Wikipedia:Reference desk/Archives/Mathematics/2012 September 7

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September 7

Hyperbolic Trig?

While it may not be quite the same, I noticed that there are articles for both Trigonometry in a Planar concept as well as spherical geometry. Is there an inverted concept to spherical geometry in Hyberbolic spaces (and this is *not* as far as I can tell related to the Hyperbolic trig functions)Naraht (talk) 12:47, 7 September 2012 (UTC)

If you take a valid formula in spherical geometry and multiply all the lengths by i, you get a valid formula in hyperbolic geometry. A trig function of an imaginary argument becomes a hyperbolic function of a real argument (multiplied by some power of i).

The easiest way for me to understand hyperbolic geometry is with vectors in Minkowski space: (it,x,y,z). Again, the formulae are analogous to those of geometry on the unit hypersphere $w^{2}+x^{2}+y^{2}+z^{2}=1$, but the hyperbolic plane is represented by a unit hyperboloid $-t^{2}+x^{2}+y^{2}+z^{2}=-1$. —Tamfang (talk) 03:37, 8 September 2012 (UTC)

Rocket fin alignment

I have a rocket that has three sets of three fins, but they are not aligned properly. So I traced the circumference of the rocket onto a piece of paper. And I think that if I trisect the circle into 3 60 degree sections, I can use that as my guide to align the fins properly. But the problem is that I cannot find the center of this circle. So how do I trisect that circle into 3 60 degree sections with a pencil, compass, protractor, and ruler at my disposal? Legolover26 (talk) 14:28, 7 September 2012 (UTC)

It is easy to construct a diameter: pick two points on the circle. Open the compass to some big enough distance. Draw two arcs centred on the chosen points. The line through the points where the arcs cross is a diameter. Do this two or three times with different points to get more diameters which of course meet at the centre.--JohnBlackburne^words_deeds 15:23, 7 September 2012 (UTC)

Thank you very much! Legolover26 (talk) 16:03, 7 September 2012 (UTC)

Also note that 3 evenly spaced fins on a 360° circle will be 120° apart, not 60°. StuRat (talk) 16:09, 7 September 2012 (UTC)

Yea, I noticed that. I don't know why i said that. Maybe because the three angles of a triangle add up to 180 degrees. — Preceding unsigned comment added by Legolover26 (talk • contribs) 16:52, 7 September 2012 (UTC)

Here is another method you can use if you can't easily trace the rocket (let's say it has an engine sticking out the bottom):

1) Mark a spot on the edge of the rocket. This will be your first fin position.

2) Make a measuring tape by cutting a long strip of paper (if you already have a measuring tape, all the better, with the softer tailor's type being preferred over the stiff metal ones).

3) Wrap the tape around the rocket, starting at the mark, and extending until you return to the mark. Record this position on the tape, as well (or just note the measurement, if the measuring tape is numbered).

4) Then divide that by three, using the ruler to measure, if necessary (a ruler divided into tenths of an inch or mm is better than one divided into eights of an inch, as that makes for some ugly math).

5) Return the measuring tape to the rocket and mark those 2 additional fin locations.

One hint for using any ruler or measuring tape is to start at 1 inch (or cm), not at 0, since the end is often less accurate. Just remember to subtract 1 from your measurement). Some professional rulers don't start right at the end, to avoid this problem. StuRat (talk) 19:36, 7 September 2012 (UTC)

Axiom of empty set

The Zermelo-Fraenkel axioms do not include the axiom of empty set; instead the article contains the line:

The axiom of specification can be used to prove the existence of the empty set ... once the existence of at least one set is established (see above).

The axiom of infinity in the same article uses the fact that the empty set has already been defined. Is it because we implicitly assume that the domain of discourse contains at least 1 set and so by the axiom of specification the empty set?-Shahab (talk) 15:09, 7 September 2012 (UTC)

The axiom of infinity can be rephrased so as not to actually assume that any of the sets named actually exist. For example, to assert that the empty set is in X, simply say that for every y, if y has no elements, then y is an element of X. Then the axiom of infinity a priori only asserts the existence of a set meeting this description, without actually asserting that it contains anything. Then the existence of a set implies the existence of the empty set by the axiom of specification. --COVIZAPIBETEFOKY (talk) 16:25, 7 September 2012 (UTC)

Name of set

The Abel-Ruffini theorem article says "The theorem says that not all solutions of higher-degree equations can be obtained by starting with the equation's coefficients and rational constants, and repeatedly forming sums, differences, products, quotients, and radicals (n-th roots, for some integer n) of previously obtained numbers."

What do you call the set of numbers that are statable by a finite number of sums, differences, products, quotients and radicals of integers? Is there a common name for this? RJFJR (talk) 16:37, 7 September 2012 (UTC)

"Numbers defined by radicals"? — Quondum 18:20, 7 September 2012 (UTC)

Sounds like irrational numbers, to me, unless we include roots of negative numbers, in which case we have complex numbers. StuRat (talk) 18:29, 7 September 2012 (UTC)

They will be irrational, but not all irrational numbers fit that description. The square root of two, for instance, is irrational but can be described in terms of radicals (I just described it that way). --Tango (talk) 22:30, 8 September 2012 (UTC)

Shouldn't your example be an irrational number which can't be described in terms of "repeatedly forming sums, differences, products, quotients, and radicals" to prove your point ? I would think any irrational number could be defined in that way, if we allow for an infinite series. However, when adding the restriction that it be a finite series, then you may well be right. StuRat (talk) 00:40, 9 September 2012 (UTC)

There are more irrationals (aleph1) than there are formulas constructable in a finite number of symbols. I was wondering if they might be related to the computable numbers. RJFJR (talk) 18:44, 7 September 2012 (UTC)

Just have to jump in here — there are $2^{\aleph _{0}}$ irrationals. This may or may not equal $\aleph _{1}$ — the statement that these are equal is the continuum hypothesis, and its truth value is not known.

(But it is certainly true that there are at least $\aleph _{1}$ irrationals, and that's a possible reading of your claim, so apologies if that's what you meant.) --Trovatore (talk) 20:32, 7 September 2012 (UTC)

Nope. Pi is a computable number, but not a number defined by radicals (the latter set is presumably a proper subset of the former). — Quondum 19:05, 7 September 2012 (UTC)

You're right. And I now see where you liked Numbers defined by radicals to and that is the answer. Thank you. RJFJR (talk) 20:24, 7 September 2012 (UTC)

"The maximal solvable extension of Q", or "Q^solv" are names. You might have been thinking of constructible numbers which form a subfield, not computable numbers.John Z (talk) 20:55, 7 September 2012 (UTC)

I suspect that "Numbers defined by radicals" is a strictly informal term (perhaps "Numbers contructible from radicals" would be better). John's suggestion would be unambiguous (though I'd have to rely on him that it is in fact the same set; for example, there are solutions to the cubic and quartic equations that I'd assume are in Q^solv but are not constructible from only real radicals of real rumbers and integers). — Quondum 06:33, 8 September 2012 (UTC)

I thought it was called the surd field? Huh, that's red. I know I've seen it somewhere. --Trovatore (talk) 08:40, 8 September 2012 (UTC)

Surd field sounds like a good, snappy, short name, but a quick google seems to show that it used for what we are calling constructible numbers, and should probably redirect there. Quondum, I think you are referring to the fact that the solutions to say a rational cubic with 3 real roots may involve radicals of complex numbers, but by the cubic & quartic formulas they sit in Q^solv. Abel-Ruffini just says that Q^solv is strictly smaller than Q, the algebraic closure of the rationals, the Algebraic numbers, and in particular the roots of most equations with degree bigger than 4 are not in Q^solv. Galois theory identifies extensions with solvable Galois group with ones definable by (iterated) radicals.John Z (talk) 21:00, 8 September 2012 (UTC)

John, yes, that is what I meant. I'm a bit out of my depth here, but it sounds like you're saying that a cubic with 3 real roots is expressible using radicals (by which we mean here real radicals of real numbers, starting with integers), which I was under the impression is impossible in general. I suspect that solvability is critically dependent on the field (R or C). For example, x² + 1 is an irreducible element over R, but is factorizable over C, and related subtleties may affect the interpretation of Galois theory. Or something like that – as I said, this is a bit beyond me. — Quondum 23:20, 8 September 2012 (UTC)

No, you are right, you can't always get all the roots of a cubic with three real roots by only using real radicals of real numbers. You may have to take roots of complex numbers, but that is OK since Q[i] for instance is an extension with a solvable (abelian, even cyclic) galois group; you get it by taking radicals. When we have enough roots of unity in the base, cyclic extensions - ones with cyclic galois group - correspond to taking an nth root of something in the base. Solvable extensions, ones with solvable galois groups correspond to taking iterated radicals. The maximal solvable extension is built up by taking all the iterated radicals, as RJFJR wants. Take all the roots of stuff in Q. Adjoin them to Q & you get some field. Then take all the roots of stuff in there, and go on ad infinitum.

So to get the solution of such a cubic, you can think of taking an extension adjoining i, or the square root of some negative number (given in the cubic formula) & then take some cube root of something in that extension to get the solution of some cubic using the cubic formula. The imaginary parts can cancel out and you land back in R, but to write the solution using radicals, taking them in fields not embeddable in R is necessary.

Another interesting & even more important, probably, subfield of Q^solv is Q^ab, the maximal abelian extension of Q, the biggest extension of Q with abelian galois group. It is a real, non-obvious theorem, the Kronecker–Weber theorem, that this is the same field as that gotten by adjoining all roots of 1, the same field as taking all Cyclotomic fields together, what our Class field theory calls $\mathbb {Q} (\mu _{\infty })$.John Z (talk) 00:05, 9 September 2012 (UTC)

Oops, I see where you are coming from, Quondum. I misread the question, as if RJFJR was asking for the name of the field gotten from taking radicals, then "radicals of previously obtained numbers" etc., as are used in the cubic & quartic formula. But he says "radicals of integers" only, which eliminates many fields generated by solutions of cubics & quartics using radicals. Hmm, not sure of a name for that, though have an idea I need to think about more, with less beer. Googling finds a sloppy statement in a reference work & a book with an incorrect definition that mistakenly says what you (reasonably) thought I was saying, that "solvable by radicals" is the same as "solvable by radicals of rationals/integers", ignoring the dependence on the base field you rightly observed, and thus states an incorrect theorem that implies there is a cubic formula using only radicals of rational numbers! To err is human, but the works of Cardano, Tartaglia, Abel, Galois, Artin etc are divine.John Z (talk) 08:01, 9 September 2012 (UTC)

I didn't notice the "integer" part either. For a while there was an article called radical integer that might have provided an answer to this, but it was AfD'd (disclosure: I filed the AfD) on the grounds that the term was not standard and the claimed result was not reliably sourced. Anyway the claimed theorem, if I have it right, is that if a number is obtainable by radicals starting with the rationals, and is also an algebraic integer, then it's obtainable by radicals starting with the integers. Rich Schroeppel coined the term radical integer for (I think) the latter concept.

It would be very nice to know if this result has been published and confirmed. I would be happy to have the content back, once it's no longer sourced to Mathworld or Eric Weisstein. --Trovatore (talk) 08:22, 9 September 2012 (UTC)

Interesting AfD. From it, "Radical number" or "Radical integer" seems to refer to the iterated radicals, solvable field extension case. Although "radical number" in Maple seems to be more at what RJFJR is getting at: Radicals of integers along with i (but what about other roots of unity?). The ref for the article is both confused & confusing, and I think the correct decision was taken. The claimed theorem sounds plausible, but it is hard to be sure we're even getting the statement right.

The only thing I think I can observe about the original question is that this field of radical numbers sits in between two fields with accepted names. If you call it Q^rad, then Q < Q^ab < Q^rad < (Q^ab)^ab < Q^solv. The galois groups of a radical extension of Q (radicals of integers) with enough roots of 1 added to make it galois are metabelian - solvable, with derived length two, group extensions of abelian by abelian. By Kronecker-Weber, the structure of finite abelian groups and galois theory, a partial converse is that an extension of Q with metabelian galois group comes from cyclic, radical extensions of a cyclotomic field, where we take radicals of the base, cyclotomic field, however. (Q^ab)^ab is the "maximal metabelian extension of Q" - the biggest extension of Q with metabelian galois group, and is the maximal abelian extension of Q^ab, and comes from taking radicals of all elements of Q^ab = $\mathbb {Q} (\mu _{\infty })$. However, the Casus irreducibilis, the necessity of using radicals of complex numbers to solve cubic (& higher) equations, that Quondum noted, (I didn't notice our article on it before) appears to show that Q^rad, while being an extension of Q^ab is strictly smaller than (Q^ab)^ab.John Z (talk) 08:50, 11 September 2012 (UTC)