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@bear.bzky.team

same idea but as the proof I was trying to write:
1) f⁴ = id so it's a homeomorphism , in particular proper
2) by functorality it induces an iso f^* H^*_c(R¹) -> H^*_c(R¹) squaring to -1
3) but H^*_c(R¹) is infinite cyclic so its endomorphism ring Z has no squareroot of -1

November 7, 2026 at 9:12 PM

wessie

@wessie.bsky.social

We've invented a square carrot. We call it Squarrot. No wait, SquareRoot. No wait, SQRT. Find it on the Vegify app!

October 20, 2026 at 12:29 AM

Valde

@valdefin.bsky.social

All these math lessons, and all I ever needed in my adulthood has been plus, minus, divide and multiply. I've been on renovation and nobody even knows what a squareroot is.

October 16, 2026 at 6:01 AM

🔞Patry🎉

@valixri.bsky.social

Yesterday I drew my squareroot sona, Nibbler. First three are the old drawing. | #osc #objectshow

October 2, 2026 at 9:24 PM

Thanushan Sivapatham

@thanushan95.bsky.social

🌱 Why do plants hate math?

Because it gives them square roots. 😄

A simple geometric trick reveals why
√s ≈ a + b/(2a)

Read the full article 👇

thanushansiva100.medium.com/the-babyloni...

#Mathematics #Math #SquareRoot #Geometry #Learning

The Babylonian Method for Finding Square Roots

How geometry turns a simple guess into an accurate square root.

thanushansiva100.medium.com

September 18, 2026 at 10:06 AM

NorseMap

@norsemap.bsky.social

🚀 Countdown to Launch: Spotlight on Our App Development Partner, SquareRoot
As we count down to the official launch of NorseMap and our photo-capturing app on the 5th, we'd like to extend our sincere thanks to Square Root for their dedication support and hard work throughout the development process.

August 28, 2026 at 12:37 PM

mattbarrs.bsky.social

@mattbarrs.bsky.social

Squareroot beer.

August 26, 2026 at 3:35 AM

Oï 🧩 | Comms are still opened! :gleeful:

@slinkoutlawed.bsky.social

Dave Algebra Class with the squareroot of infinity on his shirt who also wEARS GLASSES :rabies: :rabies: :rabies:

August 2, 2025 at 9:35 PM

dotnetbot

@dotnetbot.bsky.social

[wasm][interp] Add Store intrinsics (114711) https://github.com/dotnet/runtime/pull/114711

[wasm][interp] Add Store intrinsics by lewing · Pull Request #114711 · dotnet/runtime

Add Store aliased intrinsics and add the System.Numerics style SquareRoot to Sqrt

github.com

April 15, 2025 at 8:39 PM

Airwindows

@airwindows.com

Posted a good building block plugin, SquareRoot, thanks to herbie.uwplse.org which is a great project that helps many of my plugins sound their best through using floating-point math most accurately. Which is much less trivial than it sounds! :)

Herbie: Automatically Improving Floating Point Accuracy

herbie.uwplse.org

February 16, 2025 at 10:52 PM

YrrOi

@yrroi.bsky.social

Notes from the sqrt(2) proofs stream from 2025 08 03. Check out the 2 min preview for free, and the original 1hr 20min VOD available for patreon members in the quoted post below. #math #notes #proof #squareroot #root2 #irrational #rationalnumbers

$A rational number can be expressed as a fraction or ratio of integers: a over b, for any a and b that are integers. Prove sqrt(2) is irrational (proof from discover maths): Assume the opposite. Fundamental theory of algebra is that every natural number larger than one can be expressed uniquely as a prime factorization. assume sqrt(2) = a/b is irreducible for any a and b larger than zero. The expression above can be squared on both sides, and reduced to 2b^2 = a^2. Because a/b must be irreducible a and b are both prime, meaning we have two different prime factorizations of the same number. Proof by contradiction, therefore sqrt(2) must be irrational. Second proof (from vihart.): Consider a right angle triangle where the legs are the same length. By the Pythagorean Theorem, the hypotenuse (hyp) must be sqrt(2). If the leg is an odd length, the hyp must be odd, but if the hyp is odd then it cannot be split into the two equal leg lengths. If the hyp is even, then two times the leg length squared must also be even, meaning the leg must be even, which means the triangle can be further reduced until the leg length is odd. By this logic the hyp cannot be odd, nor can it be even, therefore it must be irrational.$

Third proof (from Vsauce ding!): If the number 2x is even, then we can be certain that the number 2x+1 is odd. An even number squared is even. An odd number squared is odd. Assume sqrt(2) = a/b. Squaring that equation gives 2b^2 = a^2. If we assume a is even, then we can write a as 2 times some other number c, and we can write the previous equation as 2c^2 = b^2. From this b^2 must be even, and b must also be even. If a and b are both even, then a/b is not in its simplest form, breaking our original assumption and proving that sqrt(2) cannot be written as any a/b, and is therefore irrational.

Fourth proof (from tipping point math): If a and b are any two integers, then a^2 has an even number of 2's as its factors, and 2b^2 has an odd number of 2's as its factors. From this argument, the absolute value of a^2 - 2b^2 must be larger than or equal to one. Absolute value of sqrt(2)- (a/b), multiplied by the absolute value of sqrt(2) + (a/b) over itself (equating to a multiply by 1/1). Multiply both the numerator and denominator by (b^2/b^2). Reduce this expression to the absolute value of (a^2 - 2b^2) over the absolute value of (b(sqrt(2)b + a). Reduce to the absolute value of( one over b(sqrt(2)b + a)), and realize this expression must be larger than 0, proving sqrt(2) cannot be written as a/b for any integers a and b, as originally assumed.

$Fifth proof (also from tipping point math): Consider a right isosceles triangle where both legs are the same length. if sqrt(2) = x/y then there must be some a and b integers which can describe the hypotenuse a and the leg b in the triangle's lowest form. If a and b are integers, then a-b must also be an integer. Consider a right isosceles triangle embedded within the previous triangle, where the legs of the smaller triangle are both the integer a-b. The hypotenuse of this smaller triangle must be the remaining part of the larger leg length b, subtracted the smaller leg length a-b, simplified to 2b-a. As the smaller triangle is inscribed into the corner of the larger triangle, the ration between leg and hypotenuse must be the same, sqrt(2), and therefore the form a/b cannot be the lowest form of the triangle, therefore sqrt(2) cannot equal any fraction of integers. Sixth proof (from wrath of math, originally demonstrated in 2000 by Tom Apostol of Caltec): Assume sqrt(2) = a/b, which is the same statement as assuming that there exists an isosceles right triangle with integer sides. Consider scaling up the triangle by some factor b. Legs of length 1 will become length b. Hypotenuse of length sqrt(2) will become length a. From our first assumption, we can also assume there exists a smallest version of the isosceles right triangle with integer sides, running form point A to O to B and back to A again, with the right angle at point O. The angle AO must equal the angle BO, which we will call x. Now consider a circle drawn about the point B, with a radius of BO, sweeping from O to intersect with the hypotenuse AB at some point C. Consider a triangle drawn tangential to that circle, with a right angle at point C, drawn from C to A to the intersection of that tangent with the line AO at some point D. This is a smaller isosceles right triangle with integer sides, and we have reached the proof of absurdity.$

July 27, 2026 at 2:56 AM

Tim Shaw

@wdncnu.bsky.social

My son goes to a STEM-focused university and their swag says they are gonna own the f**k out of it.

Red college T-shirt with three math expressions:
Fd nRT/V Squareroot(-1)

April 28, 2024 at 7:00 PM

Bert

@bert73.bsky.social

I don't care how much they dreamt about it and I don't doubt their bravery - BUT they only destroyed the structures you can see above ground - they did the squareroot of fuck all to the hidden structures underground.

June 26, 2025 at 12:30 PM

Cliff Harris (Cliffski)

@cliffharris.eurosky.social

squareroot is so slow I sometimes wonder if its mining crypto half way through.

September 20, 2025 at 9:19 AM

ren - world 𝓼𝓮𝔁𝔂 ambassador

@tendoamahikos.bsky.social

i think i need to get a diagnosis for discalculia because i struggled over the easiest one ( squareroot 90-n = n)

October 23, 2024 at 11:28 AM

Knowledge Zone

@knowledgezone.bsky.social

How the #SquareRoot of -1 Came #Alive : Medium

#NASA-Funded Study Shows How Controlled #Burns Help #Sequoias Survive #Wildfires : NASA

This #ElNiño will be the #Hottest by a huge margin : New Sci

Latest #KnowledgeLinks

knowledgezone.co.in/resources/bo...

August 15, 2026 at 6:25 AM

HARU 💫

@harukyuun.bsky.social

AHHHHHH OKAY OMFG JCNDKXDITC but why is the 2 and 10 outside the squareroot ??.? :'0

December 15, 2024 at 6:40 AM

bamhurguese (🎂 11/26

@ps2romsfree.bsky.social

squidwards, what s the squareroot of 4

Every SpongeBob Frame In Order @esfio.bsky.social · Mar 23

Season 2 Episode 7 - Frame 2244 out of 2694

March 23, 2025 at 2:35 PM

hollow jokow...

@hotmanyaoi.bsky.social

heyy man i like my irrational numbers they tell me that squareroot 6342458245355646969 is a real numbres what the fuck :(
(god spare me from this hell i have 3 days left till im free)

November 20, 2024 at 5:37 PM

Orman

@orman.bsky.social

TBF "You can search an unordered list in time proportional to only the squareroot of the list's length" doesn't need much buildup to find weird

October 26, 2025 at 10:47 PM

brendan (mathematician redux era)

@bear.bzky.team

If we set N_1(V) to be the nilpotent endomorphisms of rank 1 and PN_1(V) the set of lines in N_1(V), the composite map π : PN_2 -> P(End(V)) has image PN_1! This is because rk A^2 = 1 when A = 2, and any rank 1 nilpotent matrix has a squareroot (by JNF all such matrices are similar, check for some)

March 5, 2025 at 8:42 PM