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#proof

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And over on Facebook, the usual idiots are fighting over this obvious AI slop image of a 747 on a giant treadmill.

"This settles it. A Boeing 747 CAN take off from a treadmill!!! #proof."

"Clown get a life. Even if you could get the treadmill up to rotation speed of the four-seven, it would crash through the fence at the end of the runway into the neighborhood next to the airport!!!!!!"

"Everyone knows this is true!!! I saw it on Newsmax+, only $99.95 a month."

Proof by starvation: This is a proof form in which you first prove that a counterexample to the theorem must have property X, then, using X, prove that it must also have property Y, then that it must also have property Z, ... until you have piled up so many requirements on a counterexample that everybody sees that it cannot exist.

I have done that a few times. It is a nice way to organize one's thoughts.

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@BernhardWerner My favorite alternative #proof strategy for #induction proofs are #combinatorial (counting) proofs.

I suppose the standard example might be the proof of the coefficients in the binomial theorem expansion, or for the sum of binomial coefficients being powers of 2. These can be proved by induction, of course, but I'm not sure that's common given how easier it is to do a counting proof. It is also much clearer and avoids tedious algebra.

One I like is proving that the sum 1 + 2 + 3 + ⋯ + 𝑛 is 𝑛 + 1 choose 2, the binomial coefficient \(\binom{n+1}{2}\). Bijection proof, counts the same thing in two ways. The thing being counted is the number of ways of choosing two things (distinct, without repetition) from the set {0, 1, ..., 𝑛}. By definition, it is the binomial coefficient we want. The other way to count is to fix the larger number 𝑘, the remaining choices are any of the 𝑘 numbers from 0 to 𝑘 - 1. Thus, across all possible larger numbers, we get the sum from 1 to n.

An alternative alternate proof of the same, slightly more geometric is as follows: arrange dots in a triangle, 1 on row 1, 2 on row 2, and so on up to row n, with n dots. Add a phantom row of n+1 dots below. We want to add up all dots in first n rows: ∑ 𝑖. If you think of all of this as a binary tree/DAG, then every dot has two children (imagine Pascal's triangle). If you pick any two dots in the phantom row, their common ancestor is unique. So counting dots is same as picking two dots in phantom row. Which is the binomial coefficient we want.

Benjamin and Quinn's book on combinatorial proofs is amazing for interpretations of this form (I learned the first proof from it). See also: en.wikipedia.org/wiki/Combinat

en.wikipedia.orgCombinatorial proof - Wikipedia

#Mexico and #Canada, and every nation in the #EU, should now insist on proof of vaccination against all generally preventable diseases from Americans prior to entry into their country.

If you want Bobby Brainworm in charge of your public health, it's your funeral, but you don't get to spread those germs in our countries to those of our citizens that are too vulnerable to be immunized.

Flag down on the play, B&H Photo-Video! Was repeatedly assured in-store that buying items & having them shipped would not incur any charges (I bought $200 of stuff!). So when it shipped & I saw a shipping charge, I called right away. Not like them, they're usually more honest.
#B&H #Photo #Video #Store #free #shipping #not #really #called for #credit #back #filter #holder #effects #photography #pictures #instant #color #black& #white #negatives #proof #sheet #roll #film #bellows #camera #shoot

inspired by tavis' deep field #nebulabrot #DeepZoom images on #fractal #fractals forums, I did a little shader that for each c in the complement of the #MandelbrotSet M, colours according to how often z <- z^2 + c hits a given small target disc , weighted by derivative (as a proxy for point density).

it looks as though the hit sources are distributed everywhere near the boundary of M, which i think i can prove for target discs outside a sufficiently large esape circle, but i'm not sure how for discs nearer M. intuitively, by the time any cell pair in binary decomposition of exterior escapes, it covers an annulus with radii R, R^2, so any disc outside R will be hit by some region in every cell pair.

#math#maths#proof

Over seventy percent of the #GOP wants to see the #white #supremacist #terrorists #pardoned, for shitting in the #Capitol and spreading it on the walls, for parading a #confederate #flag there for the first time, for trying to #kill the #vice #president and trying to #overthrow the #government in a #violent coup. Zero #evidence, lost over sixty cases, no #proof yet they lie still.
They were terrorists, not rioters (#rioters have legit grievances).
#never #forget #Jan6
cbsnews.com/news/poll-analysis

CBS News · 4 years later, Republicans' disapproval of Jan. 6 attack continues to soften — CBS News poll analysisBy Jennifer De Pinto

For a long time, I didn't know exactly how the Fundamental Theorem of Arithmetic (unique prime factorization) was proved from first principles.

I finally figured out the order of proofs for the fundamental theorem of arithmetic from first principles. Now that I know the sequence, it feels like it was my own fault for not properly going through a textbook and that it was probably obvious to everyone. I'm posting this mainly to reinforce my own clarity on the topic.

1. Well-Ordering Principle, this is either taken as axiomatic or one step away from axioms. This is where I'm drawing the line for "first principles".
2. Euclidean Division: Existence and Uniqueness of quotient 𝑞 and remainder 𝑟 in 𝑎 = 𝑞𝑏 + 𝑟 form. Uniqueness if 𝑟 between [0,1,...,𝑞−1]. Proof is just well-ordering principle (often framed as infinite descent).
3. Bezout's Identity. Proof starts with well-ordering to find 𝑑, then together with Euclidean division shows that 𝑑 is a divisor. A bit more Euclidean division shows that all other factors 𝑑' also divide 𝑑. So 𝑑 is the greatest divisor.
4. Euclid's Lemma: Proof says if 𝑝 doesn't 𝑎 in product 𝑎𝑏, then gcd(𝑎,𝑝)=1. This brings in the above and some rearrangement proves that 𝑏 is a factor as needed.
5. Fundamental Theorem of Arithmetic: Existence usually goes via strong induction (which can be rephrased as Well-Ordering + contradiction). Uniqueness uses Euclid's Lemma to show that the primes in the factorization must divide each other (which implies equality).