General Comments

Saturday 1 March 2003 at 2:12 pm | In Articles | 143 Comments

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  1. I don’t think I have seen maths formulae written so cleverly anywhere else on the www, I don’t know how you do it.
    I suspect it cutting edge stuff that will soon be taken up by other writers.

    Comment by Mike — Sunday 29 February 2004 1:39 pm #

  2. It uses LatexRender which is publically available at http://www.mayer.dial.pipex.com/tex.htm#latexrender

    Comment by Steve — Sunday 29 February 2004 2:15 pm #

  3. i’m not sure who to turn to for this and I thought maybe one of the readers could help me with this. I am doing some montecarlo on excel and have a set of data that is a skewed gaussian distribution. I need an output that is a function of x,y,z and a random number generator

    x = volume associated with 10% probability success
    y = volume associated with 50% probability success
    z = volume associated with 90% probability success
    rand() = excel outputs a random number which is used as the random distribution number

    i figure, once I know the gaussian formula for this skew distribution, entering the random number will provide an output of volume at that specific random number… i can do this 10,000 times and get a much better understanding of the output potential. if you can help, that would be awesome.

    thanks

    Comment by Rushi — Saturday 3 April 2004 11:38 pm #

  4. Rushi
    You could also ask on the sci.math.num-analysis newsgroup. You can find it in Google at http://groups.google.co.uk/groups?hl=en&lr=&ie=UTF-8&group=sci.math.num-analysis

    Comment by Steve — Sunday 4 April 2004 11:35 am #

  5. i kindda like the work. but there should be somebody to talk to.

    Comment by edwin — Friday 11 June 2004 1:31 pm #

  6. What about adding RSS?

    Comment by maxim — Wednesday 7 July 2004 12:04 pm #

  7. We do provide RSS! Do you need details?

    Comment by Steve — Wednesday 7 July 2004 4:04 pm #

  8. Ref: billions. Surfing the news, I read the Hindu times report of the number of deaths caused by the tsunami. The number they gave was 1.26 lakhs (1 lakh=100,000; 1 crore = 10 million).

    Comment by shaun rogers — Tuesday 4 January 2005 2:19 pm #

  9. Is [tex ] tag working in “comment”?
    Below are examples:
    This is just text but \sqrt{2} should be shown as an image and so should \frac{1}{2}

    Comment by jun — Tuesday 29 March 2005 5:04 am #

  10. How I can allow “[tex]” tag to work in “comment” using wordpress&latexrender?

    Comment by jun — Tuesday 29 March 2005 5:06 am #

  11. Tex tags work in comments only [i]after[/i] I have moderated them. Having moderated your first comment you will see the mathematics.

    In WordPress latexrender-plugin.php just remove the comment from the line that says add_filter(‘comment_text’, ‘addlatex’);

    Comment by Steve — Tuesday 29 March 2005 10:56 am #

  12. Although the deadline for my offer has been and gone, your readers may find this (http://lovely-ideas.blogspot.com/2005/03/prime-suspect.html) of interest.

    Comment by HF — Wednesday 6 April 2005 2:50 pm #

  13. Hi, nice blog. Do you have an RSS feed? I cant seem to find it anywhere.

    Comment by anonymous — Friday 20 May 2005 3:49 pm #

  14. Yes there are RSS feeds – scroll to the bottom of the right-hand menu and look for the orange buttons

    Comment by Steve — Friday 20 May 2005 5:44 pm #

  15. I am trying to come up with a formula for calculating the deterioration of a motor oil, so that if tested one can decipher how much life is left in the oil…EG: TBN (total base number) is a measure of the ability of an oil to manage the by products of combustion, namely acids. If a fresh oil has a TBN of 14 and the oil is all used up at TBN 6, and I test the oil at 5000 (y) miles and get a TBN of 11(x) what would the formula look like to tell me how many more miles I can go to go from 11 to 6 assuming a straight line of deterioration.
    Thanks, Al

    Comment by al — Friday 22 July 2005 9:28 pm #

  16. Assuming I have understood the problem correctly, if it’s a linear relationship then y = ax +b.
    When y = 0 then x = 14 and when y = 5000 then x = 11. Put these numbers in to get simultaneous equations which you can solve for a and b.
    Finally, you can then find y when x = 6

    Comment by Steve — Friday 22 July 2005 9:43 pm #

  17. Hello,
    Very nice blog, I enjoyed reading it!
    Would you like to add my math blog http://homeschoolmath.blogspot.com to your list of Maths Blogs?

    Comment by Maria — Wednesday 30 November 2005 12:49 am #

  18. Happy to add your blog to the list :)

    Comment by Steve — Wednesday 30 November 2005 9:45 am #

  19. Hi, we have a math blog in persian/english/french
    http://mathcom.blogfa.com
    please visit us and join.

    Comment by Esmailifar — Sunday 18 December 2005 1:19 pm #

  20. I have two isometric groups. Does a solution to an equation in one group map to a solution to the same equation in the second group?

    Comment by Oliver — Saturday 14 January 2006 10:56 am #

  21. Isometric? Do you mean isomorphic groups? And what do you mean by a solution to an equation?

    Comment by Steve — Saturday 14 January 2006 1:53 pm #

  22. Yes, Steve. Isomorphic. Solution to an equation is an expression involving elements and operations in the group equating to zero. I have found it is not so in my case as addition is not closed. If all the operations in the expression are closed, would the elements in the first group making the expression equal zero transform to elements in the second group making the same equation zero.

    Comment by Oliver — Wednesday 18 January 2006 10:10 am #

  23. If G and H are isomorphic groups with binary operations * and \circ and f is an isomorphism between them, then a*b=c \implies f(a) \circ f(b)=f(c) so equations do correspond.

    Groups are defined only using one binary operation so I’m not sure what you mean by all operations in the expression are closed. I’m also not sure why you say addition is not closed. Are you in fact talking about rings which have 2 binary operations and perhaps they are isomorphic as groups with one of the binary operations but not isomorphic as rings.

    A concrete example would help.

    Comment by Steve — Wednesday 18 January 2006 11:35 am #

  24. Thanks for the comments Steve. I’ve dumped this one now as being something in between definitions, and I cannot draw any conclusion on my definition of equation. Nice web site.

    Comment by Oliver — Wednesday 18 January 2006 12:21 pm #

  25. Has anybody had a look at the P vs NP Clay Mathematics Institute millenium problem? http://www.claymath.org/millenium/P_vs_np
    The maths behind this subject is way beyond me, but I was interested to program the example.
    Is he looking for 50 pairs of compatible students who can room together, or 100 students who are all compatible and can share a building?

    Comment by Oliver — Friday 20 January 2006 12:39 pm #

  26. sorry, that link should be
    http://www.claymath.org/millennium/P_vs_NP

    Good start :-)

    Comment by Oliver — Friday 20 January 2006 12:41 pm #

  27. As I read it, it says that every one of the 100 students must be compatible with every one of the others. If student A and student B are a forbidden pair you can choose one of them but not both. Also student A may occur in another forbidden pair.

    Comment by Steve — Friday 20 January 2006 2:29 pm #

  28. This algorithm would appear to be as good as anything. If at the end the number of selected students is over 90(say), it might be worth tinkering about with others. Under 90 and probably not. Now if only my maths was better … :-)

    Begin: Store all students in processing table.
    a)Calculate everybody’s compatibility index; = sum of dislikes and being disliked.
    Sort the processing table in ascending compatibility index sequence
    Store any zero compatibility index members in selected students table to max of 100, delete from the processing table.
    Test if selected students table contains 99 elements or more.
    If 99, add one from the processing table. Print names End
    Delete the highest incompatibility index and adjust compatibility indexes of processing table.
    If processing table empty, Print ‘no solution’ this algorithm; Number of selected students; End
    Go to a)

    Comment by Oliver — Friday 20 January 2006 4:35 pm #

  29. hi dear friend…
    I am a student of mathematics in Iran, and have a weblog about mathematics to express its beauty for spread mathematics in Iran society !
    I will be glad to get closer with you for sharin the Ideas…
    be successful…
    milad

    Comment by milad — Wednesday 25 January 2006 11:26 pm #

  30. Apart from Andrew Wiles 100 page indirect proof (and way over my head!) my understanding of Fermat’s last theorem is
    a) Nobody has published a proof that in any solution one of X;Y;Z; must contain n
    b) Given that 1 of X;Y;Z; contain n, nobody has published a proof that the equation has no solution.

    Is this understanding correct? Thanks

    Comment by Oliver — Thursday 2 February 2006 12:45 pm #

  31. I’m not sure what you mean by “X;Y;Z; must contain n”. The statement is about the sums of powers of integers. Have a look at The MacTutor History of Mathematics archive – Fermat’s Last Theorem for a discussion of the problem and its solution.

    Comment by Steve — Thursday 2 February 2006 9:06 pm #

  32. we have a math blog on http://mathcom.blogfa.com and want your link

    Comment by M.S — Thursday 2 February 2006 9:13 pm #

  33. Thanks for the link Steve. Have been on that site before. From the site

    Fermat’s Last Theorem splits into two cases.

    Case 1: None of x, y, z is divisible by n.
    Case 2: One and only one of x, y, z is divisible by n.

    My question was has anybody proved case 1 for all n? That is, if they are all prime to n, no solution exists.
    Also has anyone proved case 2 for all n?

    Comment by Oliver — Friday 3 February 2006 2:03 pm #

  34. I found the following question.

    The numbers 8, 5, 4, 9, 1, 7, 6, 3, 2 form a sequence. How are these numbers ordered?

    I am curious as to the answers and thoughts of professional mathematicians. Thanks.

    Comment by Doug — Tuesday 7 February 2006 9:53 pm #

  35. The On-Line Encyclopedia of Integer Sequences

    Comment by Steve — Tuesday 7 February 2006 9:56 pm #

  36. They are listed in alphabetical order.
    eight, five, four, nine, seven, six, three, two, zero.

    Comment by Joe — Thursday 16 February 2006 2:55 am #

  37. I have constructed a set of commutative rings.Please email me your email-adress.
    Email:ger_voerman@hotmail.com

    Comment by gertvoerman — Friday 14 April 2006 12:59 pm #

  38. Can ye smarty-pants please answer my query?!……..

    If one item costs £x say,

    Two items cost less than double £x

    Three items cost less than three times £x

    Etc.

    but at the same time, buying 10 say, of the items has to be cheaper than buying any combination of smaller denominations so items – so…..

    a quantity of 10 items once,

    …… must be cheaper than buying,

    A quantity of 2 items five times or,

    A quantity of 4 items twice and a quantity of 2 items once, say!

    I hope this makes sense & Thanks

    Comment by Bernard — Tuesday 23 May 2006 1:56 pm #

  39. I saw these math T-Shirts and had to share them with everyone:

    Pi
    http://www.kleargear.com/2306.html

    Math Chicks Are Hot
    http://www.kleargear.com/1506.html

    :) amy

    Comment by Amy Zeck — Friday 30 June 2006 7:06 am #

  40. Off topic, I know, and apologies, but does anybody know how the English A-Level maths standard compares to the Irish Higher Level Leaving Certificate maths standard.

    Comment by Kevin Breen — Tuesday 10 October 2006 9:28 am #

  41. I note that Wikipedia says that On June 8th, 2004 it was decided that a Leaving Certificate (higher) subject will be worth two-thirds of an A-level (UK) but that is for all subjects.

    If anyone has experience of both systems then do please tell us.

    Comment by Steve — Tuesday 10 October 2006 10:50 am #

  42. Here’s Stats+Game Theory question :

    In the 1984 movie, The Terminator, a robotic killer is sent back through time to present day Los Angeles to kill Sarah Conner, who will one day give birth to John Conner, a hero who will lead humans to victory in a future war against machines.

    After the Terminator is sent back through time by the machines to change history, John Conner sends a dedicated soldier, Kyle Reese, back through time to defend Sarah Conner.

    The situation is as follows (details do not follow the movie precisely):

    • There are 3 Sarah Conners in L.A. Neither the Terminator nor Reese knows which is the “correct” Sarah Conner.

    • In any round, the Terminator can choose to target one Sarah Conner, and Reese can choose to defend one Sarah Conner.

    • Any undefended Sarah Conner who is targeted by the Terminator is killed. If the Terminator encounters Reese, both the Terminator and Reese are destroyed (Conner survives).

    • If, at any point, the Terminator successfully kills the true Sarah Conner, Reese will disappear as there will be no John Conner to send him back in time (there will be no need for a Terminator to go back in time either, but let’s forget the usual time paradoxes and just allow me this assumption).

    Reese cares only about saving the mother of John Conner.

    What is Reese’s optimal strategy?

    Comment by Anonym — Friday 17 November 2006 12:00 am #

  43. Nice weblog ! I haven’t read much already but so far so good !

    Comment by JS — Sunday 3 December 2006 10:19 am #

  44. anonym,
    you haven’t defined what happens at the end of each round. Do they know who was targetted and defended?
    If not, then Reese’s best strategy is to stick with his round1 choice.
    If they do know, then there is no optimum strategy.

    Comment by oliver — Thursday 21 December 2006 4:32 pm #

  45. Could you please help me install mimetex, located at http://admthe.freehostia.com/mimetex.cgi, to work in my blog so I can use $\$$ signs instead of using the texer I created at http://theani.t35.com/tex/ and then copy and paste the image? Thanks.

    Comment by anirudh — Sunday 21 January 2007 11:46 pm #

  46. Answered via email

    Comment by Steve — Monday 22 January 2007 12:02 am #

  47. But they’re on different servers

    Comment by anirudh — Monday 22 January 2007 12:13 am #

  48. Am I correct in saying the residues of integers mod n**2, n an odd prime, form a ring under addition and multiplication mod n**2?

    Comment by Oliver — Wednesday 27 June 2007 6:24 am #

  49. The set of residues modulo any n forms a ring under addition and multiplication modulo n.

    Comment by Steve — Wednesday 27 June 2007 5:12 pm #

  50. Thanks Steve. I understand that. In the case I quote, I have n elements (including 0) of the set that are not invertible, the other elements forming a multiplicative group modulo n**2. The whole set still meets the definition of a ring though, doesn’t it?

    Comment by Oliver — Saturday 30 June 2007 6:35 am #

  51. It’s still a ring just not an integral domain. So there are elements a,b in the ring which are not zero but ab = 0

    Comment by Steve — Saturday 30 June 2007 10:44 am #

  52. Thanks Steve. I like the ab = 0 bit. I don’t think rings were going to help me solve my problem, but fortunately I’ve hit on another method.

    Comment by Oliver — Monday 13 August 2007 12:29 pm #

  53. I see from wikipedia that multiplication is distributive over addition in a ring.
    so (a+b)c = ac + bc
    Does the term addition is distributive over multiplication make any sense? If so, what would be required to fulfil this.

    Comment by Oliver — Saturday 25 August 2007 7:44 am #

  54. It also needs a(b+c)=ab+ac as multiplication may not be commutative.
    For distributivity of addition over multiplication just swap addition and multiplication to get (ab)+c=(a+c)(b+c) and a+(bc)=(a+b)(a+c). Now it’s up to you to find out if this is possible and, if it is, what such a ring would be like.

    Comment by Steve — Saturday 25 August 2007 11:13 am #

  55. ok Steve. You are challenging my laziness :-), and I’ll come back to your post in due course. From the phrasing I deduce that such a ring is possible (no answer required).
    Anyway, I have solved my ‘problem’ and am trying to document it as lucidly as possible. It does deal with rings, but can be explained without mentioning rings which could frighten people.

    I leave you with some further thoughts on the P vs nP problem mentioned in posts 25, 26 and 28

    Again, just dealing with the easily understood dormitory example, and my definition of incompatibility index, I have a hypothesis.
    1) If there is a solution including the person with the highest index, then there is a solution excluding him. I wouldn’t know how to go about proving that, but the converse (mathematical inverse?) is evidently true. If there is a solution excluding a person with 0 compatibility index, then there is a solution including that person.
    2) A second thought, which probably be proved/disproved with some further thought is that likes/dislikes have the same weight towards a solution, i.e. it doesn’t matter whether persons with compatibility index 2 are 2 dislikes, 2 being disliked, or 1 of each.

    With these thoughts, I would think my algorithm would hold up, so if it filled 90 places (post 28) I would tell the residence administrator that he had to leave 10 rooms empty or find another programmer.

    P vs NP is not about finding a processing algorithm, however, but by analysing the initial data and stating whether a solution could be found. This now does not seem as mathematically daunting (in the dormitory case) as first seems.
    The sum of the incompatibility indexes must be even, as each dislike is matched by a being disliked.
    If after the zero index students have been allocated a room all the other applicants have an index of one, then you can fill half that number of rooms without causing friction. So a formula readily follows.
    I feel confident that formual could be extended to cover the number of people with 2, 3, 4… n index.
    Perhaps one for your students.

    How to tackle the wider Clay challenge is still beyond me, but perhaps there is a semblance of a method here.

    Comment by Oliver — Saturday 25 August 2007 7:20 pm #

  56. The trivial ring with just one element so 0 =1

    has addition distributive over multiplication

    I can’t imagine their would be any others.

    Comment by Oliver — Wednesday 3 October 2007 3:12 pm #

  57. My other method didn’t cope with non invertible items in my problem, but I’ve managed to get round that. I’d still like to know if the case 2 question in post 33 has an answer.

    I have 2 commutative rings (Rp and RP) similar in that they have the same number of elements and the same number of invertible elements. I have a multiplicative operation m on elements on the first ring which carries an invertible element p to an invertible element P in the second, and carries a non invertible element n into the second ring but multiple elements go to zero similar to ab=o above. This leaves orphaned non invertible elements in the second ring with no corresponding parent in the first.
    My problem is to document the conditions for the sum of two of these mappings producing an element in the second ring.
    Including a non invertible element in the sum leads to a solution of form (N +ma)mb or (N+A)B
    One condition is N+A = C, an invertible element in the second ring.
    I was tempted to say that another condition
    was that N was an non orphaned element in the second ring.
    I think I can only say that if the multiplication operation m is distributive over addition, which it isnt always.
    Does that make sense?

    Comment by Oliver — Thursday 4 October 2007 2:47 pm #

  58. Forget these ramblings Steve. I’m being lazy. Anyway, I’ve managed to document the rules for the mapping with the non invertible items.

    I wouldn’t mind reading up a bit more on this type of ring, preferably with examples, so that I could answer my own questions. Any suggested book or website. Not Wikipedia, which is great for first level definition.

    Comment by Oliver — Friday 5 October 2007 2:35 pm #

  59. If you mean commutative rings then Wikipedia’s Commutative algebra page has a number of references to commutative algebra books.

    Comment by Steve — Friday 5 October 2007 2:45 pm #

  60. Thanks Steve, that lot should keep me out of mischief for a while. :-)

    Comment by Oliver — Friday 5 October 2007 4:10 pm #

  61. Here’s one for your students, Steve, or for anybody else reading this blog.

    Let n be an odd prime integer, p an odd integer,
    X;Y; integers ¬= 0 modulo n or modulo p
    Then X**p + Y**p = 0 modulo p implies X+Y = 0 modulo p and the converse is also true.
    X+Y = 0 modulo n implies X**p + Y**p = 0 modulo n but the converse is not necessarily true.
    e.g. 1**3 + 3**3 = 0 mod 7 but 1+3 ¬= 0 mod 7
    However, X**(n-2) + Y**(n-2) = 0 mod n does imply X+Y = 0 mod n. Prove this.
    I must try this latexrender :-)

    Comment by Oliver — Monday 22 October 2007 2:01 pm #

  62. I must try this latexrender :-)

    And this is what it looks like:
    Let n be an odd integer, p an odd prime,
    X,Y integers \neq 0 \bmod{n \text{ or } p}
    Then X^n + Y^n= 0 \bmod{n} \Rightarrow X+Y = 0 \bmod{n} and the converse is also true.
    X+Y = 0 \bmod{p} \Rightarrow X^n + Y^n = 0 \bmod{p} but the converse is not necessarily true.
    e.g. 1^3 + 3^3 = 0 \bmod{7} but 1+3 \not=0 \bmod{7}
    However, X^{p-2} + Y^{p-2} = 0 \bmod{p} does imply X+Y = 0 \bmod{p}. Prove this.

    Comment by steve — Monday 22 October 2007 2:30 pm #

  63. Thanks Steve, But n should be the prime :-)

    Didn’t you know what ‘contain n’ meant in post 31?

    I know mathematics is a very precise science, but perhaps we should be more understanding or perhaps forgiving when discussing ideas.

    Comment by Oliver — Monday 22 October 2007 7:38 pm #

  64. p is always used for a prime and n for an integer so I will have to swap all the n’s and p’s.

    Comment by Steve — Monday 22 October 2007 7:47 pm #

  65. Didn’t know that, and thanks. I’ll need to see if I need to rewrite my paper.

    Comment by Oliver — Tuesday 23 October 2007 8:23 am #

  66. I suppose the real beauty of mathematics is its precision so maybe I should stop whingeing and try and practise that. :-)

    In that problem I posed in post 62, for precision it should read x,y not = 0 mod p and n

    I made that mistake in a Cobol program over 40 years ago. Took me two weeks to find out why it wasn’t working. Had to go down to analysing the machine code. Looks like life has gone full cycle.

    I have submitted my paper to the local university. I hope it hasn’t gone into a black hole, as the silence is deafening. One bit of precision I will have to alter in line with post 52. I have the statement ab = 0, b~=0, therefore a=0. Whilst its ‘obvious’ what I mean, I have to change that, in effect, to b is an invertible element, therefore a = 0

    You live and learn, …… sometimes.

    This latex render. I had computer problems which stopped me downloading it, but they are resolved so I’ll set about that again. Its for web sites, isn’t it? You wouldn’t use it for writing a paper or a book, would you?

    Comment by Oliver — Friday 9 November 2007 4:25 am #

  67. Steve, I’d like to communicate something off line with you. Can you send me an e mail address?

    Comment by Oliver — Friday 9 November 2007 12:43 pm #

  68. I combine elements from a ring to create an expression E = a*b*c to describe some non zero ring element. However, if I make c=0, E = 0. I cannot make a and b = 0

    I make E from the ring in a different way.
    E = d*(f-g) where I cannot make d = 0

    Can I say f – g must be 0?

    Comment by Oliver — Monday 7 April 2008 12:15 pm #

  69. No, as you don’t have an integral domain. Put a=f-g and then you have the same situation as before. For example, multiplication modulo 4 gives 2*(3-1)=0

    Comment by Steve — Monday 7 April 2008 12:41 pm #

  70. thanks Steve. I have an integral domain, but I need to think further through my problem.

    Comment by Oliver — Monday 7 April 2008 1:34 pm #

  71. o.k. Steve. I think I may have it.

    a) If it is an integral domain, c=0 implies a-b = 0 (sorry, still haven’t implemented the latex).

    b) If a-b = 0 implies c =0, then it is an integral domain.

    Correct?

    Comment by Oliver — Monday 7 April 2008 3:23 pm #

  72. What are a,b,c and how are they related? Where does a-b=0 come from?

    An integral domain is one in which there are no zero divisors ie ab=0 implies a=0 or b=0.
    It follows that in an integral domain
    ac=bc=>ac-bc=0=>(a-b)c=0=>a-b=0 or c=0=>a=b or c=0

    Comment by Steve — Monday 7 April 2008 3:35 pm #

  73. I need to think about your post 72 Steve.

    Isomorphic groups, I read, are ‘mathematically identical’

    So I presume being ‘an integral domain’ is a mathematical property, and therefore 2 isomorphic groups are either both an integral domain, or both not an integral domain.

    Correct?

    Comment by Oliver — Monday 7 April 2008 4:13 pm #

  74. An integral domain is a ring so you need isomorphic rings not groups.

    Comment by Steve — Monday 7 April 2008 4:32 pm #

  75. many thanks Steve. I need to go and read up about the definition of isomorphic rings and whether they help with my problem.

    Comment by Oliver — Monday 7 April 2008 4:46 pm #

  76. Test to see that I have installed Latex render.

    \displaystyle\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}dx=\frac{22}{7}-\pi

    Comment by Oliver — Wednesday 19 November 2008 5:13 pm #

  77. How would I best get a copy of Zum letzten Fermat’schen Theorem, J. für Math. 136 (1909) 293-302 by Wieferich, or a detailed explanation of his proof on the first case of FLT that p had to be a Wieferich prime.

    Comment by Oliver — Wednesday 7 January 2009 12:18 pm #

  78. It’s discussed in 13 Lectures on Fermat’s Last Theorem by Paulo Ribenboim.

    Comment by steve — Wednesday 7 January 2009 1:59 pm #

  79. Thanks Steve. That looks like an interesting book. Unfortunately the free preview has the page I need on Wieferich blanked out, although it has the page before and after it! I’m not sure one page would be enough to cover Wieferich’s proof, in the detail that I think I need it, so maybe this book won’t suffice, and I’ll have to get Woeferich’s original paper. Anyway, I’ve ordered the book for some of the interesting maths in it.
    I liked somebody’s review of it. ‘This is Fermat’s Last Theorem for amateurs, not for dummies’! :-)

    Comment by Oliver — Monday 12 January 2009 11:54 am #

  80. I liked somebody’s review of it. ‘This is Fermat’s Last Theorem for amateurs, not for dummies’!

    I hope you aren’t referring to another book by Paul Ribenboim called Fermat’s Last Theorem for Amateurs as that’s not the one you want since it has much less detail.

    Comment by steve — Monday 12 January 2009 2:09 pm #

  81. Yes Steve. I’ve ordered the wrong one!:-( It just states the Wieferich condition without any proof. I’ll order up the lectures.

    Comment by Oliver — Friday 16 January 2009 1:17 pm #

  82. test \sqrt{2}

    Comment by Oliver — Thursday 19 February 2009 2:53 pm #

  83. Let me try and give the solution to the problem in post 61.2

    For xy ? 0 mod p, x^{p-2}+y^{p-2}=x^{p-2}+y^{p-2}\frac{xy}{xy}=yx^{p-1}+xy^{p-1}=frac{x+y}{xy} mod p by Fermat’s theorem.
    So x^{p-2}+y^{p-2}= 0 mod p ? x+y = 0 mod p

    Comment by Oliver — Thursday 19 February 2009 3:35 pm #

  84. oops I’ll try again

    Comment by Oliver — Thursday 19 February 2009 3:35 pm #

  85. Second go :-)
    For xy ? 0 mod p, x^{p-2}+y^{p-2}=\left(x^{p-2}+y^{p-2}\right\frac{xy}{xy}=yx^{p-1}+xy^{p-1}=\frac{x+y}{xy} mod p by Fermat’s theorem.
    So x^{p-2}+y^{p-2}= 0 mod p does imply x+y = 0 mod p

    Comment by Oliver — Thursday 19 February 2009 4:07 pm #

  86. oops, one more go

    Comment by Oliver — Thursday 19 February 2009 4:08 pm #

  87. For xy neq 0 mod p, x^{p-2}+y^{p-2}=\left(x^{p-2}+y^{p-2}\right\frac{xy}{xy}=yx^{p-1}+xy^{p-1}\frac{x+y}{xy}=\frac{x+y}{xy}  mod p by Fermat’s theorem.
    So x^{p-2}+y^{p-2}= 0 mod p /rightarrow x+y = 0 mod p

    Comment by Oliver — Thursday 19 February 2009 4:28 pm #

  88. Steve,
    does post 11 mean that you have to change(moderate) all latex render comments? i.e. I can’t input the formula directly?

    Comment by Oliver — Saturday 21 February 2009 11:31 am #

  89. Steve, does post 11 mean I can’t input Latex render statements directly? You have to alter(moderate) them?

    Comment by Oliver — Saturday 21 February 2009 11:51 am #

  90. I have allowed LaTeXed comments since then as you will have found in your posts such as number 87.

    Comment by Steve — Saturday 21 February 2009 12:54 pm #

  91. thanks Steve. I’ll work out what I’m doing wrong.

    Comment by Oliver — Monday 23 February 2009 12:55 pm #

  92. Try out your code at this Equation Editor before you post.

    Comment by Steve — Monday 23 February 2009 1:04 pm #

  93. I can get my formula to work on Equation editor. What do I do to post them here? The ‘type your equation here’ text doesn’t seem to work when copied, nor does the html code. I’ve tried adding tex statements around it, to no avail.

    Comment by Oliver — Wednesday 11 March 2009 4:24 pm #

  94. What is the tex code? Post it here and I will adjust it so it works.

    Comment by Steve — Wednesday 11 March 2009 4:46 pm #

  95. This is what I entered in equation editor. At the end I just want to switch off the rendering and enter english statements.

    (x^{p-2}+y^{p-2})\frac{xy}{xy}=\frac{(x^{p-1}y+y^{p-1}x)}{xy}=\frac{(x+y)}{xy} by Fermat’s theorem

    Comment by Oliver — Thursday 12 March 2009 9:51 am #

  96. All you needed to do was put the tex tags round the mathematics only. The text should be done outside with no tags.

    I have changed your comment accordingly.

    Comment by steve — Thursday 12 March 2009 9:56 am #

  97. So what precisely did you do Steve? Add the word tex in square brackets before the (x and add the word /tex in square brackets before the by Fermat’s theorem. It doesn’t seem to work for me, at least in the Oliver says preamble.

    Comment by Oliver — Thursday 12 March 2009 10:38 am #

  98. Yes that is what I did. However, the preview has no facility to show mathematics – it is only shown when the comment is submitted. Just use the Equation Editor to check that the syntax is correct before adding to a comment.

    Comment by Steve — Thursday 12 March 2009 11:23 am #

  99. (x^{p-2}+y^{p-2})=(x^{p-2}+y^{p-2})\frac{xy}{xy}=(\frac{yx^{p-1}+xy^{p-1}}{xy})=(\frac{x+y}{xy}) by Fermat’s theorem. Therefore for this expression to be zero modulo p implies x+y = 0 modulo p (x=0 mod p implies y = 0 mod p)

    Comment by Oliver — Thursday 12 March 2009 2:25 pm #

  100. got it. Thanks Steve.

    Comment by Oliver — Thursday 12 March 2009 2:26 pm #

  101. I didn’t buy that second Riebenboim book. Didn’t have another £30. Also the reason for wanting it disappeared. I make an educated guess that Wieferich was able to prove FLT case 1 unless two of X;Y;-Z; were equal mod p. Certainly if that is the case the Wieferich condition follows. But using another piece of work from an old number theory book over a hundred years old, the Wieferich condition could have been extended to 2^{p-1}=1 mod p^{4} (if my guess is correct). Vandiver got close with that condition but included an ‘or’ condition, but a single condition suffices.(if my guess is correct). For the Wieferich condition Fermat was able to prove 2^{p-1}=1 mod p^{k} for all positive integers k. In fact, regardless of the Wieferich condition, he could prove that for X;Y; prime to p for the Fermat equation to have a solution, X^{p-1}= Y^{p-1} mod p^{k} for all integers k. Proof to be published this year. :-)
    Do you know if that was the condition Wieferich couldn’t handle?

    Comment by Oliver — Wednesday 15 April 2009 1:35 pm #

  102. No sorry I don’t.

    Comment by Steve — Wednesday 15 April 2009 1:49 pm #

  103. Thanks for the response Steve. It irks me somewhat that I can’t find a copy of Wieferich’s limited proof on the net (for free!), but it is hidden away in some subscription mathematical journal. With the copyright around these things I would even hazard a guess that Riebenboim’s book doesn’t even detail the exact Wieferich argument but talks around it.
    Changed days from Archimedes running down the street naked shouting ‘Eureka’ :-)

    Comment by Oliver — Wednesday 15 April 2009 3:13 pm #

  104. So, any takers on this one? If you take as a given what I say Fermat could prove in post 101, how do you extend that to him proving no integer solutions to his problem existed for both case 1 and case 2?

    Comment by Oliver — Friday 17 April 2009 6:04 pm #

  105. Steve, can you possibly direct me to a professor of number theory with an open enough mind to spend 10 minutes reading a one and a half page paper on FLT. I get the feeling that none of them will touch it with a barge pole.:-(

    Comment by Oliver — Tuesday 21 April 2009 10:48 am #

  106. or alternatively give me contact details if you are prepared to review it yourself

    Comment by Oliver — Tuesday 21 April 2009 1:01 pm #

  107. Looks like I’m talking to myself here Steve.
    I’ll post the answer to 104 for case 1 on Monday if nobody else does.

    Comment by Oliver — Friday 24 April 2009 8:57 pm #

  108. It has been published that in any solution to FLT case 1 Z^{p} = (X+Y)v^{p} , so given 101 we can continue
    ? X^{p}+Y^{p}=XX^{p-1}+YY^{p-1} =(X+Y)X^{p-1}=(X+Y)Y^{p-1}=Z^{p}=(X+Y)v^{p} mod p^{k} for all k.
    ?if X;Y; are finite X;Y;Z; have a common factor, v
    v ? 1 as X^{p} + Y^{p}  > X+Y for positive integers unless X;Y; = 1 (and 2 is not the p^{th} power of a positive integer). This is a contradiction as X;Y;Z; are co-prime.

    So all that is needed is to prove the given in 101. Anybody want to finish off the case 2 with the given in 101?

    Comment by Oliver — Monday 27 April 2009 1:01 pm #

  109. Is there a definition anywhere of what symbols you can use. I used ‘therefore’ special character and ‘not equal’ special character, and they have come out ?

    Comment by Oliver — Monday 27 April 2009 1:04 pm #

  110. ignore that last post, I’ve found them on equation editor

    Comment by Oliver — Monday 27 April 2009 1:15 pm #

  111. It has been published that in any solution to FLT case 1 Z^{p} = (X+Y)v^{p} , so given 101 we can continue
    \therefore\ X^{p}+Y^{p}=XX^{p-1}+YY^{p-1} =(X+Y)X^{p-1}=(X+Y)Y^{p-1}=Z^{p}=(X+Y)v^{p} mod p^{k} for all k.
    \thereforeif X;Y; are finite X;Y;Z; have a common factor, v
    v \neq 1 \becauseX^{p} + Y^{p}  > X+Y for positive integers unless X;Y; = 1 (and 2 is not the p^{th} power of a positive integer). This is a contradiction as X;Y;Z; are co-prime.

    So all that is needed is to prove the given in 101. Anybody want to finish off the case 2 with the given in 101?

    Comment by Oliver — Monday 27 April 2009 1:23 pm #

  112. It has been published that in any solution to FLT case 1 Z^{p} = (X+Y)v^{p} , so given 101 we can continue
    \therefore\ X^{p}+Y^{p}=XX^{p-1}+YY^{p-1} =(X+Y)X^{p-1}=(X+Y)Y^{p-1}=Z^{p}=(X+Y)v^{p} mod p^{k} for all k.
    \thereforeif X;Y; are finite X;Y;Z; have a common factor, v
    v \neq\1\because\X^{p} + Y^{p}  > X+Y for positive integers unless X;Y; = 1 (and 2 is not the p^{th} power of a positive integer). This is a contradiction as X;Y;Z; are co-prime.

    So all that is needed is to prove the given in 101. Anybody want to finish off the case 2 with the given in 101?

    Comment by Oliver — Monday 27 April 2009 1:25 pm #

  113. sorry Steve.

    Comment by Oliver — Monday 27 April 2009 1:26 pm #

  114. It has been published that in any solution to FLT case 1 Z^{p} = (X+Y)v^{p} , so given 101 we can continue
    \therefore\ X^{p}+Y^{p}=XX^{p-1}+YY^{p-1} =(X+Y)X^{p-1}=(X+Y)Y^{p-1}=Z^{p}=(X+Y)v^{p} mod p^{k} for all k.
    \thereforeif X;Y; are finite X;Y;Z; have a common factor, v
    v \neq1\becauseX^{p} + Y^{p}  > X+Y for positive integers unless X;Y; = 1 (and 2 is not the p^{th} power of a positive integer). This is a contradiction as X;Y;Z; are co-prime.

    So all that is needed is to prove the given in 101. Anybody want to finish off the case 2 with the given in 101?

    Comment by Oliver — Monday 27 April 2009 1:31 pm #

  115. Is there any practice site you can practise posting the finished article with the tex statements in it?

    Comment by Oliver — Monday 27 April 2009 1:32 pm #

  116. There are lots. See Online LaTeX. The first list is for fragments, the second list is for complete documents.

    Comment by Steve — Monday 27 April 2009 1:49 pm #

  117. It has been published that in any solution to FLT case 1 Z^{p} = (X+Y)v^{p} , so given 101 we can continue
    \therefore\ X^{p}+Y^{p}=XX^{p-1}+YY^{p-1} =(X+Y)X^{p-1}=(X+Y)Y^{p-1}=Z^{p}=(X+Y)v^{p} mod p^{k} for all k.
    \thereforeif X;Y; are finite X;Y;Z; have a common factor, v
    v \neq1\because X^{p} + Y^{p}  > X+Y for positive integers unless X;Y; = 1 (and 2 is not the p^{th} power of a positive integer). This is a contradiction as X;Y;Z; are co-prime.

    Comment by Oliver — Monday 27 April 2009 2:34 pm #

  118. I suggest you get your own (free) WordPress blog at http://wordpress.com/ which has LaTeX support, then you can write and edit your own articles there.

    Comment by Steve — Monday 27 April 2009 2:49 pm #

  119. ok Steve I’ll do that. Sorry if I’ve caused any problem. You’ve been very helpful. Can I still request info here? e.g. Do you know if Euler’s proof of Fermat’s prime theorem is available on the net?
    i.e. the one that each prime can be expressed as the sum of 2 squares. Google doesn’t seem to find it.

    Comment by Oliver — Monday 27 April 2009 3:49 pm #

  120. Euler’s proof by infinite descent

    Comment by Steve — Monday 27 April 2009 5:23 pm #

  121. Superb Steve, many thanks.

    Comment by Oliver — Monday 27 April 2009 5:37 pm #

  122. I am wondering whether a professional mathematician would find my amateur’s way of expressing something confusing.
    I have a multipliication group. I have 3 members A1;A2;A3;
    The A members have a special property.
    I have a sub group S and a member s of that sub group.
    Is the following statement clear.
    Define the As coset as the set of 3 A members multiplied in turn by s

    The reason I ask is that coset is normally used when multiplying a sub group by a member of the group, so I’m wondering if that could lead to confusion.

    The set A happens to be a sub group, but I don’t particularly want to prove that.

    Comment by Oliver — Saturday 25 July 2009 9:13 am #

  123. I see the word ‘partition’ is used sometimes in describing things like these cosets.

    I’m looking for a verb. I have a set of pairs which have a common property. Each pair relates 1:1 to another set of pairs which have a second common property.
    I wish to say each first pair ‘ ‘ which second pair is used to form the second property.
    I’ve rejected ‘mandates’; ‘dictates’ and at the moment have ‘prescribes’ as my best choice. Any suggestions welcomed.

    Comment by Oliver — Monday 5 October 2009 2:32 pm #

  124. Any comments on the use in a mathemaical proof of three dots, meaning ‘therefore’, and an upside down three dots, meaning ‘because’?

    I’m told ths usage is now ‘frowned upon’.

    Comment by Oliver — Sunday 15 November 2009 4:05 pm #

  125. What is the preferred (if any) mathematical way of expressing a congruence? After the Gaussian congruence symbol should you say,
    mod p or (mod p)?
    I have come across people that use both.

    Comment by Oliver — Sunday 6 December 2009 4:38 pm #

  126. Either will do though a publisher might have a preference.

    Comment by Steve — Sunday 6 December 2009 6:03 pm #

  127. thanks Steve. Given that the less ink on paper,the better, I’ve removed the parentheses from around the mod p. I’ll worry about a publisher after I find somebody to read the paper.
    :-)

    Comment by Oliver — Tuesday 8 December 2009 2:21 am #

  128. I have a commutative ring. I can produce the value of a term A.(B+C) in two different ways.
    Multiplying the second way by unity, in the form of F multiplied by the inverse of F, I obtain a term A.(D+E), which obviously is the same value as A.(B+C). I think I can say that either B=D and C=E, or B=E and C=D, i.e. the terms being added cannot be B + delta, and C – delta (unless of course B+delta = C). Would you agree? I have an alternative way of explaining what I’m about, but this one is more succinct.

    Comment by Oliver — Thursday 14 January 2010 3:22 pm #

  129. Unless you have division ring you can’t even say D+E=B+C. Even if that were true then you can’t say anything more eg in the ring of integers 1+3=2+2.

    Comment by Steve — Thursday 14 January 2010 4:11 pm #

  130. Having failed in my global attempts to get a review of my paper, I’m looking to submit it to be refereed. Is my understanding of this correct?

    You submit a paper. If it is accepted for publication, you are asked to transfer the copyright. It is published in a journal. Only subscribers to the journal can read it, and have to pay a subscription. A bit like the Wieferich situation above.

    You can make it open access, but you yourself pay a fee. In both cases, you do not own the copyright, and cannot do things like putting it on the web for all to see.

    Comment by Oliver — Saturday 13 February 2010 11:39 am #

  131. Two requests Steve.
    1. I’m writing a book. Can I include some of your posts above in it?
    2. I submitted my paper to the Glasgow mathematical journal. More than two weeks later it still hadn’t been assigned to anybody. I spoke with the editor, who confirmed that if it was accepted it couldn’t also be published on the internet, so we agreed to withdraw the paper. Any chance of an informal review from you before I put it on the net?

    Comment by Oliver — Tuesday 9 March 2010 9:45 am #

  132. I lack your precision in 129 Steve.

    Taking the ring of integers, and a member formed by A.(b+c). A fixed and non zero. b;c; allowed to vary, but their sum kept constant. This sum may be zero. There are an infinite number of solutions for any A and any b+c.

    Is the following precise?.
    Each factorisation is unique as A is fixed, and no b+c can be factorised into another b+c.
    If precise, does that hold for b+c=0?

    Comment by Oliver — Tuesday 16 March 2010 9:36 am #

  133. One last question and I’m off. The residues of the ring of integers mod p^2 form a ring wrt modular multiplication and addition mod p^2. I’m interested in the factors of 0.
    If I take a,b, prime to p and a+b congruent to 0 mod p^2, and I take p.(c+d) congruent to 0 mod p^2 where c+d is congruent to zero mod p but not zero mod p^2, it is correct to say, isn’t it, that a+b cannot factorise to p.(c+d) because a:b: are prime to p?

    Comment by Oliver — Tuesday 6 April 2010 1:11 pm #

  134. Any a, b prime to p with a+b=18, p=3, any c,d with c+d=6. Then a+b=p(c+d).

    Generally, if a+b=0 mod p^2 then obviously a+b=p.k with p dividing k. Now let c+d=k. If p^3 doesn’t divide a+b then p^2 won’t divide c+d.

    Comment by Steve — Tuesday 6 April 2010 7:30 pm #

  135. I’m dealing with residues, Steve, so with p=3, they are all less than 9. With a=2, b=7, a+b = 0 mod 9. With c=1, d=5, 3(c+d) = 0 mod 9.
    What I’m trying to say is that I can’t factorise the elements a:b: so that they become 3(c+d). I can express their sum as that, but to me that is not factorising the elements. I can factorise a+b as 2(1+8), 4(5+4), 8(7+2), mod 9 etc. but not as 3(1+5).
    And the reason is that the residues are prime to 3. Does this make sense?

    Comment by Oliver — Wednesday 7 April 2010 12:48 am #

  136. If 0<a,b<p2 and a+b=0 mod p2 then a+b=p2 so a+b=p.p and hence c+d=p.
    So eg (2+7)=3(1+2) answers the original question.

    Comment by Steve — Wednesday 7 April 2010 11:55 am #

  137. That’s a bit of a blow Steve. You are saying that a+b can factorise as p.p even though both a and b are prime to p. And don’t you have division by a factor of zero when you say hence c+d=p? I’ll leave it there. Thanks for your response.

    Comment by Oliver — Wednesday 7 April 2010 12:57 pm #

  138. I think I’ve cracked it Steve. I have a certain type (class) of factor a+b,a:b: prime to p, which may be zero at some modular level, but eventually has to be non zero. Dealing at mod p^2 level above I cannot make my conclusion. The answer is to deal at mod p^3 level. From an a+b zero factor mod p^3 I have to derive other a+b type factors, either 0 or of form p(a+b). Analysing these factors mod p^2 it can only be that a+b = 0 mod p^2, which is enough for my problem.

    Comment by Oliver — Thursday 8 April 2010 12:28 pm #

  139. Needless to say I cannot say a+b is zero mod p^3, so that is not the answer.

    My last effort is as follows.
    I form a+b residues mod p^2 on a 1:1 basis from other residues mod p^2. I end up with 10 say mod p^2. 5 of them don’t add to zero and can be rejected. The other 5 add to zero and are formed by adding two residues mod p^2. I can factor these as p.(a+b) where a+b is one of the rejected a+b pairs.
    Since I am not factorising an integer, but a residues sum, I contend that the (a+b) factor in p.(a+b) is a residue mod p. I cannot make it an (a+b) residue mod p^2, and therefore I maintain my 1:1 relationship with a+b mod p^2.

    Comment by Oliver — Friday 16 April 2010 8:18 am #

  140. Ignore the last paragraph above. What remains is my argument. (a+b) can only be zero. I don’t want to get into Dr. Faustus and the devil, but in dealing with the devil in mathematics he has two very powerful weapons, infinity and zero. I managed to control infinity, but zero almost beat me. Thanks for being able to bounce ideas off you. Over and out.

    Comment by Oliver — Thursday 22 April 2010 4:51 pm #

  141. After a marriage and a major house refurbishment I’m back to finish off my problem.
    Having claimed victory over the devil, I find myself amusingly playing devil’s advocate to try and find a flaw in my paper. The binomial theorem works for negative integers. Does it work when one of the integers is 0?
    My problem is to analyse the sum of two pth powers of integers, p an odd prime. We will do it with residues of these integers modulo p^k. The integers :a:b: are prime to p, but their sum has a factor p, and possibly a facot with higher powers of p.
    One stratagem generally employed is to express
    the sum as ((a+b)-b)^p +b^p and expand out the first term with the binomial theorem. The b^p terms cancel out leaving a factored expression (a+b)() and you prove the () factor has a single factor p.
    Now I can prove what a and b must be for all the various combinations, but my factorisation is of the form (a+b)() where () does not have the factor p. The devil points this out, but I reply I can take my solution :a:b: mod p^k and exponentiate it using the stratagem and the result will be a residue expression mod p^k of (a+b)p(). Is this also valid if a+b=0 mod p^k, i.e. the stratagem forms (0)p() mod p^k? If not, I need to add another section.

    Comment by Oliver — Thursday 27 January 2011 1:50 pm #

  142. Forget that question Steve. It looks like the binomial theorem for (a+b)^p will handle any type of value ( that I can think of) for a and b, including (cosx+sinx). I am thefore taking it that I can create my factorisation when a+b add to zero.

    Comment by Oliver — Thursday 27 January 2011 5:15 pm #

  143. As I see it, the set of residues of integers mod p^k, k a positive integer, meet the wikipedia criteria to form a commutative ring under residue addition and multiplication mod p^k.
    I want to know if this set is a field.
    For a ring to be a field, Wikipedia defines the requirement that every non zero element be invertible. In this set, every element that is not a factor of zero is invertible, so it is slightly different from the wikipedia requirement.

    Comment by Oliver — Wednesday 13 April 2011 1:07 pm #

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