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## Mathematics
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Q. What is the current status of Fermat's last theorem?FLT was proven in 1993 working at Princeton.(There are no positive integers x , y , z and n > 2 such that x ^ n + y ^ n = z ^ n) Ans.
It is easy to see that it implies FLT asymptotically. The conjecture was motivated by a theorem, due to Mason tha essentially says the ABC conjecture is true for polynomials. The ABC conjecture also implies Szpiro's conjecture [and vice-versa] and Hall's conjecture. These results are all generally believed to be true. There is a generalization of the ABC conjecture [by Vojta] which is too technical to discuss but involves heights of points on non-singular algebraic varieties . Vojta's conjecture also imples Mordell's theorem. [already known to be true]. There are also a number of inter-twined conjectures involving heights on elliptic curves that are related to much of this stuff. For a more complete discussion, see Lang 's article. (b) conjectures arising from the study of elliptic curves and modular forms. The Taniyama-Weil-Shmimura conjecture. There is a very important and well known conjecture known as the Taniyama-Weil-Shimura conjecture that concerns elliptic curves. This conjecture has been shown by the work of Frey, Serre, Ribet, et. al. to imply FLT uniformly, not just asymptotically as with the ABC conj. The conjecture basically states that all elliptic curves can be parameterized in terms of modular forms. There is new work on the arithmetic of elliptic curves. Sha, the Tate-Shafarevich group on elliptic curves of rank 0 or 1. By the way. An interesting aspect of this work is that there is a close connection between Sha, and some of the classical work on FLT. For example, there is a classical proof that uses infinite descent to prove FLT for n = 4. It can be shown that there is an elliptic curve associated with FLT and that for n=4, Sha is trivial. It can also be shown that in the cases where Sha is non-trivial, that infinite-descent arguments do not work; that in some sense 'Sha blocks the descent'. Somewhat more technically, Sha is an obstruction to the local-global principle [e.g. the Hasse-Minkowski theorem]. (c) Conjectures arising from some conjectured inequalities involving Chern classes and some other deep results/conjectures in arithmetic algebraic gemoetry. [about which I know epsilon]. The diophantine and elliptic curve conjectures all involve deep properties of integers. Until these conjecture were tied to FLT, FLT had been regarded by most mathematicians as an isolated problem; a curiosity. Now it can be seen that it follows from some deep and fundamental properties of the integers. [not yet proven but generally believed]. This synopsis is quite brief. A full survey would run to many pages. Q. Has the Four Colour Theorem been solved?
(Every planar map with regions of simple borders can be coloured
with 4 colours in such a way that no two regions sharing
a non-zero length border have the same colour.)
Ans.
This theorem was proved with the aid of a computer in 1976.
The proof shows that if aprox. 1,936 basic forms of maps
can be coloured with four colours, then any given map can be
coloured with four colours. A computer progam coloured this
basic forms. So far nobody has been able to prove it without
using a computer. In principle it is possible to emulate the
computer proof by hand computations. Q. What are the largest prime numbers?Ans. Largest known Mersenne prime It is 2 ^ 756839 - 1. It was discovered by a Cray-2 in England in 1992. It has 227,832 digits. Largest known primeThe largest known prime was 391581 * 2 ^ 216193 - 1. See Brown, Noll, Parady, Smith, Smith, and Zarantonello, Letter to the editor, American Mathematical Monthly, vol. 97, 1990, p. 214. Now the largest known prime is the Mersenne prime described above. Largest known twin primesThe largest known twin primes are 1706595 * 2 ^ 11235 +- 1. See B. K. Parady and J. F. Smith and S. E. Zarantonello, Smith, Noll and Brown. Largest known twin primes, Mathematics of Computation, vol.55, 1990, pp. 381-382. largest Fermat number with known factorizationF11 = ( 2 ^ ( 2 ^ 11 ) ) + 1 which was factored by Brent & Morain in 1988. F9 = ( 2 ^ ( 2 ^ 9 ) ) + 1 = 2 ^ 512 + 1 was factored by A.K. Lenstra, H.W. Lenstra Jr., M.S. Manasse & J.M. Pollard in 1990. The factorization for F10 is NOT known. Current status on Mersenne primesMersenne primes are primes of the form 2 ^ p - 1. For 2 ^ p - 1 to be prime we must have that p is prime. The following Mersenne primes are known.
The way to determine if 2^p-1 is prime is to use the Lucas-Lehmer test:Lucas_Lehmer_Test(p): u := 4 for i from 3 to p do u := u ^ 2 - 2 mod 2 ^ p - 1 od if u = = 0 then 2 ^ p - 1 is prime else 2 ^ p - 1 is composite fi The following ranges have been checked completely: 2 - 355K and 430K - 520K Q. I have this complicated symbolic problem (most likely
a symbolic integral or a DE system) that I can't solve.
What should I do?Find a friend with access to a computer algebra system
like MAPLE, MACSYMA or MATHEMATICA and ask her/him to solve it.
If packages cannot solve it, then (and only then) ask the net.Ans. Q. Where can I get Symbolic Computation Package? This is not a comprehensive list. There are other
Computer Algebra packages available that may better
suit your needs.Ans. MaplePurpose: Symbolic and numeric computation, mathematical programming, and mathematical visualization. Contact: Waterloo Maple Software, 160 Columbia Street West, Waterloo, Ontario, Canada N2L 3L3 Phone: (519) 747-2373 DOE-MacsymaPurpose: Symbolic and mathematical manipulations. Contact: National Energy Software Center Argonne National Laboratory 9700 South Cass Avenue Argonne, Illinois 60439 Phone: (708) 972-7250 PariPurpose: Number-theoretic computations and simple numerical analysis. Available for Sun 3, Sun 4, generic 32-bit Unix, and Macintosh II. This is a free package, available by ftp from math.ucla.edu (128.97.64.16). Contact: questions about pari can be sent to MathematicaPurpose: Mathematical computation and visualization, symbolic programming. Contact: Wolfram Research, Inc. 100 Trade Center Drive Champaign, IL 61820-7237 Phone: 1-800-441-MATH MacsymaPurpose: Symbolic and mathematical manipulations. Contact: Symbolics, Inc. 8 New England Executive Park East Burlington, Massachusetts 01803 United States of America (617) 221-1250 MatlabPurpose: `matrix laboratory' for tasks involving matrices, graphics and general numerical computation. Contact: The MathWorks, Inc. 21 Eliot Street South Natick, MA 01760 508-653-1415 CayleyPurpose: Computation in algebraic and combinatorial structures such as groups, rings, fields, modules and graphs. Available for: SUN 3, SUN 4, IBM running AIX or VM, DEC VMS, others Contact: Computational Algebra Group University of Sydney NSW 2006 Australia Phone: (61) (02) 692 3338 Fax: (61) (02) 692 4534 Q. What is 0 to the 0 power? Read 0 ** 0 as "0 to the 0 power."Ans. Note: According to some Calculus textbooks, 0 ** 0 is an "indeterminate form". When evaluating a limit of the form 0 ** 0, then you need to know that limits of that form are called "indeterminate forms", and that you need to use a special technique such as L'Hopital's rule to evaluate them. Otherwise, 0 ** 0=1 seems to be the most useful choice for 0 ** 0. This convention allows us to extend definitions in different areas of mathematics that otherwise would require treating 0 as a special case. Notice that 0 ** 0 is a discontinuity of the function x ** y. Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f( x ) ** g( x ) approaches 1 as x approaches 0 from the right. From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik): "Some textbooks leave the quantity 0 ** 0 undefined, because the functions x ** 0 and 0 ** x have different limiting values when x decreases to 0. But this is a mistake. We must define x ** 0 = 1 for all x, if the binomial theorem is to be valid when x = 0 , y = 0, and/or x = -y. The theorem is too important to be arbitrarily restricted! By contrast, the function 0 ** x is quite unimportant." Published by Addison-Wesley, 2nd printing Dec, 1988. Q. Why is 0.9999... = 1? In modern mathematics, the string of symbols "0.9999..." is understood
to be a shorthand for "the infinite sum Ans. This in turn is shorthand for "the limit of the sequence of real numbers Using the well-known epsilon - delta definition of limit, one can easily show that this limit is 1. The statement that 0.9999... = 1 is simply an abbreviation of this fact. Choose epsilon > 0. Suppose delta = 1 / - log_10 epsilon, thus epsilon = 10 ^ ( -1 / delta ). For every m > 1 / delta we have that So by the (epsilon-delta) definition of the limit we have An "informal" argument could be given by noticing that the following sequence of "natural" operations has as a consequence 1 = 0.9999.... Therefore it's "natural" to assume 1 = 0.9999..... 10x = 9.99999.... 10x - x = 9 9x = 9 x = 1
References:E. Hewitt & K. Stromberg, Real and Abstract Analysis, Springer-Verlag, Berlin, 1965. W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976. Q. How can I find PI to extreme precision? MAPLE or MATHEMATICA can give you 10,000 digits of Pi in a blink,
and they can compute another 20,000 - 500,000 overnight (range depends
on hardware platform).Ans. It is possible to retrieve 1.25+ million digits of pi via anonymous ftp from the site wuarchive.wustl.edu, in the files pi.doc.Z and pi.dat.Z which reside in subdirectory doc/misc/pi. References :J. M. Borwein, P. B. Borwein, and D. H. Bailey, "Ramanujan, Modular Equations, and Approximations to Pi", American Mathematical Monthly, vol. 96, no. 3 (March 1989), p. 201 - 220. P. Beckman A history of pi Golem Press, CO, 1971 (fourth edition 1977) J.M. Borwein and P.B. Borwein The arithmetic-geometric mean and fast computation of elementary functions SIAM Review, Vol. 26, 1984, pp. 351-366 J.M. Borwein and P.B. Borwein More quadratically converging algorithms for pi Mathematics of Computation, Vol. 46, 1986, pp. 247-253 J.M. Borwein and P.B. Borwein Pi and the AGM - a study in analytic number theory and computational complexity Wiley, New York, 1987 Shlomo Breuer and Gideon Zwas Mathematical-educational aspects of the computation of pi Int. J. Math. Educ. Sci. Technol., Vol. 15, No. 2, 1984, pp. 231-244 Y. Kanada and Y. Tamura Calculation of pi to 10,013,395 decimal places based on the Gauss-Legendre algorithm and Gauss arctangent relation Computer Centre, University of Tokyo, 1983 Morris Newman and Daniel Shanks On a sequence arising in series for pi Mathematics of computation, Vol. 42, No. 165, Jan 1984, pp. 199-217 E. Salamin Computation of pi using arithmetic-geometric mean Mathematics of Computation, Vol. 30, 1976, pp. 565-570 D. Shanks and J.W. Wrench, Jr. Calculation of pi to 100,000 decimals Mathematics of Computation, Vol. 16, 1962, pp. 76-99 Daniel Shanks Dihedral quartic approximations and series for pi J. Number Theory, Vol. 14, 1982, pp.397-423 David Singmaster The legal values of pi The Mathematical Intelligencer, Vol. 7, No. 2, 1985 Stan Wagon Is pi normal? The Mathematical Intelligencer, Vol. 7, No. 3, 1985 J.W. Wrench, Jr. The evolution of extended decimal approximations to pi The Mathematics Teacher, Vol. 53, 1960, pp. 644-650 Q. The existence of a projective plane of order 10 has long been
an outstanding problem in discrete mathematics and finite geometry. More precisely, the question is: is it possible to define 111 sets
(lines) of 11 points each such that:Ans. for any pair of points there is precisely one line containing them both and for any pair of lines there is only one point common to them both. Analogous questions with n ^ 2 + n + 1 and n + 1 instead of 111 and 11 have been positively answered only in case n is a prime power. For n = 6 it is not possible. The n = 10 case has been settled as not possible either See Am. Math. Monthly, recent issue. As the "proof" took several years of computer search (the equivalent of 2000 hours on a Cray-1) it can be called the most time-intensive computer assisted single proof. The final steps were ready in January 1989. Q. Is there a formula to determine the day of the week, given
the month, day and year?Ans. Here is the Standard Method. JFM AMJ JAS OND 144 025 036 146 Now take the remainder when you divide by 7; 0 is Sunday, the first day of the week, 1 is Monday, and so on. Another formula is:
W = k + [ 2.6 m - 0.2 ] - 2 C + Y + [ Y / 4 ] + [ C / 4 ] mod 7k is day (1 to 31)m is month (1 = March, ..., 10 = December, 11 = Jan, 12 = Feb)Treat Jan & Feb as months of the preceding year C is century ( 1987 has C = 19)Y is year ( 1987 has Y = 87 except Y = 86 for jan & feb)W is week day (0 = Sunday, ..., 6 = Saturday)This formula is good for the Gregorian calendar (introduced 1582 in parts of Europe, adopted in 1752 in Great Britain and its colonies, and on various dates in other countries). It handles century & 400 year corrections, but there is still a 3 day / 10,000 year error which the Gregorian calendar does not take. into account. At some time such a correction will have to be done but your software will probably not last that long!
References:Winning Ways by Conway, Guy, Berlekamp is supposed to have it. Martin Gardner in "Mathematical Carnaval". Michael Keith and Tom Craver, "The Ultimate Perpetual Calendar?", Journal of Recreational Mathematics, 22:4, pp. 280-282, 1990. K. Rosen, "Elementary Number Theory", p. 156. Q. What is the formula for the Surface Area of a sphere in
Euclidean N - Space. That is, of course, the volume of the N - 1
solid which comprises the boundary of an N-Sphere.The volume of a ball is the easiest formula to remember:Ans. It's r ^ N times pi ^ ( N / 2 ) / ( N / 2 )!.The only hard part is taking the factorial of a half-integer. The real definition is that x ! = Gamma(x+1), but if you want a formula, it's: First, we note that the integral over the line of Now we change to spherical coordinates. We get the integral from 0 to infinity of where V is the surface volume of a sphere. Integrate by parts repeatedly to get the desired formula. Q. Cutting a sphere into pieces of larger volume. Is it possible
to cut a sphere into a finite number of pieces and reassemble
into a solid of twice the volume?This question has many variants and it is best answered explicitly.
Given two polygons of the same area, is it always possible to
dissect one into a finite number of pieces which can be reassembled
into a replica of the other?Ans.
Dissection theory is extensive. In such questions one needs to
specify
Some Dissection resultsThis theorem was proven by Bolyai and Gerwien independently, and has undoubtedly been independently rediscovered many times. I would not be surprised if the Greeks knew this. The Hadwiger-Glur theorem implies that any two equal-area polygons are equi-decomposable using only translations and rotations by 180 degrees. Here, for each direction v (ie, each vector on the unit circle in the plane), let PHI_P(v) be the sum of the lengths of the edges of P which are perpendicular to v, where for such an edge, its length is positive if v is an outward normal to the edge and is negative if v is an inward normal to the edge. "If P and Q are two polyhedra of equal volume, are they equi-decomposable by means of translations and rotations, by cutting into finitely many sub-polyhedra, and gluing along boundaries?" The answer is "NO" and was proven by Dehn in 1900, just a few months after the problem was posed. (Ueber raumgleiche polyeder, Goettinger Nachrichten 1900, 345-354). It was the first of Hilbert's problems to be solved. The proof is nontrivial but does "not" use the axiom of choice. "Hilbert's Third Problem", by V.G.Boltianskii, Wiley 1978. This construction is known as the "Banach-Tarski" paradox or the "Banach-Tarski-Hausdorff" paradox (Hausdorff did an early version of it). The "pieces" here are non-measurable sets, and they are assembled "disjointly" (they are not glued together along a boundary, unlike the situation in Bolyai's thm.) An excellent book on Banach-Tarski is: "The Banach-Tarski Paradox", , 1985, Cambridge University Press. Also read in the Mathematical Intelligencier an article on the Banach-Tarski Paradox. The pieces are not (Lebesgue) measurable, since measure is preserved by rigid motion. Since the pieces are non-measurable, they do not have reasonable boundaries. For example, it is likely that each piece's topological-boundary is the entire ball. The full Banach-Tarski paradox is stronger than just doubling the ball. It states: This is usually illustrated by observing that a pea can be cut up into finitely pieces and reassembled into the Earth. The easiest decomposition "paradox" was observed first by Hausdorff This result is, nowadays, trivial, and is the standard example of a non-measurable set, taught in a beginning graduate class on measure theory.
References:In addition to Wagon's book above, Boltyanskii has written at least two works on this subject. An elementary one is: "Equivalent and equidecompsable figures" in Topics in Mathematics published by D.C. HEATH AND CO., Boston. It is a translation from the 1956 work in Russian. Also, the article "Scissor Congruence" by Dubins, Hirsch and ?, which appeared about 20 years ago in the Math Monthly, has a pretty theorem on decomposition by Jordan arcs. Banach and Tarski had hoped that the physical absurdity of this theorem would encourage mathematicians to discard AC. They were dismayed when the response of the math community was `Isn't AC great? How else could we get such unintuitive results? Q.
What is the Axiom of Choice? Why is it important? Why some articles
say "such and such is provable, if you accept the axiom of choice."?
What are the arguments for and against the axiom of choice?There are several equivalent formulations:Ans. The Cartesian product of nonempty sets is nonempty, even if the product is of an infinite family of sets. Given any set S of mutually disjoint nonempty sets, there is a set C containing a single member from each element of S. C can thus be thought of as the result of "choosing" a representative from each set in S. Hence the name. Why is it important? All kinds of important theorems in analysis require it. Tychonoff's theorem and the Hahn-Banach theorem are examples. AC is equivalent to the thesis that every set can be well-ordered. Zermelo's first proof of this in 1904 I believe was the first proof in which AC was made explicit. AC is especially handy for doing infinite cardinal arithmetic, as without it the most you get is a " partial " ordering on the cardinal numbers. It also enables you to prove such interesting general facts as that n^2 = n for all infinite cardinal numbers. What are the arguments for and against the axiom of choice? The axiom of choice is independent of the other axioms of set theory and can be assumed or not as one chooses. (For) All ordinary mathematics uses it. There are a number of arguments for AC, ranging from a priori to pragmatic. The pragmatic argument (Zermelo's original approach) is that it allows you to do a lot of interesting mathematics. The more conceptual argument derives from the "iterative" conception of set according to which sets are "built up" in layers, each layer consisting of all possible sets that can be constructed out of elements in the previous layers. (The building up is of course metaphorical, and is suggested only by the idea of sets in some sense consisting of their members; you can't have a set of things without the things it's a set of). If then we consider the first layer containing a given set S of pairwise disjoint nonempty sets, the argument runs, all the elements of all the sets in S must exist at previous levels "below" the level of S. But then since each new level contains "all" the sets that can be formed from stuff in previous levels, it must be that at least by S's level all possible choice sets have already been "formed". This is more in the spirit of Zermelo's later views (c. 1930). (Against) It has some supposedly counterintuitive consequences, such as the Banach-Tarski paradox. (See next question) Arguments against AC typically target its nonconstructive character: it is a cheat because it conjures up a set without providing any sort of "procedure" for its construction--note that no "method" is assumed for picking out the members of a choice set. It is thus the platonic axiom par excellence, boldly asserting that a given set will always exist under certain circumstances in utter disregard of our ability to conceive or construct it. The axiom thus can be seen as marking a divide between two opposing camps in the philosophy of mathematics: those for whom mathematics is essentially tied to our conceptual capacities, and hence is something we in some sense "create", and those for whom mathematics is independent of any such capacities and hence is something we "discover". AC is thus of philosophical as well as mathematical significance. It should be noted that some interesting mathematics has come out of an incompatible axiom, the Axiom of Determinacy (AD). AD asserts that any two-person game without ties has a winning strategy for the first or second player. For finite games, this is an easy theorem; for infinite games with duration less than omega and move chosen from a countable set, you can prove the existence of a counter-example using AC. Jech's book "The Axiom of Choice" has a discussion. An example of such a game goes as follows. Choose in advance a set of infinite sequences of integers; call it A. Then I pick an integer, then you do, then I do, and so on forever (i.e. length \omega). When we're done, if the sequence of integers we've chosen is in A, I win; otherwise you win. AD says that one of us must have a winning strategy. Of course the strategy, and which of us has it, will depend upon A. From a philosophical/intuitive/pedagogical standpoint, I think Bertrand Russell's shoe/sock analogy has a lot to recommend it. Suppose you have an infinite collection of pairs of shoes. You want to form a set with one shoe from each pair. AC is not necessary, since you can define the set as "the set of all left shoes". (Technically, we're using the axiom of replacement, one of the basic axioms of Zermelo-Fraenkel (ZF) set theory.) If instead you want to form a set containing one sock from each pair of an infinite collection of pairs of socks, you now need AC.
References:Maddy, "Believing the Axioms, I", J. Symb. Logic, v. 53, no. 2, June 1988, pp. 490-500, and "Believing the Axioms II" in v.53, no. 3. Gregory H. Moore, Zermelo's Axiom of Choice, New York, Springer-Verlag, 1982. H. Rubin and J. E. Rubin, Equivalents of the Axiom of Choice, Amsterdam, North-Holland, 1963. A. Fraenkel, Y. Bar-Hillel, and A. Levy, Foundations of Set Theory, Amsterdam, North-Holland, 1984 (2nd edition, 2nd printing), pp. 53-86. Q. Is there a theory of quaternionic analytic functions, that is, a four-
dimensional analog to the theory of complex analytic functions?Yes. This was developed in the 1930s by the mathematician
Fueter. It is based on a generalization of the Cauchy-Riemann
equations, since the possible alternatives of power series expansions
or quaternion differentiability do not produce useful theories.
A number of useful integral theorems follow from the theory.
Sudbery provides an excellent review. Deavours covers some of the same
material less thoroughly. Brackx discusses a further generalization
to arbitrary Clifford algebras.Ans. Anthony Sudbery, Quaternionic Analysis, Proc. Camb. Phil. Soc., vol. 85, pp 199-225, 1979. Cipher A. Deavours, The Quaternion Calculus, Am. Math. Monthly, vol. 80, pp 995-1008, 1973. F. Brackx and R. Delanghe and F. Sommen, Clifford analysis, Pitman, 1983. |