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This is extracted from the article Self-fertilization of the book What if? of xkcd:

enter image description here

At first, I just simply think that the multiplier he mentions is just a convenient name of a system with arbitrary rules. However, after he says about the multiplicative identity, I think that this has a deeper root in math, and here stems the question.

Why do you have the multiplicative identity, when the product here is made from two multipliers? The multiplicative identity I know is the element $e$ satisfies $e\cdot x=x$. But from what I understand from the book, it is the $x\cdot y=e$, and it's not required to have $y=x^{-1}$. Further more, $e=1\in\mathbb{R}$, but $x$ and $y$ are not. He explicitly says that the stat is number.

So can this happens? Is is actually be true, or am I misread him?

This is a table of how the rule works (the X letter in SEX row is irrelevant, not a multiplier):

enter image description here

It is unclear why $1$, not $0$, is the identity, but yes, the product of both multiplier should be a number so that it can be multiplied with other numbers later on.

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  • $\begingroup$ It sounds like an attempt to set the rules (the evolutionary environment) so that the "two multipliers" state is penalized quite heavily without being "killed" (which a return of zero would probably do), based on some further combination of stats from multiple characteristics. I guess $1$ is the lowest value in the system, right? $\endgroup$ – Joffan Feb 24 '17 at 16:59
  • $\begingroup$ Without having the rest of the article available to understand this scheme, i cannot say why the stat would be $1$. But from what is here, it sounds like the "stat", "number", and "multiplier" are all intended to be real numbers. $\endgroup$ – Paul Sinclair Feb 24 '17 at 23:40
  • $\begingroup$ @Joffan it's unclear why it should be 1, not 0. But yes, it is the lowest value in the system. $\endgroup$ – Ooker Feb 25 '17 at 12:34
  • $\begingroup$ @PaulSinclair good point. But even when "multiplier" is real number, then in the real number group with any operation, $y$ must equal to $x^{-1}$ in other to have a product of $1$. I have added an example of how the rule works in the post $\endgroup$ – Ooker Feb 25 '17 at 12:40
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From the sample, it appears the scheme is that the value for each gamete can be a stat, or a multiplier. However both stats and multipliers are just real numbers.

My first thought is that when the gametes are combined, two stats combine to whichever stat is higher. A stat and a multiplier combine to the product of the two, and two multipliers combine to 1. The problem with this idea is that it doesn't explain why two Dexterities of 14 combine to 14, while two Intelligences of 14 combine to 15. Either there is a typo, or there is some addtional rule that treats Dexterity and Intelligence differently.

As for why two multipliers combine to 1 instead of 0. He evidently chose to do that because multipliers act by multiplying. But if there is nothing to multiply (because both entries are multipliers, not stats), then it defaults to the identity. Since these are about multiplying, it is the multiplicative identity that they default to.

In other words, it is a somewhat arbitrary choice he made to handle a weakness in his scheme. Since it is his scheme, he can make whatever choice he wants.

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  • $\begingroup$ Yes, I also think that it's just an arbitrary choice of him. But then, he doesn't need to tell that 1 is a multiplicative identity. It seems that the multiplicative identity here is not a mathematical identity in group theory $\endgroup$ – Ooker Feb 25 '17 at 14:43
  • $\begingroup$ @Ooker Actually, when the operation is taking the highest number and $1$ is the smallest possible value, then $1$ precisely becomes the identity element. $\endgroup$ – Tobias Kildetoft Feb 25 '17 at 14:44
  • $\begingroup$ The description you give is accurate (as it states further up in the picture). It looks like the two 14's combining to a 15 is an error. $\endgroup$ – Tobias Kildetoft Feb 25 '17 at 14:44
  • $\begingroup$ @TobiasKildetoft are you mentioning about the cyclic group? $\endgroup$ – Ooker Feb 25 '17 at 14:47
  • $\begingroup$ @Ooker What cyclic group? There is no mention of groups anywhere in this, especially not cyclic ones. $\endgroup$ – Tobias Kildetoft Feb 25 '17 at 14:49

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