the difference between definition and practice, it is not complete to not mention "rational", and it does not suffice to simply mention "integral" in defining a transcendental number.


i hope you can see the logic of this situation

You amuse me to no end, much like this colourful chap.

 

[velox]

haha ur prob right nath. lol.

 

[who_loves_maths]

Originally Posted by Templar
Thank you for your concerns. Now please direct them to Wolfram, as I quoted from them on the definition instead of my own thinking that it would remove any possible errors or ambiguity.

Originally Posted by withoutaface
Hahahhahahahhaha fucking OUCH!

hey, like i said, if you don't agree with me that's fine. if you can't hold an argument of your own against me and need to resort to other sources of information then that's also fine.

However, before you get too excited, you will have noticed that Wolfram uses very pedantic and accurate words to describe what it is talking about (which is why you are using it as reliable supporting evidence i presume), and this is what it says in relation to transcendentals:

"A transcendental number is a number that is not the root of any integer polynomial ... "

and, unlike our unscrupulous use of "root" as oppose to "zero" of a polynomial in our posts, when Wolfram says "root", it means ROOT:

"The roots (sometimes also called "zeros") of an equation f(x) = 0, are the values of 'x' for which the equation is satisfied."

so in other words, when Wolfram used the word "root" in relation to "root of any integer polynomial", it ALREADY assumed the polynomial in EQUATION form (ie. = 0), in which case i have agreed already too that yes when you equate a polynomial EXPRESSION P(x) to 0 to form an equation, then you can change any polynomial with rational coefficients into a polynomial equation with integer coefficients by "multiplying all you want" [as i said].
so since Wolfram ALREADY assumed the equation (not expression) form, then there is NO NEED to refer to rational polynomial, just "integer polynomial" will do.

so now you tell me, how does that refute anything i argued before?
don't forget that i was arguing from the point of view of a polynomial EXPRESSION - that when you are handed one with rational coefficients then its NOT the same as a polynomial (expression) with integer coefficients.

Wolfram doesn't contradict nor deny anything that i said. Can you imagine what Wolfram's definition of transcendental numbers would be if it was talking in relation to a polynomial expression, and not an equation? do you think it would, then, use the term "integer" or "rational" in reference to the polynomial ?


so before you both get too jumpy over this, maybe you should take a basic maths101 lesson in the difference between an equation and an expression in mathematics... maybe go back to year 10?

i've often wondered why teachers used to stress the difference between a "root" and a "zero" in junior mathematics, i guess i know why now .

 

[withoutaface]

I've emailed the director of postgraduate mathematics at USyd about this matter. I'll post his reply when I get it.

 

[velox]

jeez wlm what school do u go to?

 

[Templar]

so before you both get too jumpy over this, maybe you should take a basic maths101 lesson in the difference between an equation and an expression in mathematics... maybe go back to year 10?

It was a joke. Get over it. No need to construct pages of essay on how incorrect it is mathematically on some minor details.