DavidQD

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UsernameDavidQD
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 #11
avatar+330 
+5

Sorry Melody, I overlooked your question until now.

 

Yes, I remember the original problem. I remember too thinking this was an odd question to become a snark. After some research I found this turned into a snark question partly because of a computer program malfunction. It also became a fascinating research project on crowed behavior.

 

The original question was

“How much tax would be on 113.98 at 9%”

This is the original URL

http://web2.0calc.com/questions/math_999

This does not display the question or the answer.

 

To find the original post search for “tax” then go to page 5. Here, you can note many “Re: math” posts all appended with “999” and clicking on them will take you to either a “404 not found” page or the “1x1 anonymous” in this case all math_999 links lead to the same (but not the original) snark post. Rather funny!

 

The PhPBB (bulletin board) system was in use at that time. This system requires a subject to post and it is used as one of two index links for the PBB system . If the subject is identical to a previous subject the system software appends a sequential number to the end.

 

“Math” was the most common subject title, and by the third week of November 2013, the number reached “Math_999”. Because “999” (or a series of 9’s) is historically used as an “end of file” or end of record” marker in many programs and data bases, exceptions may occur when it shows up in a legitimate but unanticipated place. This was one of those cases. Because of the software malfunction, the index stayed at 999 for 206 additional posts, when Herr Massow manually corrected the problem around the third week of December.

 

The posters who used the subject “math” saw the tax percentage questions and answers along with additional posts appended to their question, because of the indexing error. This may have appeared absurd and humorous to many readers and it may have prompted them to participate. The snark posts decreased quickly after the malfunction was corrected.

 

~DQD~

Jan 30, 2015
 #3
avatar+330 
+11

Greetings Melody,

 

All your accolades are marvelous . . . and thank you.

 

In the past year, this forum has become an amazing, structured, learning environment, surviving troll invasions and the ubiquitous snark questions.

http://web2.0calc.com/questions/need-help-nowwwwww

(I note the still present and perpetual 2+2 now lately replaced by 10+9)

 

The most fascinating element on this forum is the social structure. I‘ve never seen this on any other math forum. Another fascinating component is the multiple answers to the same question. The different techniques truly bring out the art that is in maths. This is like listening to skilled musicians play the same composition – where at first it seems the same, but in reality it is very different.

 

Time constrains do prevent me from posting often, but not a week will pass before I peruse this forum and study the posts. These always improve my skill set for mathematics and continuously expand my understanding of physics and related sciences.

 

I’m sure, I speak for many, if not all, when I say “thank you” for your time and energy you put into this forum to make it truly great!

 

Sincerely, 

    David Q. D.

Jan 2, 2015
 #4
avatar+330 
+8

Unlike a standard decade that has a constant value the cosmological decade does not. The cosmological decade scale is base 10 logarithmic. Each successive cosmological decade represents an increase in the total age of the universe by 10 times.

 

The two most common forms of expressing a cosmological decade are:

in Log (seconds per decade) or Log (years per decade)

 

Time zero is the instant before the big bang and the first time unit is Planck time, defined as the epoch

 

 

$$\ CD (-43.2683) = 10^{-43.2683} \; \text{(Time in seconds).}$$

 

There are an infinite number of cosmological decades between the Big Bang and the Planck epoch or any other point in time. The reason for this is there is not a Log value for zero – This is undefined.

 

To convert to years per decade, simply divide the seconds by seconds per year.

The current epoch is CÐ (17.6355) seconds. The log of seconds in a year (7.4991116)

CÐ (17.6355 - 7.4991116) , or CÐ (10.1364) years.

 

The fractional portion of the current CÐ equal to a year is:

 

$$\ \frac {1} {10^{10.1364}}\; = \; 7.30466E-11$$

 

The fractional portion of the current CÐ equal to 2015 years is

 

$$\ \frac {7.30466E-11}{2015} = 1.47188899E-7$$

 

 

To place these values in a more comprehensible perspective if the universe is 24 hours old then 150 years is a millisecond. The normal length of an eye blink is 300-400 milliseconds. it will take you about 52,500 years to blink, on this time scale.

 

Does this answer your question Адриiан Арсовки?

 

 

~D~

Dec 31, 2014
 #12
avatar+330 
0

Thanks Melody.

 

This was an attempt to demonstrate that because an integer squared (x^2) (where all such numbers have an odd number of positive divisors) is equal to another number squared (y^2), plus an additional factor (5) that the value of (x) and/or (y) cannot be integers. This means it is equal to a number that has an even number of divisors, or it does not have an even number of divisors to begin with. Because of this contradiction, the number cannot be an integer. Extending on this was an attempt to explain why an irrational number had characteristics of both even and odd numbers.

 

The explanation I presented is incomplete. This has been rolling around in my head for years, and it seemed functional, until an examination in a proper light. On closer inspection, it seems it is an incomplete amalgamation of proofs by contradiction and a hint of a proof by infinite decent thrown in for good measure (Fermat would be appalled). Mostly, it is a succession of incongruous comparisons that are not made compatible mathematically. Very lame. This is worse than mixing up metaphors. So, thank you Melody, I now see the darkness at the end of the tunnel.

 

 Your proof is to the point clear. CPhill’s, detailed and elegant, proof clearly demonstrates the paradox of (even and odd) parity.

 

Somewhere, there is a direct proof (instead of proofs by contradiction) of irrational numbers. Probably the best place to look is in Andrew John Wiles’ proof of Fermat’s Last Theorem. These maths are far outside of my skill set, and always will be. Even so, after dozens of readings, I have gleaned insights from them.

 

To find a direct proof for irrational numbers could lead to maths that close the gaps on Cantor’s Alephs of infinites. Considering there are more irrational numbers than integers, the value of this proof may be enormous. Adding to this, Wiles’ proof includes proofs of the Modularity Theorem for semistable elliptic curves, with these kinds of proofs, all three of Georg Cantor’s infinites may converge. Wiles’ proof seems like an excellent place to start.

 

Noting the post on Fermat, there is one question that remains about Fermat’s Last Theorem: Did Fermat really have the remarkable proof he wrote of in the margin of his book? Considering it took advanced, late twentieth century mathematics to solve it, I seriously doubt it. However, it is an intriguing thought that seventeenth century math may hold a solution.

 

This forum is great. The mathematicians who visit and especially the ones who live here are truly extraordinary.

 

Thank you Melody and CPhill!

 

 

~~D~~

Sep 1, 2014
 #2
avatar+330 
+10

(Q-11)

This is a potential energy calculation. The energy is the same whether this is occurs in 4 seconds or 4 minutes. (Note: there is a difference in power because of this).

Here a mass is moved 0.8 meters vertically against a force of 9.81m/s2. Potential Energy = Mass x (Force of Gravity per Kg) x vertical distance 20Kg * (9.81m/s2) * 0.8M

$$E = m \times g \times h = 20 \textrm{ kg} \times 9.81 \frac{\textrm{N}}{\textrm{kg}} \times h$$

$$E = 20 \textrm{ kg} \times 9.81 \frac{\textrm{N}}{\textrm{kg}} \times 0.8m \ =\ 156.96J \\$$

---------------

(Q-21)

The answer is A.

------------

 

(Q-22) Which shown arrows represents correctly the electric field at point p.

The answer is A. Electric fields are conventionally shown as moving to a negative charge.

 

--------------

(Q-30)

Magnetic field = Permeability * Turn density * Current

Current = (Magnetic field) / (Permeability * Turn density)

The relative permeability here is assumed to be (1). (This is used as a multiplier of absolute permeability).  

Turn density = The number of “loops” of wire uniformly distributed over a length of 1 meter.

110turns/0.5Meters = 220turns/meter

Magnetic field in milliTesla’s : 2.5mT = 0.002500 T

The magnetic field inside a solenoid is proportional to the applied current and the number of turns per unit length. The field inside is considered constant and independent of the diameter of the solenoid.

The magnetic field only depends on the current (I = amps), the number of turns per unit length (N/L), and the permeability of the core (mu0). 

B= mu0(N/L) * I mu0 is  Permeability in a vacuum (or air)

I=B/(mu0(N/L)

 

$$I = \frac{B}{mu_0 \frac {N}{L}}$$

 

$$I = \frac{B}{220 *\mu_0} \ = \ 9.04 A\\$$

 

Where

 

$$\\ \mu_0 = \ 4\pi \* 10^-7} \ H M^-^1$$

 


Yields Tesla measure of the ratio between enclosed magnetic field and the current.

 

 0.002500 T corresponds to 9.04 Amps. (Answer D).

 

~~D~~

Aug 22, 2014