# 410: Math Paper

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:Imaginary numbers on the complex plan are of the form '''a''' + '''b'''''i'' where '''a''' and '''b''' are constants and ''i'' is the square root of negative 1 (an impossibility in the plane of "regular" numbers). | :Imaginary numbers on the complex plan are of the form '''a''' + '''b'''''i'' where '''a''' and '''b''' are constants and ''i'' is the square root of negative 1 (an impossibility in the plane of "regular" numbers). | ||

− | :Joel Bradbury has a wonderful explanation on his site http://joelbradbury.net/notes/friendly_numbers | + | :Joel Bradbury has a wonderful explanation of Friendly Number on his site http://joelbradbury.net/notes/friendly_numbers. The following explanation of Friendly Numbers is taken from his site: |

:What are Friendly Numbers? | :What are Friendly Numbers? | ||

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:1,2,3,4 and 5 are solitary. 6 is friendly with 28. ( σ(6)/6 = (1+2+3+6)/6 = 12/6 = 2 = 56/28 = (1+2+4+7+14+28)/28 = σ(28)/28. | :1,2,3,4 and 5 are solitary. 6 is friendly with 28. ( σ(6)/6 = (1+2+3+6)/6 = 12/6 = 2 = 56/28 = (1+2+4+7+14+28)/28 = σ(28)/28. | ||

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==Transcript== | ==Transcript== |

## Revision as of 02:53, 12 March 2013

Math paper |

Title text: That's nothing. I once lost my genetics, rocketry, and stripping licenses in a single incident. |

## Explanation

- It's all basically just a set up to use the joke about Imaginary Friends by taking "friendly numbers" into the complex (imaginary) plane.

- Imaginary numbers on the complex plan are of the form
**a**+**b***i*where**a**and**b**are constants and*i*is the square root of negative 1 (an impossibility in the plane of "regular" numbers).

- Joel Bradbury has a wonderful explanation of Friendly Number on his site http://joelbradbury.net/notes/friendly_numbers. The following explanation of Friendly Numbers is taken from his site:

- What are Friendly Numbers?
- We need first to get define a divisor function over the integers, written σ(n) if you’re so inclined. To get it first we get all the integers that divide into n. So for 3, it’s 1 and 3. For 4, it’s 1, 2, and 4, and for 5 it’s only 1 and 5.

- Now sum them to get σ(n). So σ(3) = 1 + 3 = 4, or σ(4) = 1 + 2 + 4 = 6, and so on.

- For each of these n, there is something called a characteristic ratio. Now that’s just the divisors function over the integer itself ( σ(n)/n . So the characteristic ratio where n = 6 is σ(6)/6 = 12/6 =2.

- Once you have the characteristic ratio for any integer n, any other integers that share the same chacteristic are called friendly with each other. So to put it simply a friendly number is any integer that shares its characteristic ratio with at least one other integer. The converse of that is called a solitary number, where it doesn’t share it’s characteristic with anyone else.

- 1,2,3,4 and 5 are solitary. 6 is friendly with 28. ( σ(6)/6 = (1+2+3+6)/6 = 12/6 = 2 = 56/28 = (1+2+4+7+14+28)/28 = σ(28)/28.

## Transcript

- Lecturer: In my paper, I use an extension of the divisor function over the Gaussian integers to generalize the so-called "friendly numbers" into the complex plane. [Points to equations on the board]
- Guy in room: Hold on. Is this paper simply a build-up to an "imaginary friends" pun?

[Lecturer stands speechless]

- Lecturer: It MIGHT not be.
- Guy in room: I'm sorry, we're revoking your math license.

**add a comment!**⋅

**add a topic (use sparingly)!**⋅

**refresh comments!**

# Discussion

Shouldn't it say something about the whole math licence, and that you don't actually need a licence to do math? 108.162.231.228 21:01, 31 October 2013 (UTC)Synthetica

Despite what this comic implies, the divisor function is defined over the Gaussian integers. There still is a problem, though. If a divides b, then so does -a, along with ai and -ai. The divisors will inevitably sum to zero. You could get around this by ignoring all the numbers that aren't in a given quadrant. I personally like the idea of using ones where the real part is greater than the imaginary part (although that still does become a problem with multiples of 1+i). This way, a friend of a natural number will also be a natural number (though it's only the same as what you'd get normally if all the factors are three mod four). 199.27.128.167 (talk) *(please sign your comments with ~~~~)*

- I do not agree. Here is how it should work. You should define that a divides b if and only if there is a natural number n such that an = b. This way, natural numbers don't get new divisors when you move to the Gaussian plane. Consequently the extended sigma function gives the same value as the classical one when applied to a natural number. So, natural numbers will be friends according to the new definition if and only if they are friends according to the old definition and we are indeed allowed to say that the new definition extends the old one (Burghard von Karger). 162.158.93.28 (talk)
*(please sign your comments with ~~~~)*

Does someone care to hazard a guess about the alt text? Are there any possible papers that could have been written about those three fields? 173.245.53.186 15:04, 5 January 2014 (UTC)

I think no one's noticed the "Straight man's deadpan stare" in panels 3 and 4. After being rudely interrupted in panel two, the lecturer stares silently at the person who stopped him in panel 3. Note the shape of the lecturer's head: his forehead and chin are facing towards the right as revealed by the angled features at the top and bottom of his head. In panel 4 the lecturer turns and stares at the reader and stands silent. Note the completely round shape of the head. This is the classic deadpan stare. Also, am I the only one who assumed that Cueball himself is the lecturer and the person interrupting him is likely the proffessor of the class?ExternalMonolog (talk) 20:14, 21 January 2014 (UTC)ExternalMonolog

- I don't know. In panel three DOES look straight at the commenter in the audience in a deadpan way, like he'd been caught unawares by the person's observation, but in panel four seems to look down, head tilted towards the floor as if wondering "aww shit, now what do I do?" Or like he felt slightly ashamed of being caught. 108.162.216.76 22:56, 27 February 2014 (UTC)
- I thought it was pretty plainly evident that Cueball is the presenter - going along with the XKCD theme of banning/revoking Cueball's privilege due to "inappropriately" wasting people's time at an ostensibly serious event. Also, in relative agreement with 108.162.216.76, I think that Panel 3 is his feeble attempt to continue to look serious (or straight/deadpan) while thinking of a way to respond, followed in Panel 4 with a slightly defeated shoulder/head slump (giving up with trying to perpetuate his premise). Panel 5 could be viewed as a plain admission, but I prefer to see it as a final, hail-Mary attempt to maintain the graces of the audience without
*actually lying*to them. -- Brettpeirce (talk) 16:09, 5 March 2014 (UTC)- They are probably beat panels more than anything else. -Pennpenn 108.162.250.155 00:13, 27 March 2015 (UTC)

- I thought it was pretty plainly evident that Cueball is the presenter - going along with the XKCD theme of banning/revoking Cueball's privilege due to "inappropriately" wasting people's time at an ostensibly serious event. Also, in relative agreement with 108.162.216.76, I think that Panel 3 is his feeble attempt to continue to look serious (or straight/deadpan) while thinking of a way to respond, followed in Panel 4 with a slightly defeated shoulder/head slump (giving up with trying to perpetuate his premise). Panel 5 could be viewed as a plain admission, but I prefer to see it as a final, hail-Mary attempt to maintain the graces of the audience without

There's a joke among amateur songwriters that a forced rhyme or other dubious technique might lead to revoking the writer's "poetic license." Gmcgath (talk) 22:08, 30 October 2016 (UTC)