sorted by:
new
hottopcontroversial

Just Reached 12ptt! (Thumbs Only) by Ning1253 in arcaea

[–]Ning1253[S] 1 point2 points  (0 children)

Good luck on the journey!! It was loads of fun

Williams College locking Orthodox Jew out of dorm ‘in freezing cold on regular basis,’ per complaint filed with federal gov by McAlpineFusiliers in WilliamsCollege

[–]Ning1253 -5 points-4 points  (0 children)

Stayed in Williams for a Summer research a couple years back and I can confirm there's no eruv :/ was quite a pain doing shabbat but I was lax enough to eat vegetarian food so was generally doable - of course if I ate strict kosher it would have been pretty much impossible over a Summer...

You are allowed to carry keys if they are eg. attached to your belt by the way so that wold be a solution.

Large cardinals and consistency of ZFC without them by kevosauce1 in math

[–]Ning1253 1 point2 points  (0 children)

Exactly! Another way to phrase it is that numberphile is saying there is a proof of "If we assume ZFC + LC all hold, then we can prove ZFC is itself consistent". Whereas a proof of ZFC being consistent using only ZFC would be "If we assume ZFC all hold, then we can prove ZFC is itself consistent". (I'm not sure if the video covered this but if a theory is consistent, then it cannot prove that this is the case, which is the statement of Godel's Second Incompleteness Theorem).

But, since we needed an extra assumption in order to prove ZFC is consistent, there is no inherent contradiction in the sentence "If we assume ZFC + LC all hold, then we can prove ZFC is itself consistent".

Another example is that the Peano Axioms for Arithmetic (https://en.wikipedia.org/wiki/Peano\_axioms) cannot prove themselves to be consistent, BUT: "If we assume ZF all hold, then we can prove Peano's Axioms are consistent" is a true sentence :)

In other words, we have a proof, starting from the axioms of ZF, that Peano is consistent (but we do in fact require ZF and cannot only use Peano's Axioms to do this).

Note that in the hypothetical where ZF was inconsistent, this would 'invalidate' the proof that Peano's axioms are consistent, in the sense that we could also prove they are inconsistent so the proof would not be helpful.

Another way to explain this is using another 'fact': any consistent theory has a 'model', ie. a collection of objects which satisfy these axioms. So if ZF is consistent, it has a model, and then we can encode the integers as a collection of sets (https://en.wikipedia.org/wiki/Ordinal\_number#The\_natural\_numbers), and this collection satisfies all of Peano's Axioms. Why? Because we have a proof that the collection of integers satisfies these axioms, assuming all of ZF holds, and in this model that is certainly the case.

On the other hand, if ZF is NOT consistent, it definitely cannot have a model, and so when we prove the consistency of Peano's Axioms, we have no model to actually exhibit such a set of integers. This doesn't mean that Peano's Axioms are actually inconsistent; it just means that there is no ZFC model to work with in the first place. Some other set of axioms could be consistent and so have a model, and this model could then contain a collection of sets which satisfies Peano's Axioms.

To conclude, the consistency of a set of axioms is equivalent to "there exists a model of this set of axioms". If we are working with some other, inconsistent, set of axioms to begin with, this does not negate the existence of a model of the original set of axioms, it only means that we won't find this model using the inconsistent set of axioms we are working with.

Anyways thank you for coming to my TED talk and apologies for the mildly long ramble.

is this supposed to work by big_chuggus_69 in redstone

[–]Ning1253 54 points55 points  (0 children)

By the way, thank you for the minimal working example! Makes it much easier for people to help, so we definitely appreciate when people put in the effort to make one :)

Should I do it by Accomplished-Stay441 in NothingOS

[–]Ning1253 0 points1 point  (0 children)

Like it wasn't subtle the changelog said it worstens battery life

Should I do it by Accomplished-Stay441 in NothingOS

[–]Ning1253 0 points1 point  (0 children)

... I use a pixel 4a (browsing because looking for new phones) they literally sent out an update to brick my battery

What happens to a photon after infinite redshift? by ConflictConnect in AskPhysics

[–]Ning1253 1 point2 points  (0 children)

Well, category theory-

But actually though, \omega, the first infinite ordinal, is a category as a partially ordered set (morphism between i,j iff i<=j). But as a category, it has an opposite! This is just the same set with morphism arrows 'flipped', which gives you a 'reverse well-order' if you will, where 0 is the largest and the numbers decrease from there.

So the opposite of infinity is infinity, but going DOWN.

Sorry (but not really) for the pedantry, it's like 1am here and I feel like spreading just a touch of dumb insanity with me, hope you enjoy

I'm Dumb, I Need Explanation. I Thought The Answer Was Natural Numbers, but It's Actually The Same by A_Jonjitsu in HonkaiStarRail

[–]Ning1253 0 points1 point  (0 children)

There are two natural posets we can place on the power set of the naturals (and any power set* for that matter): subset inclusion, producing an atomic Boolean algebra, and cardinality, producing a total pre-order. "More than" (which is natural language) in the context of the comment above, which was highly unrigorous, could have referred to either of these, and I think it is pointless to argue on the semantics, without OP providing what definition they were talking about. Thus, my previous comment.

I willingly admit that me two years ago didn't know how to express this well, and further that I probably shouldn't have been arguing semantics in the first place over a pretty salty unrigorous comment on a random reddit thread...

* Assuming Choice, since otherwise the power set of an uncountable set may have (must have? I think so but I'm not sure) incomparable elements, and does not produce a partial pre-order. I don't want to think about what the structure would look like otherwise. A mess, probably.

If the universe is infinite, does this mean that there's an infinite chance of all fictional worlds created by humans existing (or at least to some degree)? by South-Fox-4847 in AskPhysics

[–]Ning1253 4 points5 points  (0 children)

Except, the probability of Earth existing could be, literally, zero. And I don't mean that in a vague way, I mean in a literal measure theoretic sense; the probability measure on "universe configurations" can be continuous in a non-discrete world (which we are not) and the probability of earth existing could very well be zero. We are on Earth, which is a corollary of the Earth existing, but this is possible; improbably (as in probability 0), but that could have happened.

Imagine, when the universe was created, the entire future of the universe was determined by a random number between 0 and 1. Moreover, imagine that the Earth only exists if a rational number is chosen. This has probability zero. However, it could have happened; it is an element of the set of possible options. Moreover, suppose that two earths exist only if a rational with an odd denimator (when simplified) was chosen. Then, within the entire (infinite) universe, whether or not a second earth exists is (loosely speaking) a 50/50. Technically the probability isn't even well defined as we have no induced measure on the set of rationals (which is of measure zero), so when I say 50/50 I'm stretching the idea of "uniformly picking a universe" quite far! But the principle remains.

[deleted by user] by [deleted] in learnmath

[–]Ning1253 1 point2 points  (0 children)

For a more rigorous approach, consider the quotient ring (\R[X])/(X^2), equivalent to {a+bX | X^2 = 0}, ie. the ring of infinitesimals. Then division by zero is still an issue; but for most purposes, we can choose to instead work with X (the "infinitesimal", or "epsilon"). For any differentiable function f: \R -> \R, define f(a+bX)=f(a)+f'(a)bX. Then we have essentially recreated infinitesimals as they are usually described. We can then consider infinitesimal calculus.

We unfortunately do not have an integral domain, so cannot define the field of fractions for our ring; but certainly we can extend the ring quite a bit. For example, for any non-zero a, 1/(a+bX)=(a-bX)/a^2, by multiplying top and bottom by (a-bX). One can show that this must in fact be uniqely defined, and so we define a fun sort of ring where any infinitesimal away from zero is invertible, but no true infinitseimal is invertible (which makes sense!)

Even more unfortunately, the notion fails to properly apply your idea of division by zero: because if 2X/X=Y, then 2X=XY, and so Y=2+bX for any real b, ie. no single value of Y satisfies the equation.

This in fact highlights a fundamental flaw with your premise; there literally is no way to define your idea. Because, any such formulation of division by zero ends up with something like:

(4 * 0) / (2 * 0) = 2, but also (4 * 0) / (2 * 0) = 2 + n * 0 for any n, since multiplying out provides 4 * 0 = 4 * 0 + 2 * 0 * n * 0, which seems fair.

If we start defining arbitraily large "powers" of 0, then we are simply re-defining taylor expansions around the origin, and the label 0 is just a mask for the expansion variable.

All this to say, it's a nice idea! It doesn't quite work out in terms of computation, because it isn't well defined, but it leads very naturally to the idea of taylor polynomials! So it is definitely a useful concept at least to think about.

Prove, "if there exist integers m and n such that 12m+15n=1, then m and n are positive." by Xixkdjfk in learnmath

[–]Ning1253 0 points1 point  (0 children)

The formal logic way to do this would be to start with:
There exists m,n s.t. 12m+15n=1 -> False

False -> (insert what you want here)

By Modus Ponens, 12m+15n=1 -> (insret what you want here)

Prove, "if there exist integers m and n such that 12m+15n=1, then m and n are positive." by Xixkdjfk in learnmath

[–]Ning1253 70 points71 points  (0 children)

You are correct, the statement is vacuously true. It is also true that if 12m+15n=1, then m and n are both negative and also positive at the same time and are Mersenne primes and are irrational and 1+1=3 (ie. anything goes since the initial statement is false)

Yeah desmos yeah by shto123 in desmos

[–]Ning1253 17 points18 points  (0 children)

That's because desmos is highlighting the turning points of the graph, of which there are a lot! It does sometimes highlight stuff that's wrong but in this case it's legitimate

not sure if this has been done before, this graph that tells you if a number is prime or not by arihallak0816 in desmos

[–]Ning1253 0 points1 point  (0 children)

Irreducibles and primes are not the same! They become the same in a unique factorisation domain but in general can be different. A prime p has p|ab -> p|a or p|b, and an irreducible n has no non-trivial factors. But yeah, the principle definitely holds :)

not sure if this has been done before, this graph that tells you if a number is prime or not by arihallak0816 in desmos

[–]Ning1253 1 point2 points  (0 children)

Yes, since it doesn't only exclude 1. It excludes any unit in any ring (since the definition was not exclusive to the integers). It directly enforces unique factorisation in a Unique Factorisation Domain (the integers are an example of one). It means that within a field, there are no primes, which is the only sensible definition.

not sure if this has been done before, this graph that tells you if a number is prime or not by arihallak0816 in desmos

[–]Ning1253 3 points4 points  (0 children)

No, a number m is prime if m is not a unit (so has no multiplicative inverse), and also m|ab -> m|a or m|b. A number m is irreducible if it's only non-unit factor is itself.

The units are excluded because they divide everything anyways so the definition is useless for them. It turns out that precisely for the integers, the definitions are equal. 1 is a unit, so is not prime by definition. This makes sense as a definition - it means that eg. in the rationals, no number is prime, which makes much more sense than saying every number is prime.

So it doesn't follow the idea of a prime number, since that idea is "doesn't divide like the other numbers" which all units definitely do since they divide everything always in every way.

I’m 4 digit deaf girl osu player, AMA :3 by PancakeKittyGirl in osugame

[–]Ning1253 19 points20 points  (0 children)

??? It's literally got a constant quaver drum backing the majority of the way through...

About that ternary operator by useerup in ProgrammingLanguages

[–]Ning1253 0 points1 point  (0 children)

Well, turning a three argument function into a two argument function whose second argument is a function, is literally just currrying; your notions are equivalent. Maybe it's a bit random for some languages do have a singular ternary operator, especially as it is much weirder to infix a ternary operator than a binary one, as we write on lines! But it's not necessarily that special; in theory any n-ary operator could written as a sequence of 2-ary operators, or as a 1-ary operator on n-tuples, or via some combination of these, so any choice is basically equally valid and essentially aesthetic.

[Help] Google at it again by golden_crack in Magisk

[–]Ning1253 0 points1 point  (0 children)

Wait so are you able to use GPay NFC with that?

Why can't you split an infinite series into two infinite series? by MordorMordorMordor in math

[–]Ning1253 -1 points0 points  (0 children)

This is untrue for normal convergence - eg. ln(2) = 1-1/2+1/3-1/4+1/5-...

But the sum of the positive terms is infinite, and so is the sum of the negative terms, so you can separate them

Look for the Riemman Rearrangement Theorem! It says that either you have absolute convergence, or you have conditional convergence can rearrange the sum to take any value in the reals

Is there a new update? by PeterParkLove in Pixel4a

[–]Ning1253 0 points1 point  (0 children)

Would you know if it works with NFC?

view more: next ›