1. Order of Kilopi
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Well then you're almost there. Proof by contradiction, those two lemma's, and a good look at the last given axiom should do the trick

2. Order of Kilopi
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Originally Posted by caveman1917
The term "totally ordered field" is the strictly correct one, since you can also have a partially ordered field. However it is not uncommon to say "ordered field" instead of "totally ordered field" and assume that if a partially ordered field is meant, it would have been specified.
There seems to be a difference between english and dutch usage here, it seems that in english the term "ordered field" is standard usage (with non-standard use of "totally ordered field"), whereas in dutch it is the equivalent of "totally ordered field" (with non-standard use of "ordered field"). My statement quoted above is how it's done in dutch, but apparently not english. That may have contributed to the initial ambiguity.

3. Originally Posted by caveman1917
There seems to be a difference between english and dutch usage here, it seems that in english the term "ordered field" is standard usage (with non-standard use of "totally ordered field"), whereas in dutch it is the equivalent of "totally ordered field" (with non-standard use of "ordered field"). My statement quoted above is how it's done in dutch, but apparently not english. That may have contributed to the initial ambiguity.
I think the issue that grapes refers to was something else. We can have a field, and we can have a total ordering on that field, such that the two additional axioms are not satisfied.

So we could have a field (like the real numbers), and a total ordering (different than the usual one) such that 0<2<1. But then the two additional axioms would not be satisfied, and it would not be a (totally) ordered field. Your problem requires that it be a totally ordered field (i.e., the additional axioms be satisfied). Without those axioms, it is impossible to prove that 1>0, because it might not be true.

4. Do we have to prove that ?

5. Let's assume it

6. Oh, I guess not, it's in one of the axioms.

Originally Posted by caveman1917
This notation gives me headaches Making the appropriate substitutions and adding some decoration (parentheses), it is

Then using the second lemma,

Since is prohibited by axiom,

Then by the first lemma,

But , and if and , then . But this is precluded by . So this is a contradiction, and we know is not true. Then , and by the first lemma, .

Any weak points?
Last edited by Sarawak; 2012-Sep-25 at 07:54 AM. Reason: Never mind

7. Order of Kilopi
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Originally Posted by Coelacanth
It's meant to signify the axioms in a general sense, not just applied to a specific field such as the real numbers.

Any weak points?
It's correct, though this bit is not necessary:

But , and if and , then . But this is precluded by .
because

You can do it somewhat faster by starting with rather than

8. Originally Posted by caveman1917
It's meant to signify the axioms in a general sense, not just applied to a specific field such as the real numbers.
I know, I just find it easier to think in terms of the specific example Of course, thinking in terms of the concrete case, it becomes relatively easy to make statements which are true but not proven, because we are so familiar with them

I thought a bit about whether the non-equality of the two identity elements needs to be axiomatic. We could have the rather trivial field with a single element in it, which violates this one. I haven't come up with a more interesting example so far (although I haven't tried very hard).

So my question:

A marathon is sometimes defined as a race of 26 miles and 385 yards, or 42.195 kilometres. However, these two distances are not the same, being 12 mm apart. So the question is, what is shortest possible race whose distance can be expressed in an integer number of yards and also an integer number of metres?

For those not familiar, 1 mile is 1760 yards, 1 yard is 36 inches, and 1 inch is 2.54 centimetres.

9. Since an inch is exactly 2.54 cm, one yard is exactly 36*.0254, or .9144 meters, which as a reduced fraction is 1143/1250. So, 1250 yards is exactly equal to 1143 meters.

BTW, one mile is exactly 1.609344 km. :)

10. Originally Posted by grapes
Since an inch is exactly 2.54 cm, one yard is exactly 36*.0254, or .9144 meters, which as a reduced fraction is 1143/1250. So, 1250 yards is exactly equal to 1143 meters.

BTW, one mile is exactly 1.609344 km. :)
I agree with everything I see, but I was hoping someone would answer "zero"

11. Ok, also an easy one, to get us warmed up. Four points on a circle, A, B, C, D, are such that AC is perpendicular to BD and they intersect at the center of the circle. Two chords from A are divided into segments by BD, in ratios 1:2 and 1:3. What is the ratio of the longest chord to the shortest chord?

And, I was going to say zero, but then thought that wasn't much of a race. :)

12. Looks good, you're next

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Originally Posted by Coelacanth
I thought a bit about whether the non-equality of the two identity elements needs to be axiomatic. We could have the rather trivial field with a single element in it, which violates this one.
There is something known as the trivial ring, note that the ring axioms don't state the distinctiveness between zero and one. A field is in essence a commutative ring where every non-zero element has a multiplicative inverse. As such your "trivial field" is equivalent to the trivial ring, which is obviously commutative and every non-zero element has a multiplicative inverse. So that is one reason to exclude it from the field axioms, since it doesn't bring anything new. There are other reasons to exclude this from the field axioms though, for example a vector space over the trivial field would have an undefinable dimension. Also the notion that the subset of non-zero elements together with multiplication form an abelian group would be broken since the empty set is not a group.

I haven't come up with a more interesting example so far (although I haven't tried very hard).
You won't be able to. If you want you can try to tackle the proof for that, that there exists only a unique (up to isomorphism) "pseudofield" where 0=1, which is the "trivial field". It's not much harder than the proof you gave for 1 > 0.

14. I hope I have set this one up correctly

Three-dimensional Euclidean space, use (x,y,z) coordinates. Consider the solid object specified by

.

What is the ratio of its volume to its surface area?

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Originally Posted by Coelacanth
I hope I have set this one up correctly

Three-dimensional Euclidean space, use (x,y,z) coordinates. Consider the solid object specified by

.

What is the ratio of its volume to its surface area?
It is zero.

There's an obvious but hard way to do it (with triple integrals) and an easier one where we don't have to calculate integrals or stuff like that if we reason through this.

We can see the cylindrical symmetry (slices of constant z are disks), the object is a sort of double cone.

Using the well-known formulae for cylinders we get the volume of a slice of thickness as

and the surface area as

Since the solid is also symmetric about the z=0 plane (invariant under ), we look at

For getting the volume and surface area up to a height we would integrate to get functions of the form

which we know exist by the fundamental theorem of calculus and noting that and are continuous.

But we are interested in the ratio of total volume/surface area, so we would then have to take

Since both the numerator and the denominator goes to infinity in that limit we can use l'Hôpital and get

which works out to

Last edited by caveman1917; 2012-Sep-27 at 10:28 PM. Reason: completely changed from first version

16. Originally Posted by caveman1917
It is zero.
That was indeed my design objective, a container which holds a finite amount of paint, but whose inner infinitely large surface would be covered by that paint However, regarding the surface area

shouldn't we have another factor there, because the sides of this object are not vertical (in the z direction), making the surface area even larger?

17. Originally Posted by Coelacanth
That was indeed my design objective, a container which holds a finite amount of paint, but whose inner infinitely large surface would be covered by that paint
Well, sorta! :)

Of course the outer surface would take an infinite amount of paint--the paradox is resolved because the container soon necks down to where a single paint molecule won't fit inside, and there is still most of the volume (infinite) below that point.

18. Originally Posted by grapes
Well, sorta! :)

Of course the outer surface would take an infinite amount of paint--the paradox is resolved because the container soon necks down to where a single paint molecule won't fit inside, and there is still most of the volume (infinite) below that point.
I'm not using real, discrete paint, but abstract, continuous paint

19. OK, but then you could cover the outer surface with the same paint! :)

20. Originally Posted by grapes
OK, but then you could cover the outer surface with the same paint! :)
It would have to escape from my theoretically perfect and non-porous container

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Of course a finite amount continuous abstract paint can paint an infinite surface. Start anywhere and use half the paint to paint the first meter in all directions. Then use half of what remains to paint the next meter out. Then use half of what remains to paint the next meter out from that. Continue forever. You'll never run out of paint that way so every point on the surface will be under a positive thickness of paint. Hire an infinite number of painters and assign each square meter of surface to one painter so you'll finish in finite time.

What will be the average thickness of the paint?

22. Originally Posted by Chuck
Of course a finite amount continuous abstract paint can paint an infinite surface. Start anywhere and use half the paint to paint the first meter in all directions. Then use half of what remains to paint the next meter out. Then use half of what remains to paint the next meter out from that. Continue forever. You'll never run out of paint that way so every point on the surface will be under a positive thickness of paint. Hire an infinite number of painters and assign each square meter of surface to one painter so you'll finish in finite time.

What will be the average thickness of the paint?
Averaged across area, or volume of paint?

23. Originally Posted by Chuck
Hire an infinite number of painters and assign each square meter of surface to one painter so you'll finish in finite time.
BTW, a bunch of the people over here seem to be struggling with that one a bit . . .

(Warning - your head might explode if you spend too much time there. And not because it will be swollen with huge volumes of brilliance acquired there.)

Edited to add - it appears that discussion of that particular topic is prohibited at this board. Politics, religion, and 0.999....

24. Back to my problem, I am getting a volume of .

25. Order of Kilopi
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Originally Posted by Coelacanth
That was indeed my design objective, a container which holds a finite amount of paint, but whose inner infinitely large surface would be covered by that paint However, regarding the surface area

shouldn't we have another factor there, because the sides of this object are not vertical (in the z direction), making the surface area even larger?
True, but since it goes to infinity it doesn't matter that we underestimated the surface area. If it were a finite surface area it would matter, but not now. In any case, in the limit of z to infinity the cone goes towards being vertical and it's only that limit we are interested in. You only asked about the ratio so we can use a couple of shortcuts

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Originally Posted by Coelacanth
Back to my problem, I am getting a volume of .
That's correct. I wonder how the store employee would react if we asked a container of paint of liters

27. Originally Posted by caveman1917
True, but since it goes to infinity it doesn't matter that we underestimated the surface area. If it were a finite surface area it would matter, but not now. In any case, in the limit of z to infinity the cone goes towards being vertical and it's only that limit we are interested in. You only asked about the ratio so we can use a couple of shortcuts
Yes, but we have a symbol "<=" to use instead of "=" in that case

28. Honestly, I'm so math challenged I probably can't even play this game.

29. Originally Posted by Buttercup
Honestly, I'm so math challenged I probably can't even play this game.
For each of us, there was a time when we knew none of the math in this thread!

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