Wikipedia:Reference desk/Archives/Mathematics/2012 August 1

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August 1

Axioms for Euclidean geometry

I'd like to explore what axioms of perpendicularity need to be added to the affine structure of R² in order to obtain Euclidean geometry.

More precisely, I'll assume we have a real affine plane P, that is, a set on which a two-dimensional real vector space V acts simply and transitively. Now I would like to add axioms concerning the relation of perpendicularity between lines in order to reach Euclidean geometry (with a scalar product that is well-defined up to a constant multiple).

I've discovered that the following axioms work to get us to the final stage in Euclidean geometry (unless I'm mistaken), but I'm not entirely satisfied with them for reasons I'll explain.

1. The set of lines of V is partitioned into classes of cardinality 2. In each class we say that one line is perpendicular to the other. The relation of perpendicularity is then extended in the obvious way to lines of P. In particular, this allows us to define the "perpendicular bisector" of any segment.

2. The perpendicular bisectors of the three sides of a nondegenerate triangle meet at a single point.

The reason I'm not satisfied with 2 is that the proof I know relies on the notion of "congruence" of segments. That is, there is no way I know to make Axiom 2 intuitively clear without an elaborate argument appealing to the concept of distance.

I'd like to find a set of axioms for perpendicularity that are visually obvious, that don't rely for their intuitive justification on the additional concept of congruence (beyond what is present already in the affine structure), and that give us all of Euclidean geometry. Can anyone help? 96.46.200.71 (talk) 03:38, 1 August 2012 (UTC)

I think perhaps your terminology is confused. You don't have to add any axioms to the affine structure of R² to get Euclidean geometry -- you can define perpendicularity in terms of coordinates. Also your axiom 1 seems to be equivalent to a statement that for each line there is precisely one other line that is perpendicular to it, which is obviously false. Looie496 (talk) 04:51, 1 August 2012 (UTC)

On your first point, you'd have to pick a system of coordinates first. In a vector space V, there is not necessarily a privileged basis. (I didn't mean that V actually had to be R², I just meant that it was isomorphic to R², but without a particular isomorphism singled out.) So your axioms would look like this: pick two vectors which we decide are perpendicular and have length 1. Then define perpendicularity numerically in that basis. That's logically consistent but quite ugly. There is no reason to have two privileged vectors that way. Also, Axiom 1 pertains to the associated vector plane, not the affine plane. In a vector space, "lines" are subspaces and so always go through the origin. Axiom 1 is only about lines through the origin of the vector plane, so it's true that every line has only one line perpendicular to it. 67.224.85.67 (talk) 23:46, 1 August 2012 (UTC)

Factorization

Find the factors of (factorize):
x² + 4y² + 4y - 4xy - 2x - 8 106.199.10.161 (talk) 09:44, 1 August 2012 (UTC)

Try doing little bits, the bits with two variables then the bits with one variable and you should see it get simpler. After that personally I'd add and subtract a 1 rather than invoking the formula for solving quadratics, can you see why? Dmcq (talk) 10:00, 1 August 2012 (UTC)

This is as far as I get:

x² + 4y² + 4y - 4xy - 2x - 8

x² + 4y(y+1-x) - 2x - 8

x² - 2x - 8 + 4y(y+1-x)

(x-4)(x+2) + 4y(y+1-x)

StuRat (talk) 10:23, 1 August 2012 (UTC)

(x-2y+2)(x-2y-4). Bo Jacoby (talk) 10:39, 1 August 2012 (UTC).

You can keep y as a parameter and solve the quadratic equation for x. The solutions are 2y+4 and 2y-2, which gives the factorization Bo Jacoby posted a minute before I finished calculating. —Kusma (t·c) 10:43, 1 August 2012 (UTC)

[1] :-). Bo Jacoby (talk) 10:48, 1 August 2012 (UTC).

The bits with two variables were x² + 4y² - 4xy, the bits with one variable were 4y - 2x. Kusma's method is the more general though. Dmcq (talk) 09:32, 2 August 2012 (UTC)

simple toy axiomatic system that something is true in?

So, I am still back at square 0, trying to understand "entailment". Could you give me a simple axiomatic system that something is true in, but it is so simple (e.g. a toy system) that it can't prove anything at all? For example, maybe there are enough axioms to start counting 1, 2, 3, 4, 5, but the system is so simple that it can't prove that by continuing it won't ever get back to 1 again. (Which is nevertheless "true" in the system). Or any other example you want. Thank you. 84.3.160.86 (talk) 12:41, 1 August 2012 (UTC)

In other words, my example is such that from "inside" it would seem independent. Well, if our only guarantee is that if we take the next number we always get a number, we never run out, that doesn't proven the whole thing isn't cyclical. But in fact that IS the truth. (This is just my foolish laical attempt to come up with something where something is trivially true to an outsider. 84.3.160.86 (talk) 12:56, 1 August 2012 (UTC)

What you ask for isn't possible, by the theorem. If we weaken the logical part rather than the mathematical part, we can do this: I'm not sure if this is so trivial it will just confuse matters, but here goes. Consider the axiomatic system with no axioms (and no deduction rules) [including not even standard axioms saying what equality is]. Then $(\forall x)(x=x)$ is true in all models of the theory (because that's reflexivity of equality) but cannot be proved because we don't have anything to prove it with. Note that we can't take 0=0 as such a statement because 0 doesn't currently make sense.

Your example is listed as an axiom (well, two - 7 and 8) in Peano axioms. If we removed axiom seven (for simplicity we'll also ignore addition and multiplication) then taking the set {0, 1} with 1 as the successor of 0 and 0 as the successor of 1 would still satisfy the remaining axioms. It wouldn't be ordinary arithmetic but it would not be entailed by the remaining axioms that there are infinitely many numbers (i.e. this wouldn't be "true"). Does that clarify anything? Straightontillmorning (talk) 16:03, 1 August 2012 (UTC)

Yes, it does. But why do you start by saying what I ask isn't possible, and then proceed to give an example? Whether there is ANY x for which x = x is independent under an axiomatic system with no axioms; nevertheless, if you pick the value, "False" (so that your only axiom is "THere is no x such that x = x") then I suppose you could proceed to prove false things, but you will be consistent. Do I have this right? 84.3.160.86 (talk) 16:15, 1 August 2012 (UTC)

Well, you were asking for something in what I am thinking of as the "content" part, and we have to weaken the foundations. It's unusual that you have to give axioms for equality if the question is "give an axiomatisation of the theory of groups" (for instance.)

On a point of pedantry, the negation of "for every x, x=x" is "there is an x such that x does not equal x". If we take this as a lone axiom, this is consistent (because you can't prove the statement "false" - in the absence of deduction rules like modus ponens you can't prove anything other than the axiom). But it does not have a model (which is the semantic analogue of being consistent, in the same way that entailment is analogous to proof) because in any model, there cannot be anything that does not equal itself, so in some sense is "false". I suppose the way to consider this is that the original statement is a tautology (true in all models) so its negation is false in all models. Straightontillmorning (talk) 17:06, 1 August 2012 (UTC)

"Enough axioms to start counting" doesn't make sense. You can't count within an axiomatic system; you can only prove things. "True in the system" also doesn't make sense. Sentences can only be true under some interpretation as statements about the real world. The real world includes the Platonic realm, which includes the system itself, but there isn't much you can say about a formal axiomatic system except what you can and can't prove in it. -- BenRG (talk) 22:21, 1 August 2012 (UTC)

Mathematical terms in English

Sorry, my English is not perfect.

F: X -> Y. The name of the functor what give me the X, when I give F to it, is "domain". But what is the sign of this? "domain F", or for example "dom F", or other? The names of the pair of this, what give me F(X) what is a subset of Y is: "codomain", "image", "target", "range". But which is the best word? And what is the sign of this? For example, a formula: "(domain F) union (image F)", or "(dom F) union (img F)" or other?

What is the better expression: "sequences with(out) repetition", or "permutation with(out) repetition".

In combinatorics, what is the name of this: A, B are sets, |A| >= |B|, we search the size of the set of A->B functions: the whole: |B|^|A|, injectives only: |B|!/(|B|-|A|)!. In word-for-word translation from Hungarian: "variation with(out) repetition". But I did not find this in the English Wikipedia in article "Elementary Combinatorics".

For a function, which is the better expression: a function does: "assign something to something" or "map something on(to) something".

For example in Phisics, if a number is greater than 1, the unit is in plural: "2 meters". If the number is between 1 and 2, the unit is in plural or in singular? "1.5 meter" or "1,5 meters".

If I use noninteger ordinal numbers, which is the correct form: "1.1st" or "1.1th". And if I say not 1st but "1.0st": "1.0st" or "1.0th" (because of 0 uses -th: zeroth).

Thank you. — Preceding unsigned comment added by 94.21.140.73 (talk) 14:45, 1 August 2012 (UTC)

I can answer some of these. Plural is used for all values except for exactly 1, so "1.5 meters", "0.3 meters", "0 meters", "-1 meters". If you're using fractions you can write "3/5 meters" but another common construction is "3/5 of a meter" (said as "three fifths of a meter").

For noninteger ordinal numbers, I think I would use the ending for whatever the last digit is so "1.1st", "1.2nd", "1.0th", etc since that's the only reasonable way you could say these things in speech. However I don't think I've ever encountered this, so there may be another convention.

For a function, "assign", "map" and also "send" are all acceptable verbs. For all of them I would use "maps/assigns/sends something to something". "On" isn't used, and "onto" is usually reserved for surjective maps (in fact "onto" is sometimes used as a synonym of "surjective"). Rckrone (talk) 17:51, 1 August 2012 (UTC)

See Function (mathematics)#Function_spaces. See English_numerals#Ordinal_numbers: Noninteger ordinal numbers are not (commonly) used. See SI: Symbols of units are not pluralised; e.g., "25 kg", not "25 kgs". Bo Jacoby (talk) 21:05, 1 August 2012 (UTC).