Have we discovered or invented maths?

[honestrosewater]

Quite. And there are different ways of arriving at things (in this case reals)
that follow the same rules. But because they do folllow the same rules
we say they are all different construction of the same objects. The point being
that we are not free to construct 'new' reals that follow different rules form the old ones -- the new objects just wouldn't be reals. Hence maths is not
'created' in an artistic sense (which does not mean it is discovered in a Platonic sense; discovery vs invention is not a genuine dichotomy).

Would you agree that, say, poems are "created" in an artistic sense? Many poems, though constructed of different words and having different structures, express the same meaning, so they are semantically the same and follow the same semantical rules.

 

[matt grime]

Quite. And there are different ways of arriving at things (in this case reals)

that follow the same rules. But because they do folllow the same rules
we say they are all different construction of the same objects. The point being
that we are not free to construct 'new' reals that follow different rules form the old ones -- the new objects just wouldn't be reals. Hence maths is not
'created' in an artistic sense (which does not mean it is discovered in a Platonic sense; discovery vs invention is not a genuine dichotomy).

But the reason why we are "not free to construct new reals that follow different rules" is because we have made the rules that they satisfy such that there is only one complete totally ordered field.


If we just take the axioms of a field, there are infinitely many non-isomorphic examples of them.

It is a convention that once we have fixed the rules we don't change them as this avoids confusion, but if you look at cutting edge research then you'll see many different definitions with the same name competing to see which one is the one we ought to adopt.

And remember we chose the rules that the objects satisfy. Just like we can choose to write poems in tetrameter or pentameter, or maybe make it a haiku.

 

[C0nfused]

Hi everybody,
I had some time to check this thread and I am surprised from the number of answers. By reading all of them, I first of all saw that there are many different interpretations of mathematics. I come to think that each person has a unique "understanding" of mathematics, and has interpreted math concepts in such a way that he can be more efficient in studying and solving math problems. I also agree with this:

We discover the consequences of our inventions.

One more thought: there can't be right or wrong in this issue as there wasn't one person that "created" or "invented" maths so there is not any specific and "formal"-strict definition of it. However i don't agree with this

I think the premise of this thread has no point. Whether math is discovered or invented, you're still going to use it. That's because the formulas and postulations can be stacked up against arbitrary amounts of data and still remain consistent in their conclusions. Thus they have become axiomatic and factual in nature.

Answer:
I think that trying to understand the nature of mathematics will help to make you understand maths better. And why do you think that maths are definitely going to be used for ever(that's what you imply by saying that "you're still going to use it")? I know that maths has proved it's importance all these thousands of years but this doen't mean that it will stay there for ever. Or does it? (I have just remembered something i think Hardy(is this written this way?) said, that Archimides will be remembered when Aeschylus will have been forgotten,because languages "die" while mathematical truths are eternal ( or something like that!))

 

[Doctordick]

Matt: does it make sense to say that, e.g real numbers could be constructed differently and still be (in the standard) sense real numbers?

That depends on what is meant by the phrase "and still be (in the standard ) sense real numbers? If the term "real numbers" is no more than a tag for entities which obey the set of rules divined to be "the rules of real numbers", what does the term differently mean?

I think you are just stirring that pot of vague terms with the hope that, if you stir it enough, they will stop being vague.

Have fun -- Dick

 

[Tournesol]

That depends on what is meant by the phrase "and still be (in the standard ) sense real numbers? If the term "real numbers" is no more than a tag for entities which obey the set of rules divined to be "the rules of real numbers", what does the term differently mean?

I was trying to edge Matt towards the conclusion that "differently" has nothing to refer to in this case.

 

[mathwonk]

name dropping:
gee. i drove professor penrose to the airport a couple weeks ago, but didn't ask any of these things. he was extremely charming however. he remarked it took him about 8 years to write "road to reality", but not having read it, i could not question him intelligently.

commentary:
as to math: I suggest theorems are discovered; proofs are invented.

i.e. the phenomena are built into the structure of the universe, but we invent tools to discover them, and verify them to our senses.

so there.

of course i could be wrong, i was once or twice before.

 

[metrictensor]

The origin of mathematics is the mind's ability to discriminate. When I take one apple and another I say I have two apples and have defined addition. The question is how do we relate the abstract truth of 1+1=2 to the particular cases.

 

[metrictensor]

name dropping:
gee. i drove professor penrose to the airport a couple weeks ago, but didn't ask any of these things. he was extremely charming however. he remarked it took him about 8 years to write "road to reality", but not having read it, i could not question him intelligently.

commentary:
as to math: I suggest theorems are discovered; proofs are invented.

i.e. the phenomena are built into the structure of the universe, but we invent tools to discover them, and verify them to our senses.

so there.

of course i could be wrong, i was once or twice before.

There is nothing "out there" in the sense you are thinking. No structure built into the universe. All of physics is just a conceptual imputation on the physical world. Apples fall from trees without knowing how fast they are going or that they are accelerating at 9.8 m/s^2. There are no patterns out there in the sense of being independent of the observer.

 

[metrictensor]

If you want to understand the question "are mathematical truths discovered or invented?" answer the question "are the rules for the combination of words [i.e. syntax] discovered or invented?"

 

[eNathan]

Wow, and all this is theoritcal. On another note, mathematics do not describe everything in exisence because we are then left to conclude that we are not conscious beings.

 

[metrictensor]

Wow, and all this is theoritcal. On another note, mathematics do not describe everything in exisence because we are then left to conclude that we are not conscious beings.

Do you have a serious comment?

 

[mathwonk]

i have a question for metrictensor: have you discovered (or invented) a theorem of your own, or proved one? i.e. have you published a research paper in mathematics? if so, what was your favorite? and how did you think of it?

 

[metrictensor]

i have a question for metrictensor: have you discovered (or invented) a theorem of your own, or proved one? i.e. have you published a research paper in mathematics? if so, what was your favorite? and how did you think of it?

If you want to address my views please do. Otherwise, I have nothing to say to you.

 

[mathwonk]

i just wondered how you came to your views. I.e. I wondered how you "found" your mathematical results if there is nothing out there to find? On the other hand if you have never discovered any results, it becomes easier to understand how you could feel there is nothing to discover.

You seem to trying hard to be as rude as possible, and as dogmatic. I.e. your views are stated as blunt "take it or leave it" assertions, without any background, evidence or argument, so it is hard to appreciate them fully.

If you do not wish to answer me I don't mind, as so far you have done nothing but insult people.

 

[metrictensor]

i just wondered how you came to your views. I.e. I wondered how you "found" your mathematical results if there is nothing out there to find? On the other hand if you have never discovered any results, it becomes easier to understand how you could feel there is nothing to discover.

There are results and I did find them. For example, if you fit a sphere inside a cube the ratios of the surface areas are equal to the ratio of the volumes. How did I do this? Using the formulas for volume and surface area I followed the rules of algebra and came to the result. Did I discover something? Was there something there that existed but was not known previously? No. Does a poet discover something new when they write a poem combining words in a never before used way?

 

[mathwonk]

thank you. that's a cute little result, by the way. I did not know that.

you seem to have a talent for geometric observation and experimentation.

Now if you could prove it without knowing the formula for area of a sphere in advance, it would give a way to produce that formula. that would be nice.

in that spirit let me ask another one, if i may. Is the same thing true for a sphere inscribed in a regular tetrahedron?

what about a sphere inscribed in any regular polyhedron?

 

[metrictensor]

thank you. that's a cute little result, by the way. I did not know that.

you seem to have a talent for geometric observation and experimentation.

Now if you could prove it without knowing the formula for area of a sphere in advance, it would give a way to produce that formula. that would be nice.

in that spirit let me ask another one, if i may. Is the same thing true for a sphere inscribed in a regular tetrahedron?

what about a sphere inscribed in any regular polyhedron?

I have thought about what you said that if you did it w/o knowing the area and volume formulas but I am not sure you could uniquely determine the four equations. As far as the others I don't know.

What is your background?

 

[mathwonk]

After a brief stint as a meat lugger recorded elsewhere here, I got my phD in algebraic geometry and became slowly a professor of mathematics.

I suspect what you have discovered here is a general phenomenon that holds for all ways of inscribing a sphere in any symmetric solid.

I just meant the obvious fact that if you knew three of the four volume formulas, and your ratio principle, then you could get the fourth formula.

I.e. if someone knew the volume of a sphere, and the area and volume of a cube, he could immediately get the area of a sphere. It thus cuts the work in half. that is a big help and labor saver, and I think it is the way Archimedes worked.

I.e. he figured out which things had a constant ratio to each other, and only afterwards concerned himself with evaluating that constant. for instance look at a circle inscribed in a square.

then your principle holds again, the ratios both beign 4/pi this time, instead of 6/pi for a sphere in a cube.

say i wonder if we went up to 4 dimensions, if the ratios would be 8/pi?

 

[metrictensor]

I tried it for a circle/squre as you did and noticed the pattern as well. The difference is that I am taking the sphere/cube instead so I get Pi/4, Pi/8, etc.

After that I thought about the N-dimentional case and found out that if you take the ratio of the volumes and the ratio of the spheres in N-dimentions the ratios are in fact the same. The tast now is to find a pattern since it looks like the denominator is increasing by some kind of pattern. The general formula for the ratios (with sphere in the numerator) is

\frac{\pi^{N/2}}{2^N(N/2)!}

where the half factorial is evaluated using

(n+\frac{1}{2})! = \sqrt{\pi} \frac{(2n+2)!}{(n+1)!4^{n+1}}

If we can work with this perhaps a general pattern involving \pi can be found.

 

[mathwonk]

wow! cool! you are way ahead of me. i found the formulas for dimension 4,5 with ratios 32/pi^2, 60/pi^2.

this is neat!

 

[metrictensor]

wow! cool! you are way ahead of me. i found the formulas for dimension 4,5 with ratios 32/pi^2, 60/pi^2.

this is neat!

Yeah, I wrote you I realized that every two terms the \pi in the result goes up by a power. This is obvious when we look at the general solution given by the formula. BTW, how did you get your results for n=4 and n=5?

yeah, this is cool. I wonder if the sum of general equation I gave converges as N -> infinity. I demonstrated that the demononator is always bigger than the numerator so at least we know it doesn't blow up. Let's see where all this goes.

 

[mathwonk]

(i looked on the web for some formulas) what if we look at you ration a different way?

i.e. instead of V1/V2 = A1/A2, look at it as V1/A1 = V2/A2. i.e. look at the ratio of the volume/area for various figures. it seems you get something simpler then.

i.e. in dimension n you seem to get V/A = r/n. do you get this?


I did not see offhand the limit of your formulas, but they should be interesting.

 

[Berislav]

i.e. in dimension n you seem to get V/A = r/n. do you get this?

If I may add...
It seems to me that that would simply follow from V/V'=A/A'. That is it would follow if we define a n-dimensional cube to be a n-dimensional space bounded by 2n n-1 dimensional boarders (i.e, two boarders for each dimension, which bind it in every other possible direction, hence dimension n-1).
The volume of the n-cube would be simply
V_n=a^n

and the surface area should be:

A_n=2na^{n-1}

and a=2r.

 

[metrictensor]

(i looked on the web for some formulas) what if we look at you ration a different way?

i.e. instead of V1/V2 = A1/A2, look at it as V1/A1 = V2/A2. i.e. look at the ratio of the volume/area for various figures. it seems you get something simpler then.

i.e. in dimension n you seem to get V/A = r/n. do you get this?


I did not see offhand the limit of your formulas, but they should be interesting.

I did get r/n. The formulas I used for the four volumes/surface areas are (I looked these up on the internet)

<br /> V_s=\frac{R^N \pi^{N/2}}{(N/2)!}

A_s=\frac{NR^{N-1}\pi^{N/2}}{(N/2)!}

V_c=2^NR^N

A_c=2^NNR^{N-1}

The one thing I am not 100% certain of is the condition that L=2R for all N. I think this is true but can not visualize or determine how I would prove this for N>3.

 

[mathwonk]

well lengths should be measurable by slicing down to lower dimensions, i.e. for a cube we can slice and get a square and still measure the length of the square. so the length should be the same as in lower dimensions i.e. L = 2R.

I.e. R is the distance from the center to the side, and this should be measurable on a slice? I also agree with Berislav's computation for a cube. of course the point of metrictensor's result is that the same ratio holds for other figures as well, such as spheres, and possibly other shapes. (I checked it for a triangle, and am pretty confident the shape does not affect the ratio in general.)

I.e. the simpler ration r/n still leaves one obliged to calculate r for a complicated shape. it is only easy for cubes and spheres.

for instance, in a tetrahedron, r is the distance from the center of the tetrahedron to the center of one face, i.e. its the radius of an inscribed sphere as metrictensor originally said.

this same slicing principle seems to apply also to spheres. for example one obtains the area of a unit circle by integral calculus, from the length of a slice.

i.e. for a half circle, one integrates, for x going from 0 to 1, the length of a slice lying a distance x away from the center of the circle. by pythagoras that length is 2sqrt(1 -x^2).

the antiderivative is then (after some cazlculation)

xsqrt(1-x^2) + arcsin(x), which evaluated from 0 to 1 gives pi/2, or the area of 1/2 a semicircle.


one gets the volume of a sphere the same way, integrating the area formual for a circle, i.e. of a slice of a unit sphere, where again the radius of this circular slice is sqrt(1-x^2), so the area of this slice is pi(1-x^2).

this is far easier to integrate, the antiderivative being pi(x - x^3/3), and hence the integral from x = 0 to x=1 equals pi(2/3), the volume of the unit hemisphere.

to do the volume of a 4 dimensional sphere one then considers half of it again, the radius of the spherical splice at x being again sqrt(1-x^2), but now the volume of this slice (itself a three dimensional sphere) is (4pi/3)(1-x^2)^(3/2), i.e. 4pi/3 R^3 where R = sqrt(1-x^2).

to integrate this i needed to find the antiderivative of cos^4, which i was unable to quite do in my head in the car on the way home. let me see if i can get it now with pencil and paper... ok, i got ther ight answer (pi)^2/2, but it was a pain to do the antiderivative of cos^4.

i got the antiderivative of (1-x^2)^(3/2) to be something like:


xsqrt(1-x^2)(1-2x^2)/8 + xsqrt(1-x^2)/2 + 3arcsin(x)/8. yaak! but it gave the right answer.

[i hope i remember not to assign this to a class, or perhaps I SHOULD assign the sort of unscripted thing one actually runs into in "real life"?]

I did this because it feels somehow unfulfilling to look these up on the internet as I did yesterday, and frankly I don't fully trust formulas (even my own) given without proof.

 

[mathwonk]

well anyway, i did notice that the area of a circle is r/2 times the circumference formula, but I do not recall noticing before that this also holds for a square, nor that the area formula for a sphere is 3/r times the volume formula, much less that this also holds for the cube, and conjecturally every other regular solid.

say since there are only a few (5?) regular polyhedra in three dimensions, i wonder how many there are in 4 dimensions? [tetrahedron, cube, dodecahedron, icosahedron, octahedron, in 3 dimensions. by duality there should be an even number, but the tetrahedron is "self dual". octahedron - cube, dodecahedron - icosahedron, tetrahedron - tetrahedron, where faces are interchanged with vertices.]

and what about non regular solids? what is the ratio there?

say this holds aalso for any triangle, not just a regular one. and it reminds me of a famous theorem i never learned in school, due to Heron?

well Heron's theorem is a little different but seems to follow from this formula plus Pythagoras.

i.e. this principle shows that the area of any triangle equals the radius of an inscribed circle times the semi perimeter, i.e. half the perimeter times the radius...well anyway.

the key to the ratio principle seems to be not strict regularity but that there exist an inscribed sphere, i.e. a sphere inside which is tangent to every face, not like arectangle for example, but any triangle works. ?

 

[mathwonk]

here is a "proof" of the basic phenomenon found by metrictensor, and shows it is a very fundamental relation, essentially the most fundamental fact underlying most computations of area and volume, namely the rate of change of these quantities, as codified in calculus.

i.e. an n - dimensional volume formula for a figure with a center, has form V = Cr^n, where C is some constant.

if we assume that the distance from the center to each face is the same, [as is true whenever there exists a truly inscribed sphere], then the method of derivatives, shows that the area formula for this figure is the drivative of the volume formula wrt the radius, i.e. A = nCr^(n-1). hence the ratio V/A = r/n.

since this holds for all such symmetrical shapes, we also have V/A = V'/A' for sphere and cube, and hence also V/V' = A/A', as metrictensor originally observed, and the same holds in all dimensions.

metrictensor, i think this is one of the most basic facts about areas and volumes anyone has ever shown me. i.e. it holds over a very wide variety of cases, it is simple, and it lies at the foundation of the subject. also it has computational uses to render actual calculations easier.

congratulations! that's really nice! in many years of teaching, very few people have ever shown me something simultaneously this simple and interesting, and which held up in so many new cases stretching so far from its beginnings.

I think you have a gift.

best wishes

 

[metrictensor]

My next problem is to see if there is a relationship between an ellipse in a rectangle.

 

[mathwonk]

good idea!

if you find a simple relationship here it will be even more interesting, as there is no simple formula for the arclength of an ellipse.

i.e. in calculus books you will notice that the area formulas for an ellipse are given but not the length formulas, although they may refrain from saying why not.

 

[metrictensor]

good idea!

if you find a simple relationship here it will be even more interesting, as there is no simple formula for the arclength of an ellipse.

i.e. in calculus books you will notice that the area formulas for an ellipse are given but not the length formulas, although they may refrain from saying why not.

Yeah, I have been looking into this and found out that you can only approximate the arc length of an ellipse. There is an infinite series but I assume it does not converge. The strange thing is that if you look at an ellipse it certainly has an arc length so why can't we find a closed solution?