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#101 Re: This is Cool » 2424[b]2[/b]4242 in [math]\pi[/math] » 2007-03-07 04:39:25
if I recall correctly, there is a string of six 9s in the first thousand digits
#102 Re: Help Me ! » hey! » 2007-03-02 13:36:07
Is that some kind of maths club dance?
#103 Re: This is Cool » INFINITE 0.9 <> 1 PROOF : COMPLETE : By,Anthony.R.Brown,15/02/07 » 2007-03-02 13:35:05
I believe it is commonly accepted that 1/infinity = 0, due to that limit.
No, it isn't, it's indeterminate.
#104 Re: This is Cool » Permanent Moon Base » 2007-03-02 06:49:30
This is ancient news
#105 Re: This is Cool » Ramanujan's "Mock Theta Functions" solved » 2007-03-01 07:55:36
Quote:
" and it makes no sense to my mortal brain. "
Truth is Truth is Wonder full!...............................................................................................
You thougth I wasn't mortal?
#106 Re: Dark Discussions at Cafe Infinity » Optical Illusions, et al » 2007-03-01 03:19:03
For some reason 'of' and 'the' are hard to count because people normally skim over them when reading normally.
Actually that isn't the correct reason. The brain does process every word, but the brain does not register the word 'of' as containing an f, because it's pronounced "ov".
If you read it backwards, you'll catch every single one, so it's not a case of "skimming".
#107 Re: Help Me ! » help needed » 2007-02-28 12:17:15
Whoops, yeah Z19 has no solutions.
#108 Re: Dark Discussions at Cafe Infinity » Bullying-reasons or whims? » 2007-02-28 08:29:54
When you're older things get worse, like idiots trashing your back garden and breaking into your house.
Yes, I had a bad weekend
#109 Re: Help Me ! » help needed » 2007-02-28 07:55:53
For Z17 : 6^2 = 2
For Z19: 18^2 = 2
That was by inspection, but I'm not sure how to do it otherwise.
Edit: Z17 has a second solution: 11^2 = 2
#110 Re: This is Cool » Ramanujan's "Mock Theta Functions" solved » 2007-02-27 13:30:19
Haha, the topic from the thread list reads
Ramanujan's "Mock Theta Functions" solved by MathsIsFun
#111 Re: Help Me ! » Algebra Help Needed Please » 2007-02-27 10:09:48
The question I have is this:
In Z13, calculate 2/3+3/5.
*Z being the symbol for integer and 13 is the subscript to it
Thanks in advance!!!
Uh... 2/3 and 3/5 definately do not belong to the integers
#112 Re: This is Cool » Ramanujan's "Mock Theta Functions" solved » 2007-02-27 10:08:21
Man, this dude really was on a completely different plane of existance to everybody else. I'm reading the wikipedia article, and it makes no sense to my mortal brain.
Nice find
#113 Re: Dark Discussions at Cafe Infinity » What is your favorite kind of Math? » 2007-02-27 08:23:31
topology, galois theory, anything that lets me play around with abstract algebra.
#114 Re: Help Me ! » sigma » 2007-02-27 08:21:50
maybe you can split that sum into two sums: 'i' odd and 'i' even. Then you just have to sum two expressions not involving (-1)^i-1
Like sigma i=1 to 99 of some function of 2i plus i from 1 to 100 of some function of 2i+1 (possibly reversed I can't care enough to think about it)
Nah, delta's way is far superior.
But change i-1 for i+1....for the sake of sake's sake.
#115 Re: Dark Discussions at Cafe Infinity » The World of Warcraft... » 2007-02-27 08:17:21
[Edit: Deleted]
#116 Re: Formulas » Set Theory » 2007-02-27 06:54:07
You have still yet to define union in terms of a binary operation. Please, answer the question. You state that union is a binary operation, all I ask is that you define it as such. I'm not looking for examples, I'm look for you to define union as a mapping from a set AxA to A.
Are you serious? You want me to list every possible tuple in existance? You're clinically insane, I'd like to see you do that for an infinite set.
Here it is for my example above
Here we go
There you go, have fun writing it out for every possible set imaginable. My previous posts are the method in which it is defined, not by the mapping. Are you telling me scalar addition is defined by writing out every single possible tuple and a permutation? Of course not, it's defined by a couple of predicate rules.
#117 Re: Formulas » Set Theory » 2007-02-23 13:15:40
or better, why not just consider the power set of the universal set under both operators, and I believe that forms a boolean lattice.
Well, any set under both operators will be a boolean lattice, I believe, unless you can think of an axiom that fails.
#118 Re: Formulas » Set Theory » 2007-02-23 13:06:49
hence A and B are both elements of P|{1,2,3,4,5,6}| and no generality is lost, Z x Z -> Z necessarily for some Z = P|{1,2,3,4,5,6}|
Operators are associated with algebras to form more composite algebras, if I take a set, I can find an operator to form an algebra. Likewise if I find an operator and element I wish to perform, I can find a set containing them such that the operator will hold an algebra.
#119 Re: Formulas » Set Theory » 2007-02-23 12:23:28
Given any set, both intersection and union map P|A| x P|A| -> P|A|
Union an intersection each take two different sets. Not the same set, though they can be.
So you're saying any binary operator can only ever take the same element for it's operands? 2 + 3 = 5 last time I heard.
The mapping maps all elements of the set cart set to another element of the set, and they will forever be included in the power set of any given set, thusly.
Let A = {1,2,3}
P|A| = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
Now take the union and intersection of all possible cartesian tuples of the power set, and the yield will ALWAYS be an element of the power set, hence P|A| x P|A| -> P|A|
#120 Re: Dark Discussions at Cafe Infinity » Future job; what do you want to do? » 2007-02-23 01:10:32
[Edit: Deleted]
#121 Re: Introductions » Deleted » 2007-02-22 19:59:58
Sekky wrote:...a ring whose set was the set of all functions, and whereby function composition was the additive operation...
In a ring (R, +, ·), (R, +) is an Abelian group. However, function composition is in general not commutative. (Even if (R, +, ·) were a semiring, (R, +) would be a commutative monoid; we cannot escape the commutativity.)
Oh yeah, (R,+) is Abelian....my mistake
So we try it in reverse, make function composition the multiplicative operation and find something over which is distributes, which is never going to happen.
#122 Re: Help Me ! » making t the subject » 2007-02-22 11:49:15
Hi guys!!!
how do I rearrange the following equations to make t the subject???????
a) y = t - 4
b) x = 2t - 4 + 8
c) y = ( -3t + 2/3) to the power of 2
Basically that you're looking to do, is have one side of the equation such that t is on it's own, then you have a surefire way of knowing what t equals
a) simply add 4 to both sides:
b) you can compose the 8 and -4, there's no need to have them separate, so simply subtract 4 from both sides and then divide boths sides by 2
c) I'll rewrite this in latex
take the square root of both sides
subtract two thirds from both sides, and then divide by negative three
it all just consists of elementary arithmetic, perform operations in the order necessary to make t singular. I'm probably gonna get flamed for doing somebody's homework for them, so as a contingency plan, you're going to solve this one:
#123 Re: Introductions » Deleted » 2007-02-22 11:12:12
I'd like to note that "Problem 1" on your website, although very slightly different than my statement, was introduced to Delta by me, which in turn was presented to me by a colleague. I just thought I'd tell you this since you state that you "feel obliged to ask" before posting problems.
Here is the thread:
http://www.mathsisfun.com/forum/viewtopic.php?id=5833
Also, if you haven't solved it yet, you probably never will unless in a hallucinogenic state or if intense insight strikes. I only know one other person who has solved it.
Well the comment I wrote underneath the problem sums up my theory, the function can't possibly be functional, although we could limit it.
Myself and a friend came up with the idea of having a ring whose set was the set of all functions, and whereby function composition was the additive operation (with an automorphic identity, obviously), such that IF we could prove that the group under addition contained an idempotent element (or that we could form a cyclic subgroup) we could find an element such that g is equal to its own inverse, so the problem is finding a monoid operation that distributes over function composition, which isn't easy.
I'd like to see a proof.
#124 Maths Teaching Resources » First Order Linear ODE Primer » 2007-02-22 08:17:49
- Sekky
- Replies: 1
http://www.stylebucket.co.uk/primers/Fi … Linear.pdf
To back up my Bernoulli Primer I wrote yesterday, I wrote another short primer on solving first order linear ODEs.
I had no idea I was going to write this until about an hour or so ago, so I'm still looking for suggestions, or tomorrow I may write nothing! 
Anyway, criticism and suggestions/changes are welcome and noted.
Dan
#125 Re: Formulas » Set Theory » 2007-02-22 05:29:33
Sekky, this is a big world with many, many, mathematicians. Is it really that much of a surprise that two different mathematicians don't use exactly the same symbols?
Symbols are arbitrary. They can mean whatever you want them to. There are standards, but different people may conform to different standards.
Actually this list should really be in the algebra formulas thread, since a set is just a degenerate algebra.
Can you please define "degenerate algebra"? Wikipedia, MathWorld, and myself have no idea what you mean when you say that. Also, if you're going to go with that notion, then every thread in the formula section should be put into Sets, as sets are pretty much the basis of all math.
He's using the same symbol to denote both the relation and the opposite relation, it makes no sense. The subset relation means the same back to front providing the relatives are switched, as does the less than symbol, as do any mathematical relations.