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[PSGarak]

i = sqrt(-1) = sqrt(1/-1) = 1/sqrt(-1) = 1/i

Since we've shown i = 1/i, we conclude that i^2 = 1

Thus 1 = -1. 

Whoops, guess 1 is negative after all.

Argh, where is the contradiction?

I think the contradiction is that the operator sqrt() isn't well-defined; sqrt(–1) can evaluate to either i or –i. So

i = 1/sqrt(–1)

is true, but only if the sqrt(–1) on the right side there is –i, not i.

The problem is definitely with the Square-Root operation and the fact that it can produce more than one result. Since mathematicians don't like the phrase "isn't well-defined," we've cooked up a bunch of technicalities that make it not only well-defined, but so that everything works out to the correct answer.

The usual approach involves something called "branch cuts"; basically, the input to the function is not the normal complex plane, but rather a pair of complex planes "sewn together" along a seam (the result looks kinda like spiral-cut ham). Then, rather than one input having two possible results, we say that there are two possible inputs (one from each plane), each of which has its own result.

Having a one-to-one mapping of input and output ("injectivity") makes the function perfectly reversible, which prevents the paradox above. However, the the details of this are a giant pain the ass to work through because you have to do a lot of bookkeeping of which plane you're on. The details also depend on where exactly you place your branch cut. Also don't forget to define how the function f(z) = 1/z interacts with your branch cut, since that's sort of a big part of the paradox as well (this probably involves more branch cuts).

[ftl]

Okay... but the problem is, (well first of all, who cares, we don't live in some fairy world of different mathematics),

I think you're trying to think about the mathematics in too applied a fashion.

Mathematics is/(can be) about abstract concepts, not about "the real world" which you're contrasting to the "fairy tale world of different mathematics". I think geometry is a fine common example of how that makes sense.

Euclidean geometry is a self-consistent geometry. You can prove lots of stuff! There's a whole geometrical world of interesting things! But there also non-Euclidean geometries. Which have axioms which differ, and thus you can derive entirely different things. There isn't a need to pick one of those sets of axioms and declare it "right" and the others "wrong". We live in a universe where you can formulate both euclidean and non-euclidean geometries and have them each be internally consistent, though obviously not in agreement with each other. They'll describe different abstract concepts. They may or may not be useful for describing the physical world. Both Euclidean and non-euclidean geometries are reasonable fields of study.

(As it turns out, in practice, our universe is fundamentally non-euclidean, but of course euclidean geometry is still useful.)

So if you could make up a different number system which was internally consistent and which violated some obvious "intuitive" properties while keeping others, that would probably be an interesting endeavor! And knowing how mathematics goes, it would stay an esoteric branch of math with everybody looking at it askance and thinking "Why the F does anyone care about this weird thing?" until 100 years later it ends up being the cornerstone of some field of physics or something.

Here's another example - modular arithmetic. Imagine, if you will, a different set of axioms about numbers - ones that say the number line wraps around. 0,1,2,...,N,0,1,2,... I mean, that certainly fits your description of "a fairy world of different mathematics" - if you take 6 apples and add more apples to them, you'll never end up with 0 apples. But yet you can make a number system like that, and prove a lot of things about its properties, how addition and multiplication and division work in this weird system. None of which would be at all useful for counting apples, but so what? Since when is apple-counting so important?

...and then you find that this system is incredibly useful in the modern world. It's how all computers do arithmetic at a fundamental level. Also it's the foundation of all of cryptography. And a bunch of other things, which I don't have the math expertise to understand.

The history of mathematics is filled with ideas which seemed weird and counterintuitive and "The real world doesn't work like that so why do you even care?" which turn out, in retrospect, to be super-important and super-useful.  There's nothing wrong with formulating a self-consistent system of axioms and just seeing what it turns out to be, even if it's not immediately evident how the axioms relate to apple-counting.

[sudgy]

So if you could make up a different number system which was internally consistent and which violated some obvious "intuitive" properties while keeping others, that would probably be an interesting endeavor! And knowing how mathematics goes, it would stay an esoteric branch of math with everybody looking at it askance and thinking "Why the F does anyone care about this weird thing?" until 100 years later it ends up being the cornerstone of some field of physics or something.

Actually, I've realized that also some things the universe does isn't intuitive (you mean this particle is here and there at once?!?), so maybe the fundamental math behind it isn't either.

[ftl]

Well, all of quantum mechanics does basically run on imaginary numbers.

[sudgy]

Well, all of quantum mechanics does basically run on imaginary numbers.

Well, imaginary numbers aren't really intuitive...

[SirPeebles]

Sometimes I see people think that studying the geometry of a five dimensional curved space is ridiculous, since the physical space in which we move around is so clearly Euclidean and 3-dimensional (or absolutely close enough for most anyone's daily life).  As though mathematicians are taking unwarranted precautions in case it is revealed that we're living on a Klein bottle.

Suppose we would like to study the motion of single small object.  With three numbers, we are able to specify it location.  With three more numbers, we are able to specify its momentum (or velocity, if you prefer).  So in sum, to understand all of the possible states of motion for this object, we need to track these six numbers.  These numbers can each take any value (here I'm brushing aside issues of the speed of light or finite extent of the universal.  I'm trying to keep this down to earth).  So you can think of each possible state of this object as being the coordinates of some point in 6-dimensional Euclidean space.

But now, suppose you add in some sort of constraint, like tether your object to some point by means of either a rigid pole or a loose rope.  Each of those deform the region of this 6-dimensional space whose points correspond to admissible states.  I haven't studied these situations since 2004, so I can't think of a particularly compelling example on the spot, but this is where curved higher dimensional spaces often come up:  the points in the space correspond to "good" or "admissible" states of data subject to certain constraints.  Understanding the geometry can help you predict how the data will change, whether the data can be changed from one state to another without passing through the inadmissible region, and how this could be done most efficiently.

[PSGarak]

The most accessible example of non-Euclidean spaces is just the surface of the Earth. It lives in three dimensions, but the surface is effectively (locally) two-dimensional because you only need two coordinates to specify it (say, lat & long). This is a curved space living inside a normal one.

And, as you say, finding optimal paths isn't entirely obvious. Airplanes want to fly in the shortest possible path, because fuel is stupid expensive. We're used to straight lines being the shortest path, but that doesn't work here. A straight line on a map isn't close to being the shortest path on a globe, and a straight line in three-dimensional space tunnels underground, which airplanes are currently unsuited for. But by studying the surface of the globe as a non-Euclidean space, we can find the geodesics that minimize fuel costs.

It gets cooler when you start modeling things like e.g. cooler temperatures further north, so that one mile in one location costs more than a mile in another location. Now space is stretched in addition to curved.

[Schneau]

Re: Alchemy with the other sets numbers

[sudgy]

Re: Alchemy with the other sets numbers

FTFY

[Watno]

I thought the OP meant sets of numbers?

[PSGarak]

I thought the OP meant sets of numbers?

Absolutely. Alchemy with ℕ, Alchemy with ℤ, Alchemy with ℝ, Alchemy with ℂ. Unfortunately, Dominion doesn't seem to enable the construction of transfinite numbers, but it doesn't explicitly disallow them either.

[Witherweaver]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

[Witherweaver]

There is no way to make the Complex Numbers into an Ordered Field (with the standard field operators), which is why there's no canonical total order. However, if you don't care about relating the ordering to the field algebra, then you're free to use whichever order you choose. I'm partial lexicographic orderings, myself. I mean, if you're not going to relate it to the field, might as go all the way.

My first homework problem in my first graduate math course was to prove that 1 is positive.

Wait, couldn't you just say that 1 > 0, so it is positive?

Most people could; someone in a graduate math course can't.

From this, I logically deduce that entering a graduate math course makes you less intelligent



Seriously, this kind of thing is what I hate about some of this formal math - you come up with some obscure, strange 'proof' that 1>0, but what it rests on is not at all more intuitive than the conclusion 1>0 which it is supposedly providing a basis for.

This seems like it's missing the point entirely.  The "1" and "0" are the multiplicative and additive identities of the ordered field.  They are only your usual counting 0 and 1 in the case of the field of real numbers (under the usual operations).  This need not have anything to do with real numbers, but you still want to have your collection of things that you're dealing with be well defined and have structure.  So that's why it has an axiomatic definition.  It's a set of stuff with certain properties.  With those properties you can prove more things,  yes, and you could have taken those things in the definition, but why do so?

I don't do anything with algebra, but maybe someone has an example of an ordered field where the ordering is not intuitive and the fact that a certain element is positive is important yet not intuitive.  But even if there isn't such an example, these things are still important because often you deal with these structures in a non-concrete  way.  You may need to use properties of a field in your argument, or deal with a generic one, so you can't just rely on what you find intuitive about real numbers.

Edit: Another reason not to include a lot of provable properties in definitions is because those properties may not be universal to generalized objects.  There are a bunch of things that are true about commutative groups, so you could make the definition of a commutative group be the usual definition plus all those things.  But then the definition of a group and a commutative group are vastly different.  One has a list of things it satisfies, and one has a much larger list.  It's more palatable to define them in a more minimal way and focus on the key difference being ab=ba holds for one and not the other. 

[SirPeebles]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

That would be another valid proof, although you didn't explain why 0y=0.

If you consider all of that together to be trivial, then yes it is trivial.  I mean, it was problem 1 of assignment 1 of real analysis 1.  It isn't supposed to be a huge strain.

[Witherweaver]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

That would be another valid proof, although you didn't explain why 0y=0.

If you consider all of that together to be trivial, then yes it is trivial.  I mean, it was problem 1 of assignment 1 of real analysis 1.  It isn't supposed to be a huge strain.

Oh I see, given the amount of discussion that followed I had thought at first that it was something major   And yeah I guess 0y=0 isn't part of the definition since it follows from 0=e-e and ey=y.

By the way, did you go for your PhD?  If so, what focus?

[SirPeebles]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

That would be another valid proof, although you didn't explain why 0y=0.

If you consider all of that together to be trivial, then yes it is trivial.  I mean, it was problem 1 of assignment 1 of real analysis 1.  It isn't supposed to be a huge strain.

Oh I see, given the amount of discussion that followed I had thought at first that it was something major   And yeah I guess 0y=0 isn't part of the definition since it follows from 0=e-e and ey=y.

By the way, did you go for your PhD?  If so, what focus?

Most crucial in proving 0y=0 is the distributive law.  You see, 0y=0 is rather special because you are asserting a multiplicative property about 0, which is itself defined by an additive property.  Thus you absolutely must invoke distributivity since that is the only field axiom which imposes any relationship between addition and multiplication.

I wrote my dissertation on homological algebra.

[Witherweaver]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

That would be another valid proof, although you didn't explain why 0y=0.

If you consider all of that together to be trivial, then yes it is trivial.  I mean, it was problem 1 of assignment 1 of real analysis 1.  It isn't supposed to be a huge strain.

Oh I see, given the amount of discussion that followed I had thought at first that it was something major   And yeah I guess 0y=0 isn't part of the definition since it follows from 0=e-e and ey=y.

By the way, did you go for your PhD?  If so, what focus?

Most crucial in proving 0y=0 is the distributive law.  You see, 0y=0 is rather special because you are asserting a multiplicative property about 0, which is itself defined by an additive property.  Thus you absolutely must invoke distributivity since that is the only field axiom which imposes any relationship between addition and multiplication.

I wrote my dissertation on homological algebra.

Right, I was using the distributive law.  That's a good point that the distributive property connects the two operations.

I did my dissertation on nonlinear boundary value problems.  I wonder if there are a lot of math people on this forum.  I have a feeling Dominion would attract more of the combinatorics and algebra types.

[Watno]

There are.

[Drab Emordnilap]

This is all alchemy to me. 

[sudgy]

You are just given that you have an ordered field.

A field means you have an addition and multiplication, and they obey the usual algebraic properties:  associativity, commutativity, distributivity, identities (0 and 1), and inverses (except that the additive identity 0 has no multiplicative inverse, i.e. you can't divide by zero).

An ordered field means that there is a total ordering which is compatible with the algebraic structure in the following sense:  There is a subset of positive elements.  This collection of positives is closed under both addition and multiplication, and given any element, either that element is positive, its additive inverse positive, or it is the additive identity.

tl;dr

You can add, multiply, subtract and divide as usual.  In particular, 1 and 0 do their thing.  Positive + Positive = Positive.  Positive x Positive = Positive.  Given any x, then either x is positive, x is zero, or -x is positive.

Edit:  Technically, you need to also specify that 0 is not equal to 1.

This is why Analysis is way cooler than Algebra.

Edit: By the way, why isn't this trivial?

Let x be a positive element in the field. (Such an element must exist, because a field must have a zero element {0} and a multiplicative identity {e}.  Either e is positive or its additive inverse (-e) is positive.)  Let e be the multiplicative identity element.  Then ex = x.  If e were negative, then ex=x would be negative as well, which contradicts the positivity of x.

Or is it not clear that (negative)*(positive)=(negative)?  This should be clear since x being negative means -x is positive.  Suppose we have a negative x and a positive y.  We want to show -(xy) is positive.  Now -(xy) is the additive inverse of (xy), but so is (-x)y by the distributive law: (xy) + (-x)y = (x+-x)y = 0y = 0.  Thus, -(xy) = (-x)y.  As -x is positive and y is positive, (-x)y = -(xy) is positive, and so xy is negative.

That would be another valid proof, although you didn't explain why 0y=0.

If you consider all of that together to be trivial, then yes it is trivial.  I mean, it was problem 1 of assignment 1 of real analysis 1.  It isn't supposed to be a huge strain.

Oh I see, given the amount of discussion that followed I had thought at first that it was something major   And yeah I guess 0y=0 isn't part of the definition since it follows from 0=e-e and ey=y.

By the way, did you go for your PhD?  If so, what focus?

Most crucial in proving 0y=0 is the distributive law.  You see, 0y=0 is rather special because you are asserting a multiplicative property about 0, which is itself defined by an additive property.  Thus you absolutely must invoke distributivity since that is the only field axiom which imposes any relationship between addition and multiplication.

I wrote my dissertation on homological algebra.

Right, I was using the distributive law.  That's a good point that the distributive property connects the two operations.

I did my dissertation on nonlinear boundary value problems.  I wonder if there are a lot of math people on this forum.  I have a feeling Dominion would attract more of the combinatorics and algebra types.

I'm working my way towards that, I'm doing advanced algebra now and am doing Calculus next year.

[WalrusMcFishSr]

Since you guys apparently know way more about this than I...

Would it be correct to describe the poset of coins and potions as a lattice?

Is it possible to describe the order type of a poset using an extension of the ordinal numbers? Like maybe some sort of sweet ordinal matrix or something?

I apologize in advance if these questions are stupid.

[florrat]

Since you guys apparently know way more about this than I...

Would it be correct to describe the poset of coins and potions as a lattice?

Is it possible to describe the order type of a poset using an extension of the ordinal numbers? Like maybe some sort of sweet ordinal matrix or something?

I apologize in advance if these questions are stupid.

Warning: complicated math ahead (but if you need this warning, why are you still in this thread?)

Yes, the costs of cards is a lattice, because you can take suprema and infima of costs. For example the supremum of 3P and 6 is 6P and the infimum is 3. Taking suprema and infima of card costs is never used in Dominion (because apart from alchemy it would just be taking maxima/minima). Actually the order is the product order of N and N (or N and the 2-element lattice, if you prefer), and the product order of lattices is always a lattice.

Your other question is interesting, and my answer is that I have no idea. It depends on what you exactly mean. Of course you can take the class of all posets, and call two posets equivalent if they are isomorphic. Then you can take the equivalence classes of this relation (it is possible to do this on a proper class, but one should be careful). Then this class contains all order types of all posets, in particular of all ordinal numbers. You can now try to define some operations from the ordinals on this class, but I have no idea how interesting this is (I make this up on the spot, so I have no idea if this idea is silly).

You can define addition of two posets by taking the disjoint union and then define the lexicographic order on that (putting one poset above the other).
You can probably also define an order on this class of posets. I'm not sure what the "correct" definition of this would be, but maybe one of the following:
* A <= B iff there exists an injective order-preserving function from A to B.
* A <= B iff there exists an injective order-preserving function f from A to B satisfying: for all a in im(f) and b in B we have b<=a implies b in im(f).
The first one is simple, but the second one states that the image is a "initial segment", which is often useful for well-orders. Both relations will turn the class of posets into a (proper) poset (not a linearly ordered one, of course).

If you're interested in extensions of the ordinal numbers, make sure to check out the Surreal numbers. This is an extension of the ordinals, containing all real numbers, infinitesimals, and which is in fact an ordered field! The addition and multiplication on the surreal numbers are of course different than those on the ordinal numbers, because they have to be commutative. But the fact that such an extension exists really blowed my mind. Think of things like 1/omega, omega-1 or the square root of omega (where omega is the smallest infinite ordinal). They are all defined in the surreal numbers!

Hopefully somebody came this far through my post. Please ask questions (or use wikipedia, wikipedia is great for definitions in mathematics, imho) if something is not clear.

[SirPeebles]

Some of you may find Tarski's High School Problem interesting.

[Witherweaver]

Some of you may find Tarski's High School Problem interesting.

Sweet, I have a friend that teaches high school algebra/calc.  I'll get him to assign this for the midterm!

[florrat]

Some of you may find Tarski's High School Problem interesting.

Sweet, I have a friend that teaches high school algebra/calc.  I'll get him to assign this for the midterm!

Poor students...