studiot

Reading through this thread it seems to me to be a discussion between North Americans about the system in the US. I can't see the relevence of this comment about Scotland, which is so short as to be misleading about Scottish education. What exactly did you mean please ?

First congratulations for looking at alternative proposals and trying to evaluate them. +1 Thre are many factors in play here so you probably haven't considered all of them. After all the experts tend to only consider a subset of the total facts. However some that you should give consideration to. The Earth is ball shaped not flat. So imagine a plate the size of North America moving. What direction is it moving in, on the curved surface of the ball ? You cannot say that Alaska, Washington and California are all moving bodily in the same direction or in a linear fashion. And what about Maryland on the other side ? What trajectory is it moving in? Yet that is so often what is depicted in the simple sections showing subduction, or else the mind of the viewer wrongly assumes this. Such motion as actually occurs induces sideways motion and rotation along much of the interface. These two facts alone make the task of reconstructing the eventual motion over hundreds of millions of years quite difficult to put it mildly. Much of the rheology of the processes are just our best guesses. Remember also the history of the development of the theories.Plate tectonics grew out of measured inconsistencies in continental drift, which itself had not long been confirmed.

You ae right that there are graphical methods that can solve polynomials, and other equations, to any required degree of accuracy. For polynomials Lills (1867) method is one such. But please get your definition of a polynomial expression. It is of the form anXn + a(n-1)X(n-1)+ a(n-2)X(n-2) ...........a0X + b Where the an, b are coefficients , not all zero Conventionally you would generate an equation by setting this polynomial expression equal to zero. Lills method was originally in French, An English version appears in Theory of Equations, Turnbull, 1939 along with other methods of dealing with polynomials eg that of Horner's reduction of of degree. Lills method is also dealt with at great length by Cremona Graphical Calculus and Reciprocal figures. This also deals with a whole raft of graphical methods. The original 1888 translation book was in Italian but has been translated to English in 1977 by Beare. Finally Ewart (1919) Elements of Graphic Dynamics also deals with this and other useful subjects such as graphical integration and graphical differentiation.

That's a shame, they have obviously taken them down very recently. I went to the website and copy/pasted the link in from my address bar make my last post. Edit No they are still there try again, though 1 has gone awol. (I see I somehow lost an R) http://www.umsl.edu/~chickosj/c365/lectureNMR2.pdf

Here are a series of lectures you can donwload in pdf. Be warned it's the full monty - pretty advanced stuff www.umsl.edu/~chickosj/c365/lectureNM2.pdf just change the NM2 to NM3 , 4 etc up to 7.

I would be very grateful if you would review / explain this sentence.

Then you will have to have another one. It doesn't hurt. Hundreds of millions now have the ordinary flu vaccine annually without problem, but to general benefit.

I do hope you are not creating unneccessary work for others and confusing 'laid back' with 'lazy'.

In what way is this Modern and Theoretical Physics, not a speculation on your part ? What Modern and Theoretical Physics is used ?

I agree and indeed suggested such a thing in my first post in this thead. I also agreethat wtf is best placed to analyse and Sensei to program answers. The question has similar vagaries as the simpler one I posted and this is only a four digit number. The one here is a nine digit number so I would expect candidate numbers for addition to be of the form xxxxxxxxx + yyyyyyyyy Which has a lot more combinations.

Interesting way to put it, I don't know any even primes other than two

I don't see any justification for the first statement, (Which is think is wrong because of carries as below) I don't understand the second statement at all. The following beginning analysis may help, Start with two nine digit numbers XXXXXXXXX and YYYYYYYYY. Allow all digits 0 through 9 including leading zeros, which will take care of the possibility of the actual numbers having less than 9 digits. The last digit is a zero so consider all possible pairs that can result in a zero. Strike out those that have a factor from the disallowed list. [math]\begin{array}{*{20}{c}} {Pair} \hfill & {Divisor} \hfill \\ {\left\{ {0,0} \right\}} \hfill & 2 \hfill \\ {\left\{ {1,9} \right\}} \hfill & - \hfill \\ {\left\{ {2,8} \right\}} \hfill & 2 \hfill \\ {\left\{ {3,7} \right\}} \hfill & - \hfill \\ {\left\{ {4,6} \right\}} \hfill & 2 \hfill \\ {\left\{ {5,5} \right\}} \hfill & 5 \hfill \\ \end{array}[/math] Hence the last digit (of each number) can be chosen in 2 possible ways (as Tina says ) but However since both add up to 10 there must be a carry to the next digit. Also all possible last digits are now odd so from no on no number will be divisible by 2. Since there is a carry, the two digits must add to 6 or 16 Listing these as before [math]\begin{array}{*{20}{c}} {Pair} \hfill & {Divisor} \hfill \\ {\left\{ {0,6} \right\}} \hfill & - \hfill \\ {\left\{ {1,5} \right\}} \hfill & - \hfill \\ {\left\{ {2,4} \right\}} \hfill & - \hfill \\ {\left\{ {3,3} \right\}} \hfill & - \hfill \\ {\left\{ {8,8} \right\}} \hfill & - \hfill \\ {\left\{ {7,9} \right\}} \hfill & - \hfill \\ \end{array}[/math] Which yields 6 possible pairs, not 3 as Tina hopes I will leave it to you to decide if any further divisors can be eliminated

The point is that prime no prime number greater than 23 is divisible by any of the first 9 primes. So each prime number and its pair which adds to make 223092870 will be candidates for inclusion and so must be checked. and there are a great many prime numbers up to 9 digits long. But added to these are numbers which are not prime yet still not divisible by any of the first 9 primes such as the example 2419 I gave. All of these must also be checked.

Why are you suprised? Yes, you made a valid comment that the problem is ill defined. But you also made an invalid one about irrational numbers in your examples. Yes I also made an invalid one, which joigus picked up and I immediately corrected. We all make mistakes; we all need to admit them. Well this is one very long way to do it since that is only a beginning, not a solution. There are plenty of primes less than your criterion for example the 1,000th prime is only 7919 Further there are other non prime numbers not divisible by your set eg 2419 is not prime nor divisible by your set.

Counterexample: π is irrational; 1−π is irrational too. but π+1−π=1 . Nice one. +1

I did look at this one, but I am sorry to say that all of us have parts of our subject we like more than other parts. And amongst my personally least liked parts are combinatoric and number theory. There seem to be several problems of this type going around at the moment. On another (maths) forum they have been debating this simpler one for a couple of weeks now. and this is only a four digit number. The one here is a nine digit number so I would expect candidate numbers for addition to be of the form xxxxxxxxx + yyyyyyyyy Which has a lot more combinations. This really is problem that lends itself to a computer solution so I am suprised at Sensei's response, which is incorrect. since the last digit is a zero, there must be a carry so possible choices for pairs in the last positions are [0,0}, {9,1} , {8,2}. {7,3), {6,4} and {5,5} So {x,y} can be selected in 6 different ways. But any number ending in a 5 is divisible by five so the last one can be discounted. Further reduction can be had by noting that any pair containing an even number can be discounted as divisible by 2, another orignal factor. You can work through the digit pairs to find out how many possibilities there are for each other pair. Then multiply these together an see the final result, using the laws of combinatorics. Tina please note this first result and compare with your answer. I don't think there are any two irrational numbers that can be added together to make a rational one. In any case none of these are irrational.

Trying to post a Cayley table [math]\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math] [math]\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math] [math]\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math]

I can see my table in the list in the activity tab, but I can't see it in the thread itself. This is even after refreshing the thread in the normal manner for MathML. [math]\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math] [math]\left( {\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}} \right)[/math] [math]\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math] [math]\left( {\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}} \right)[/math]

[math]\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\\end{array}[/math]

Here is what is inside the front cover of a typical computer book. I assume since you have your book you can find something similar. It not only details the rights and restrictions for the reader, but also acknowledges where the authors used software to generate example pages for the book and example screens from real life.

This is the last thing you should do. Wiley or its subsidiaries are under no obligation to reply to you. Period. You, however, owe the actual copyright holder a duty of diligence to respect that copyright. Copyright arises automatically at the moment of creation of the work. There is no requirement on the creator to announce his or her copyright. If they chooose to do so or hand it to a publisher or other person they may do so. Wiley will certainly have stated the rules of use or restriction on the copyright at the very beginning of the book. However with textbooks that state their aim is to tell you how do do something, their is implied tranfer of enough copyright to perform whatever they tell you. So in a maths book if they said "to add 3027 to 5144 place one above the other and draw a line under the stack. Then add each vertical pair of digits, with carries, starting from the right hand side to form the sum placed under the line," you would be OK to do this as often as you wish. The same if they showed you how to code a macro in Word or use the command line editor. If however the author states in the preface something like "there are new results, never before published" you should at the very least write to them explaing your interest and offering to include an acknowledgement as the originator. Remember you can't copyright an actual idea, only a specific presentation of it.

+1

So you thought that the rules of both politeness and this forum don't apply to you ? A response typical of those ignorant of the subject they are espousing.

My dog changes from somnolent to very intelligent and communicates most successfully with me every mealtime and walkies time.

I would suggest the way the question is phrased misses a trick from our own historical experience. You suggest one language, and each language on Earth is unique because it doesn't need to be otherwise. Yet it is known both from deciphering codes and the experience with the Rosetta Stone that having more than one different representation of the sme thing in the single most useful thing to have. Consequently our communication attempt should contain not one, but many, 'languages' for comparison. Edit cross posted this this. +1