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Did anyone miss me?
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She Turned Me Into A Newt
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PostPosted: Fri Jul 02, 2010 6:18 am    Post subject: Did anyone miss me? Reply with quote

I offer a hearty Ni! to you Loonies!
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PostPosted: Fri Jul 02, 2010 8:58 am    Post subject: Reply with quote

...and to you, you who spells LooNies correctly!
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PostPosted: Fri Jul 02, 2010 7:06 pm    Post subject: Reply with quote

AHA! Yet ANOTHER East Tennesseean that can't SPELL!
LOONEYS, my lad... T'is OK tho, considering your age and such.
Especially the In-Breeding! (Excuuuse me, 'Pure-Breeding') #ni-2 #ni-1 #ni-2
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PostPosted: Sat Jul 03, 2010 9:09 am    Post subject: Reply with quote

nah. not really but... wb anyway Wink
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PostPosted: Sun Jul 04, 2010 10:06 am    Post subject: Reply with quote

But the question how long are you going to stay around this time.

Ni
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PostPosted: Sun Jul 04, 2010 3:16 pm    Post subject: Reply with quote

Depends on how many PGGB's I consume...

#ni-1
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PostPosted: Sun Jul 04, 2010 4:30 pm    Post subject: Reply with quote

Using M theory I had calculated your return to fall on November 7th, the fact that you did is yet another proof of the 11 dimensional Euclidian Reality.

Ni! Ni! Ni!

#ni-1
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PostPosted: Sun Jul 04, 2010 6:09 pm    Post subject: Reply with quote

Well since everyday is November 7th that is not a hard prediction.

Ni
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PostPosted: Sun Jul 04, 2010 8:10 pm    Post subject: Reply with quote

KD0711 wrote:
Using M theory I had calculated your return to fall on November 7th, the fact that you did is yet another proof of the 11 dimensional Euclidian Reality.

Ni! Ni! Ni!

#ni-1


Did you consider that due to the Bott's periodicity, we must study at first the generic D=3 Euclidean supersymmetry case? The role of complex and quaternionic structures for D=3 and D=11 Euclidean supersymmetry is thereby elucidated.
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PostPosted: Mon Jul 05, 2010 5:29 am    Post subject: Reply with quote

PhastPhred wrote:
KD0711 wrote:
Using M theory I had calculated your return to fall on November 7th, the fact that you did is yet another proof of the 11 dimensional Euclidian Reality.

Ni! Ni! Ni!

#ni-1


Did you consider that due to the Bott's periodicity, we must study at first the generic D=3 Euclidean supersymmetry case? The role of complex and quaternionic structures for D=3 and D=11 Euclidean supersymmetry is thereby elucidated.


Your Princely Highness,

I did indeed consider and study Bott's periodicity in connection with the untimely arrival of Sir Newt.

As we all know Bott's periodicity deal mainly with K-theory, and in as much lay merely a foundation upon which the Euclidean relativistic aspects can be computed.

M-theory as we all are conversant with, is an extension of string theory in which 11 dimensions are identified.

Baed on the necessary consolidation of five separate string theories to produce an 11th unifying dimension and a consistent string theory.

Because the dimensionality exceeds the dimensionality of five superstring theories in 10 dimensions, it is known that the 11-dimensional theory unites all string theories (and supersedes them).

Though a full description of the theory is not yet known, the low-entropy dynamics are known to be supergravity interacting with 2- and 5-dimensional membranes.

5 and 2 being noteworthy, as the sum 7 entail that the M-theory is in fact a 711 dogmatism.

Clarifyingly yours.

Ni! Ni! Ni!


#ni-1
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PostPosted: Mon Jul 05, 2010 5:40 am    Post subject: Reply with quote

Well, YEAH, but is that any reason to get DOGMATIC about it all?

Laughing

(BOY this is taking me a long time to get my 1M!)
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PostPosted: Tue Jul 06, 2010 7:20 am    Post subject: Reply with quote

Newt!

Let's have a fine PGGB Welcome back party at Camelot! I'll bring in the PGGB's from the Royal PGGB Brewery, may you bring in the lutefisk or bocalhau?

A hearty cheers!

HoC
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PostPosted: Tue Jul 06, 2010 10:07 am    Post subject: Reply with quote

I shall bring them both!!! Who is bringing the dancing Alpacas?

Embarassed I mean girls!

As for this November 7th non-sense... Any fool can tell you that it has been mis-calculated, and should be October 6th.
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PostPosted: Tue Jul 06, 2010 12:22 pm    Post subject: Reply with quote

She Turned Me Into A Newt wrote:
I shall bring them both!!! Who is bringing the dancing Alpacas?

Embarassed I mean girls!

As for this November 7th non-sense... Any fool can tell you that it has been mis-calculated, and should be October 6th.


**cough**

Proof of validity

The validity of the Euclidean relativity can be proven by a two-step argument.

In the first step, the final nonzero remainder rN−1 is shown to divide both a and b. Since it is a common divisor, it must be less than or equal to the greatest common divisor g. In the second step, it is shown that any common divisor of a and b, including g, must divide rN−1; therefore, g must be less than or equal to rN−1.

These two conclusions are inconsistent unless rN−1 = g.

To demonstrate that rN−1 divides both a and b (the first step), rN−1 divides its predecessor rN−2

rN−2 = qN rN−1

Since the final remainder rN is zero. rN−1 also divides its next predecessor rN−3

rN−3 = qN−1 rN−2 + rN−1

Because it divides both terms on the right-hand side of the equation. Iterating the same argument, rN−1 divides all the preceding remainders, including a and b.

None of the preceding remainders rN−2, rN−3, etc. divide a and b, since they leave a remainder. Since rN−1 is a common divisor of a and b, rN−1 ≤ g.

In the second step, any natural number c that divides both a and b (in other words, any common divisor of a and b) divides the remainders rk.

By definition, a and b can be written as multiples of c: a = mc and b = nc, where m and n are natural numbers.

Therefore, c divides the initial remainder r0, since r0 = a − q0b = mc − q0nc = (m − q0n)c.

An analogous argument shows that c also divides the subsequent remainders r1, r2, etc.

Therefore, the greatest common divisor g must divide rN−1, which implies that g ≤ rN−1.

Since the first part of the argument showed the reverse (rN−1 ≤ g), it follows that g = rN−1. Thus, g is the greatest common divisor of all the succeeding pairs:

g = GCD(a, b) = GCD(b, r0) = GCD(r0, r1) = … = GCD(rN−2, rN−1) = rN−1

I appologize for my brief explanation, however I am preparing for the PGGB shindig at Camelot.

Ni! Ni! Ni!

#ni-1
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PostPosted: Tue Jul 06, 2010 4:39 pm    Post subject: Reply with quote

After reading the post above HoC sit back with a tremedous headache trying to fight the beast with PGGB's ....... without luck so far .......

Shocked Laughing Wink

HoC
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PostPosted: Tue Jul 06, 2010 4:53 pm    Post subject: Reply with quote

KD0711^711 converges to DSM with probability 1.0 Very Happy

ni! i!u
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PostPosted: Tue Jul 06, 2010 6:35 pm    Post subject: Reply with quote

Newt and HoC, you two are a sight for old eyes.

@Newt, Raquel, I think that's what she said her name was, was here looking for you and she said to tell you the kids are grown and its safe to come home.

@HoC, I bet the kiddies are as big as you, how you been and send me some jpeg of the children.

Good to see you both.
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PostPosted: Wed Jul 07, 2010 12:32 am    Post subject: Reply with quote

You forgot the Penguin Army Quiotent.

x^2-1+f-4=dlfdsjgmasfksdkfg

Ni
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PostPosted: Wed Jul 07, 2010 7:59 am    Post subject: Reply with quote

Perhaps you were not aware that mathematics reawakened in Western Europe in the 13th century. At that time works in mathematics were translated from the Arabic into Latin allowing Western European scholars to learn about the medieval Arabic-language mathematics and the older Greek mathematics, such as Euclid's Elements. In all this mathematics, only positive numbers were considered to be numbers. Negative numbers were not yet accepted as entities. (Some ancient cultures, including that of China and India, accepted negative numbers, but not the ones mentioned above.)
Solution of quadratics. With negative numbers we understand that every quadratic equation in the variable x can be written in the form


ax2 + bx + c = 0,
where a, b, and c are constants. We also know that the general solution is given by the quadratic formula:


x = –b ±√(b2 – 4ac)
--------------------------------------------------------------------------------
2a

where there are two distinct real solutions if the discriminant b2 – 4ac is positive, one double real solution if the discriminant is 0, and no real solutions if the discriminant is negative.

Back in the 15th century, this was not understood. Instead, quadratic equations were classified into four different kinds depending on the signs of the coefficients a, b, and c. Since the leading coefficient a is not zero in a quadratic equation, you can always divide by it to get an equivalent quadratic equation where a equals 1, that is, x2 + bx + c = 0. This one form gives rise to four forms when you move the negative terms to the other side of the equation and when you drop zero terms from the equation:

x2 = c
x2 + bx = c
x2 + c = bx
x2 = bx + c

There are other forms, but either they have no solutions among the positive numbers or else they can be reduced to linear equations. Each of these forms required a different form of a solution. With hindsight, we see that the 15th century solutions are just special cases of the quadratic formula. One would think that the consolidation of four cases into one might be enough justification for accepting negative numbers, but apparently it wasn't. It seems to take a lot of time before people will extend their concept of number to include new entities.
Solution of cubics. Equations of the third degree are called cubic equations. The general form of a cubic is, after dividing by the leading coefficient,


x3 + bx2 + cx + d = 0,
As with the quadratic equation, there are several forms for the cubic when negative terms are moved to the other side of the equation and zero terms dropped.

Back in the 16th century it was a big deal to solve cubic equations. There was a great controversy in Italy between Cardano (1501-1576) and Tartaglia (1499-1557) about who should get credit for solving the cubic equation. Any book on the history of mathematics will go into the details of this fascinating controversy. What's interesting to us, though, is that negative numbers were becoming legitimatized, a deeper insight into equations was developed, and the first inkling of a complex number appeared. Incidentally, at this time symbolic algebra had not been developed, so all the equations were written in words instead of symbols!

Cardano, in his Ars Magna, finds negative solutions to equations, and he called these numbers "fictitious". He also noted an important fact connecting solutions of a cubic equation to its coefficients, namely, the sum of the solutions is the negation of b, the coefficient of the x2 term. At one other point, he mentions that the problem of dividing 10 into two parts so that their product is 40 would have to be 5 + √(–15) and 5 – √(–15).

Cardano did not go further into what later became to be called complex numbers than that observation, but a few years later Bombelli (1526-1572) gave several examples involving these new beasts. Here's one example. One of Cardano's cubic formulas gives the solution to the equation x3 = cx + d as


x = 3√(d/2 + √e) + 3√(d/2 – √e)
where e = (d/2)2 – (c/3)3). Bombelli used this to solve the equation x3 = 15x + 4 to get the solution


x = 3√(2 + √–121) + 3√(2 – √–121)
Now, the square root of –121 is not a real number; it's neither positive, negative, nor zero. Bombelli continued to work with this expression until he found equations that lead him to the solution 4. He determined that


√(2 + √–121) = 2 + √–1
√(2 – √–121) = 2 – √–1

and, therefore, the solution x = 4. This example is not given to show that Bombelli knew everything there is to know about complex numbers, rather to indicate that he was starting to understand them.

October 6th!
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PostPosted: Wed Jul 07, 2010 9:41 am    Post subject: Reply with quote

Waaaaaaaaah run away, run away ............

I'll better stay in the Royal Brewery as my math brain must have been wiped out from the PGGB vapours ......

Cheers

HoC
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