Ad4mWayn3

joined 2 years ago
[–] Ad4mWayn3@sh.itjust.works 2 points 3 weeks ago (1 children)

It gets deeper. It's also the same as the 0-k-vector, the 0-k-blade, the 0-multivector, the only number that is its own square besides 1, etc...

[–] Ad4mWayn3@sh.itjust.works 5 points 3 weeks ago

0! = 1 isn't an exception.

Factorial is one of the solutions of the recurrence relationship f(x+1) = x * f(x). If one states that f(1) = 1, then it only follows from the recurrence that f(0) = 1 too, and in fact f(x) is undefined for negative integers, as it is with any function that has the property.

It would be more of an exception to say f(0) != 1, since it explicitly denies the rule, and instead would need some special case so that its defined in 0.

[–] Ad4mWayn3@sh.itjust.works 1 points 3 weeks ago

Who did ever say that? Not a single article that I've read about Terryology has praised it. I guess the Joe Rogan podcast helped it gather some followers?

[–] Ad4mWayn3@sh.itjust.works 2 points 3 weeks ago

Multiplication order in current mathematics standards should happen the other way around when it’s in a non-commutative algebra.

The good thing about multiplication being commutative and associative is that you can think about it either way (e.g. 3x2 can be thought of as "add two three times). The "benefit" of carrying this idea to higher-order operations is that they become left-associative (meaning they can be evaluated from left to right), which is slightly more intuitive. For instance in lambda calculus, a sequence of church numerals n~1~ n~2~ ... n~K~ mean n~K~ ^ n~K-1~ ^ ... ^ n~1~ in traditional notation.

For example, we can’t write 2ω for the next transfinite ordinal because 2ω is just ω again on account of transfinite and backwards multiplication weirdness, and we have to write ω·2 or ω×2 instead like we’re back at primary school.

I'd say the deeper issue with ordinal arithmetic is that Knuth's up-arrow notation with its recursive definition becomes useless to define ordinals bigger than ε~0~, because something like ω^(ω^^ω) = ω^ε0^ = ε~0~. I don't understand the exact notion deeply yet, but I suspect there's some guilt in the fact that hyperoperations are fundamentally right-associative.

[–] Ad4mWayn3@sh.itjust.works 7 points 3 weeks ago (1 children)

They are still real numbers. Specifically uncomputable, normal numbers. Which means their rational expansion contain every natural number.

[–] Ad4mWayn3@sh.itjust.works 13 points 3 weeks ago

Zero is the absence of a quantity, but it is still a number.

 

For example: I don't believe in the axiom of choice nor in the continuum hypothesis.

Not stuff like "math is useless" or "people hate math because it's not well taught", those are opinions about math.

I'll start: exponentiation should be left-associative, which means a^b should mean b×b×...×b } a times.