Mathematics desk
< October 28	<< Sep \| October \| Nov >>	October 30 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 29

Introductions to number theory, geometry

I'm looking for rigorous and comprehensive introductions to number theory and geometry at the undergraduate level. The ideal thing would include many problems, be suitable for independent (recreational) study, and have a somewhat conversational tone (I'm not a huge fan of the relentless theorem-proof style). I wouldn't mind having non-Euclidean geometry as a later feature, but I feel I should have a strong hold of Euclidean geometry first (no?). Suggestions I've already had for number theory:

An introduction to the theory of numbers (6th ed.) by Hardy and Wright
An Introduction to the Theory of Numbers (5th ed.) by Niven, Zuckerman, and Montgomery.

And for geometry:

Euclidean and non-Euclidean Geometries (4th ed.) by Greenberg
Introduction to geometry (2nd ed.) by Coxeter.

Some advice on which of these is best, or other suggestions, would be a great help. Thanks in advance. —Anonymous Dissident^Talk 01:06, 29 October 2011 (UTC)[reply]

As regards number theory, Hardy & Wright is indisputably a classic text, but it's style is more "theorem-proof" than "conversational" - you can see for yourself here. For a less formal approach, you could try Recreations in the Theory of Numbers by Albert H. Beiler, or What Is Mathematics? by Courant and Robbins (this has a broader scope but goes into less depth). For self-study, I recommend A pathway into number theory by R.P. Burn - this unusual book consists almost entirely of exercises for the reader (with answers) - you can get a taste of it here. Gandalf61 (talk) 13:47, 29 October 2011 (UTC)[reply]

Thanks for the input. A pathway to number theory looks interesting. After the leading exercises does he prove results in generality? The context is that I'd like to improve my knowledge enough to attempt International Mathematical Olympiad problems; would either one be sufficient for that (if you're familiar with the IMO)? —Anonymous Dissident^Talk 05:37, 30 October 2011 (UTC)[reply]

Burn's exercises build up step by step from specific examples to general proofs. If you work through the whole book then you will have proved several key theorems in number theory, including the fundamental theorem of arithmetic (Chapter 1), the law of quadratic reciprocity (Chapter 4), Lagrange's four-square theorem (Chapter 6) and Liouville's theorem on Diophantine approximations (Chapter 11). IMO problems usually require some degree of lateral thinking in addition to a sound knowledge of the theory, as they often require applying the theory in a less-than-obvious way. The best way to get a feel for the style of IMO problems is to look at past problems and solutions - there are a whole range of "handbooks" and other training materials published, like this one. Gandalf61 (talk) 10:22, 30 October 2011 (UTC)[reply]

Do Hardy & Wright include many exercises? And yeah, IMO problems do seem to require lateral thinking, but beforehand it will be necessary to learn the right theory – right now, looking at past solutions doesn't mean much to me because of this deficit. Would either Burn or Hardy & Wright cover all the number theory you'd need? Thanks again. —Anonymous Dissident^Talk 21:19, 1 November 2011 (UTC)[reply]

rational and irrational numbers

hi. I am a high school math teacher and today I introduced complex numbers and the idea that they can be thought of as vectors (with real and imaginary components). One of my students (one of the sharper ones, in my opinion, though I shouldn't judge ;) asked since we can represent complex numbers as vectors why we cannot do the same for real numbers, with rational and irrational components. I gave her several very good reasons we do not, i.e. that irrational numbers thus constructed do not follow normal component addition (√a+√b generally cannot be expressed √c for rational a,b,c), reals do not have the vector-like quality of complex numbers, any construction would be hard-pressed to cover the transcendental numbers in any more useful way, but I was intrigued because I cannot think of any definitive indisputable reason that we cannot. If there is one, it would have to come from the field axioms or some such, no? Is there such a reason? thank you. — Preceding unsigned comment added by 24.92.85.35 (talk) 04:30, 29 October 2011 (UTC)[reply]

You can think of real numbers as vectors in a space of dimension 1 (i.e. a line) just as you can consider complex numbers as vectors in a space of dimension 2 (a plane). But while it is possible there is the question of whether it's useful or interesting. You lose many of the interesting properties of numbers when you convert them to vectors, for example vectors are usually not compared in terms of 'greater than' or 'less than', but real number are. The language of vectors can be used in spaces of any dimension including 1, but it's more useful when talking about spaces of dimension 2 or more since the language of numbers can be used to talk about a line.--RDBury (talk) 05:27, 29 October 2011 (UTC)[reply]

If you have a field extension L over a base field K, then L has a K vector space structure. This includes R over Q. You can check that R satisfies all the properties here of a Q vector space. Note that the dimension of the vector space (the degree of the field extension) is infinite, and even uncountable, so it's a bit harder to get your hands around then for C over R. Rckrone (talk) 06:47, 29 October 2011 (UTC)[reply]

The split of a real number into rational and irrational components is not unique. 0+√2=1+(√2−1)=1.41421+(√2−1.41421). And the sum of two irrational numbers is not necessarily irrational even if it is nonzero. (1+√2)+(3−√2)=4. But the split of a complex number into real and imaginary components is unique, and the sum of two imaginary numbers is imaginary unless it is zero. Bo Jacoby (talk) 06:52, 29 October 2011 (UTC).[reply]

Yeah, comparing the irrationals in R/Q to the imaginary numbers in C/R is really apples to oranges. The irrationals are all the reals that are not rational, so the analogous elements for C/R are the complex numbers that are not real, which also don't form a subspace. If we pick a subspace of R/Q, for example the rational multiples of π, it will also be closed under addition and scalar multiplication. With the axiom of choice, there is a basis for R/Q which would let you uniquely decompose any real into components, just as you can compose a complex number into real and imaginary parts. Rckrone (talk) 07:20, 29 October 2011 (UTC)[reply]

Bo - isn't that just due to choice of basis ? A complex number has a unique representation relative to the "standard" basis (1, i), but it has different representations relative to alternative bases, such as (1, -i) or (1, 1+i) or (1+i, 1-i). Similarly, a member of Q(√2) has a unique representation relative to the basis (1,√2), but it has different representations relative to the bases (1, -√2), (1, √2-1), (1, √2-1.41421) etc. And the analogy in Q(√2) of the sum of two imaginary numbers is the sum of two rational multiple of √2, which is always irrational. 24.92.85.35 - your student has asked an excellent question - the field of mathematics that develops these ideas (much) further is called algebraic number theory. Gandalf61 (talk) 09:56, 29 October 2011 (UTC)[reply]

Gandalf - you can uniquely split an element a+b√2 of Q(√2) into a rational number a and a rational multiple of √2, b√2 , but you cannot uniquely split an element of R into a rational and an irrational part, which I think was what the OP was requesting. Bo Jacoby (talk) 12:54, 29 October 2011 (UTC).[reply]

Bo - I think the OP is asking how they should respond to their student's question. Your approach seems to lead to the response "No, of course that doesn't work", which completely discourages any further exploration. I suggest they should say "That's a good question. It doesn't work in exactly the same way, but some subsets of the real numbers can be thought of as vector spaces over the rational numbers". If the student is interested, this could open their eyes to the whole area of fields, field extensions and the rest of algebraic number theory. Gandalf61 (talk) 13:20, 29 October 2011 (UTC)[reply]

Gandalf - the student's question was "since we can represent complex numbers as vectors why we cannot do the same for real numbers, with rational and irrational components"? That is a good question! And the answer is why not. It is true that the infinite dimensional vector space (R,+,Q) has finite dimensional subspaces such as Q(√2).Bo Jacoby (talk) 21:45, 29 October 2011 (UTC).[reply]

As I think Rckrone was saying above it is possible to choose an infinite basis for the real numbers using rationals for the coefficients, however it requires using the axiom of choice to choose the basis elements. This is called a Hamel basis, unfortunately however we don't seem to have an article about it - that just redirects to Basis (linear algebra). Dmcq (talk) 22:57, 29 October 2011 (UTC)[reply]

The fact that Hamel bases for (R,+,Q) do exist does not imply that anybody can actually choose one. Bo Jacoby (talk) 10:38, 30 October 2011 (UTC).[reply]

Well yes and there is the little problem that we don't even know if

π

and e for instance are linearly independent under the rationals. Dmcq (talk) 14:05, 30 October 2011 (UTC)[reply]

Deal or no Deal

Probabilty question, what is the chance of a contestant picking the jackpot box? what is the chance of him not picking the jackpot box after 20 tries? — Preceding unsigned comment added by 203.112.82.1 (talk) 22:32, 29 October 2011 (UTC)[reply]

Assuming the jackpot is in a random case, the probability of picking the jackpot case on any particular guess is 1/n where n is the number of cases left (assuming the jackpot hasn't already been picked). If there are 26 cases in total, after 20 guesses there are 6 cases left. The chance the jackpot was in one of those 6 is 6/26. Rckrone (talk) 23:18, 29 October 2011 (UTC)[reply]