this post was submitted on 18 May 2025
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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.

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[–] [email protected] 5 points 1 day ago

I believe that the polar plot of prime numbers that reveals spirals and rays and when extended out to millions of numbers shows deeper fractals and geometric patterns is a glimpse into the structure of something we haven't yet discovered.

Maybe it's the higher dimensional structure of a photon, maybe it's something we don't even know about, but the fact that math describes everything in our universe EXCEPT prime numbers sounds like nonsense. There's something staring us right in the face that we can't see yet.

[–] [email protected] 2 points 1 day ago (1 children)

1 should be a prime number.

[–] [email protected] 4 points 1 day ago (1 children)

It was once, but got kicked out due to new discoveries and equations I'm not smart/mathematically trained enough to understand.

[–] [email protected] 2 points 21 hours ago (1 children)
[–] [email protected] 2 points 21 hours ago

I prefer the term "got Ceresed", but yes. That's happened.

[–] [email protected] 20 points 2 days ago (3 children)

Mixed numbers fraction syntax [1] is the dumbest funking thing ever. Juxtaposition of a number in front of any expression implies multiplication! Addition? Fucking addition? What the fuck is wrong with you?

[1] https://en.m.wikipedia.org/wiki/Fraction#Mixed_numbers

[–] [email protected] 3 points 2 days ago

I have never made that connection before but I think you're 100% right!

[–] [email protected] 6 points 2 days ago

Amen. Pick a lane either they're both additive or multiplicative. Maybe a different symbol.

[–] [email protected] 2 points 2 days ago
[–] [email protected] 33 points 3 days ago

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

Interesting. Why?

[–] [email protected] 25 points 2 days ago

Everyone keeps talking about pi r²

This doesn't make any sense because pies are round. Brownies are square

[–] [email protected] 13 points 2 days ago (3 children)

The exceptions including the number 1. Like it not being a prime number, or being 1 the result of any number to the 0 power. Or 0! equals 1.

I know 1 is a very special number, and I know these things are demonstrable, but something always feels off to me with these rules that include 1.

[–] [email protected] 14 points 2 days ago* (last edited 2 days ago) (2 children)

X^0 and 0! aren't actually special cases though, you can reach them logically from things which are obvious.

For X^0: you can get from X^(n) to X^(n-1) by dividing by X. That works for all n, so we can say for example that 2³ is 2⁴/2, which is 16/2 which is 8. Similarly, 2¹/2 is 2⁰, but it's also obviously 1.

The argument for 0! is basically the same. 3! is 1x2x3, and to go to 2! you divide it by 3. You can go from 1! to 0! by dividing 1 by 1.

In both cases the only thing which is special about 1 is that any number divided by itself is 1, just like any number subtracted from itself is 0

[–] [email protected] 2 points 1 day ago

It's been a few years since my math lectures at university and I don't remember these two being explained so simple and straightforward (probably because I wasn't used to the syntax in math at the time) so thanks for that! This'll definitely stick in my brain for now

[–] [email protected] 1 points 2 days ago (2 children)

The numbers shouldn't change to make nice patterns, though, rather the patterns that don't fit the numbers don't fit them. Sure, the pattern with division of powers wouldn't be nice, but also 1 multiplied by itself 0 times is not 1, or at least, not only 1.

[–] [email protected] 8 points 2 days ago

1 multiplied by itself 0 times is not 1, or at least, not only 1.

Sure it is. 1 is the multiplicative identity, the number you start at when you multiply anything. 2^2 is really 1x2x2. 2^1 is 1x2. So 2^0 is... just 1.

[–] [email protected] 5 points 2 days ago (1 children)

We make mathematical definitions to do math. We can define 0! any way we want but we defined it to be equal to 1 because it fits in nicely with the way the factorial function works on other numbers.

Literally the only reason why mathematicians define stuff is because it’s easier to work with definitions than to do everything from elementary tools. What the elementary tools are is also subjective. Mathematics isn’t some objective truth, it’s just human made structures that we can expand and better understand through applying logic in the form of proofs. Sometimes we can even apply them to real world situations!

[–] [email protected] 5 points 2 days ago

Honestly I think it's misleading to describe it as being "defined" as 1, precisely because it makes it sounds like someone was trying to squeeze the definition into a convenient shape.

I say, rather, that it naturally turns out to be that way because of the nature of the sequence. You can't really choose anything else

[–] [email protected] 6 points 2 days ago (1 children)

Or 0! equals 1.

x factorial is the number of ways you can arrange x different things. There's only one way to arrange zero things.

[–] [email protected] 2 points 2 days ago (1 children)

I could still debate the proposition that zero things can be arranged in any way.

[–] [email protected] 3 points 2 days ago (1 children)

That sounds like a philosophical position, not a mathematical one.

[–] [email protected] 2 points 2 days ago

You are right but that is a dangerous proposal because math is just applied philosophy :)

[–] [email protected] 3 points 2 days 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.

[–] [email protected] 8 points 2 days ago (1 children)

“Terryology may have some merits and deserves consideration.”

I don’t hold this opinion, but I can guarantee you it’s unpopular.

[–] [email protected] 1 points 2 days 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?

[–] [email protected] 7 points 2 days ago

I don’t think ‘I don’t believe in the axiom of choice’ is an opinion, it’s kind of a weird statement to make because the axiom exists. You can have an opinion on whether mathematicians should use it given the fact that it’s an unprovable statement, but that’s true for all axioms.

Any math that needs the axiom of choice has no real life application so I do think it’s kind of silly that so much research is done on math that uses it. At that point mathematics basically becomes art but it’s art that’s only understood by some mathematicians so its value is debatable in my opinion. <- I suppose that opinion is controversial among mathematicians.

[–] [email protected] 8 points 2 days ago

Some of the more complex proofs might be wrong just because so few understand them, and the ones who do might have made mistakes.

Hell, I’ll trust a math result much more if it’s backed up by empirical evidence from eg. engineering or physics.

Don’t know if that counts as being ”in math” by OPs definition.

[–] [email protected] 2 points 2 days ago (1 children)
  1. I have this odd, perhaps part troll, feeling that there are two, and only two, roots of the Riemann Zeta function that aren't on the critical line, and are instead mirrors of each other at either side of it, like some weird pair of complex conjugates. Further, while I really want their real parts to be 1/4 and 3/4, the actual variance from 1/2 will be some inexplicable irrational number.

  2. Multiplication order in current mathematics standards should happen the other way around when it's in a non-commutative algebra. I think this because transfinite multiplication apparently requires the transfinite part to go before any finite part to prevent collapse of meaning. 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.

[–] [email protected] 2 points 2 days 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.

[–] [email protected] 3 points 2 days ago

Most real-world phenomena would be better represented as regular Directed Graphs than Directed Acyclic Graphs, even ones that are traditionally abstracted as DAGs.

[–] [email protected] 3 points 2 days ago

P vs NP can be solved, and is within P. Good luck proving it though, I'm not smart enough

[–] [email protected] 1 points 2 days ago

Matrix multiplication should be the other way around, i.e. not like cascading functions. Oh and function cabbages should also be the other way around :) i prefer to read it not like a manga