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#51 Re: Help Me ! » Proving! » 2010-11-30 04:53:03
OMG!! What is going on? ampohmeows proof for #1 is correct!
Is my proof for #1 correct?
Yes. You have proved that when a(−b) is added to ab, the result is 0. This shows that a(−b) is the additive inverse of ab (since additive inverses are unique). As the additive inverse of ab is −(ab) by definition, you have .
That is all.
How to show that if 2a = 0, then a = 0, i don't have here any theorem or axioms that will support the statement.
Oh yes, you do. From the order axioms of the integers, you should be able to get a theorem that the integers form an integral domain in other words, it has no nonzero divisors of 0. Hence, from 2a = 0, you can conclude that a = 0.
#52 Re: Euler Avenue » Permutations » 2010-11-29 02:13:44
Okay, here is another definition of even and odd permutations.
[align=center]
[/align]Lets look at the expression on the right-hand side. The denominator is a product of
factors, and so is the numerator. Let be a factor in the denominator. Either or . In the former case is a factor in the numberator, while in the latter case is a factor in the numerator. Conversely, if is a factor in the numerator, then (as ) either is a factor in the denominator (if ) or is a factor in the denominator (if ).So we see that if
is a factor in the denominator then either or is a factor in the numerator and conversely if is a factor in the numerator then either or is a factor in the denominator. It follows that is always either +1 or −1. We define to be even if is and odd if .#53 Re: Help Me ! » Proving! » 2010-11-29 01:40:11
You are making a big mess there. What you need to do is start with the expression
You want to show that it is equal to 0. Using the distributive law gives
Can you carry on from here? What is b + (−b)?
#54 Re: Exercises » Janes Exercises III » 2010-11-29 00:33:15
OMG! Its been such a while since I posted the problem that Ive forgotten the solution myself! 
Ive glanced through your solutions, and I dont think they look right. Give me some time to remember the solution. 
#55 Re: Help Me ! » Proving! » 2010-11-29 00:28:46
#56 Dark Discussions at Cafe Infinity » Haiku 2 » 2010-11-29 00:26:04
- JaneFairfax
- Replies: 4
Here are some haiku poems I recently wrote.
Christmas is coming
Again. Which can mean one thing:
Its winter again.I have had enough.
He is more changeable than
The British weather.What goes round comes round
In times eternal cycle.
Cant wait to come round.Time is a cycle,
Camusian absurdity,
Existential jail.How fast does time fly?
No faster, and no slower.
dt/dt = 1.
#57 Re: This is Cool » CauchySchwarzBunyakovsky inequality » 2010-11-26 00:46:50
...hi jane
.........but you assumed that "lambda" was real but by setting discriminant to less than zero ur making lambda value imaginary:)
#58 Re: This is Cool » Geometry problem » 2010-11-22 23:28:09
Hi, Ronnie and Bob.
#59 Re: Introductions » Hi from Finland! » 2010-11-22 23:18:25
Hyvää päivää, zonesey! Tervetuloa foorumiin! 
#60 Euler Avenue » Homeomorphism of Euclidean plane to sphere with point removed » 2010-10-29 00:39:29
- JaneFairfax
- Replies: 2
Let us take the unit sphere centred at
, given by , and let us remove the north pole . Call this pointless sphere . Then is the disjoint union of circles formed by the intersection of with the plane as t varies from 0 to 2 (including 0 but not 2).Now consider
itself. This is the disjoint union of origin-centred circles of all possible non-negative radii (counting as a circle of radius 0). Let be any continuous bijection (e.g. or ).Then if we define
by and for[align=center]
[/align]where , we have a homeomorphism!
I was thinking about this last night. Thinking about math problems is a great way to pass the time when youre having insomnia. 
#61 Re: Help Me ! » (Dis)Proof pi = 2 (!) » 2010-10-09 01:55:09
I believe I have a partial solution to the problem. Consider this. With one semicircle, the curve touches the straight line at two points (the end points). With two semicircles, it touches the straight line at three points (the end points and the midpoint). With four semicircles, it touches the line at five points. And so on
Now if you let the process tend to infinity, it is clear that the endpoints of the semicircular arcs will only touch the straight line at only countably many points of the straight line. It may be impossible to visualize what the infinite curve will be like as those countably many points will be densely spread over the straight line (like the rational numbers on the real line). However, the straight line itself contains uncountably many points! Hence, it is a false assumption that in the infinite limit, the curve will become the straight line itself. It wont it will remain longer than the straight line.
#62 Re: Puzzles and Games » Make 1000 using 8 8s » 2010-09-22 00:15:38
#63 Re: Help Me ! » Convergence of Sequence "Does {An^2} converges => {An} converges? » 2010-09-09 02:01:29
I kinda think it is false, but couldnt think of any counterexample to directly proof it.
#64 Re: Puzzles and Games » Four People Travel » 2010-09-09 01:51:04
Well, this particular puzzle definitely does not have a unique solution. More clues are needed.
#65 Re: Puzzles and Games » Four People Travel » 2010-09-08 06:59:12
or can anyone explain why there are not lots of solutions:
There are at most 4! = 24 solutions and 24 is not a lot, is it?
#66 Re: Introductions » New here.. » 2010-09-07 23:42:32
Welcome, soccorss12!
#67 Maths Is Fun - Suggestions and Comments » Site URL » 2010-09-04 23:05:45
- JaneFairfax
- Replies: 1
A moment ago, while trying to access the forums, I wrongly typed the URL as www.mathsisfun.com (with an extra s). I got a white page saying nothing to see here move on.
Wouldnt it be more convenient if the slightly wrong URL could be made to redirect to the correct one instead? It would save retyping.
#68 Re: Puzzles and Games » Complete the series.... » 2010-08-31 00:41:28
To use hide tags:
produces
#69 Re: Puzzles and Games » Complete the series.... » 2010-08-31 00:06:23
Is it dedelute ?
#71 Re: Puzzles and Games » Complete the series.... » 2010-08-28 23:20:21
(xlvii) Liverpool, Arsenal, Liverpool, Arsenal, ___
#74 Re: Jai Ganesh's Puzzles » 10 second questions » 2010-08-25 00:50:29
Welcome back JaneFairFax!
The thought of once again being able to interact with the legendary JFF has me ecstatic as well!
Thanks guys! Its nice to feel welcome. 
#2348. Without using long division, short division, calculator, computer program, Microsoft Excel or any other form of software or technology, calculate 35999999/5999 mentally.
#75 Re: Help Me ! » group theory (simple groups) » 2010-08-20 07:08:38
Sorry, my mistake. 
The Sylow p-subgroup would be unique if p > qr (which was what I was probably thinking of), otherwise it need not be. The other Sylow subgroups also need not be unique.