Yet Another Trans Girl

We don’t have any of the large predators we used to (black bears, mountain lions, or gray wolves) but still retain large deer populations. If nothing is removing animals, they’ll quickly overload the carrying capacity of the environment and have massive losses to starvation and disease that can also pass on to livestock. Human hunters replace the large predators that our landscape can no longer support.

It’s kinder to euthanize an un-releasable hawk rather than try to find it a permanent home with humans. Wildlife rehabs have extremely limited space and resources and are usually run entirely on donated money and volunteer time. Only a few are large and stable enough to care for permanent residents long-term, and those spots are few and far between.

An invasive species poses a danger to threatened native wildlife. I will admit- Australian possums are adorable. But not in New Zealand, where they’re an invasive species that eats the eggs of ground-dwelling birds that previously had no such predators. The landowners I worked with replanting native bush, all native Maori, had no qualms about setting the dogs on them.

Golp: a roundel purpure.

Roundels are circles in heraldry. They are named according to their color, which also has its own lingo. Let’s meet them!

Bezant: roundel or (gold) 🟡

Plate: roundel argent (silver) ⚪️

Torteau: roundel gules (red) 🔴

Pomme: roundel vert (green) 🟢

Hurt: roundel azure (blue) 🔵

Golp: roundel purpure (purple) 🟣

Pellet: roundel sable (black) ⚫️

This is a given.

If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.

If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it's 3, 6, or 9.

Like the rule for 2, but check if the last two digits are divisible by 4.

If it ends in 5 or 0.

If the rules for both 2 and 3 apply.

No shortcut. Alas.

Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)

Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.

For example, 294

29 4

4×2=8

29-8=21

21 is divisible by 7

Second example, 1946

194 6

6×2=12

194-12=182

182 is still large, so let do it again

18 2

2×2=4

18-4=14

“Paned?” (Pron: pan-edd) - you’re in Wales, the question is rhetorical. You don’t have a choice, you’re having tea.

“Brew?” Or “Fancy a brew?” - Northern and you’re either a friend, familiar or been in the house long enough to warrant one.

You will also be judged for the amount of milk you have (clue: more is worse, because you’re weak, like your tea), and how much sugar you have (none is best, one is suspect, two is fine especially if a service man, above that will inspire disgust and horror).

Like, I’m not necessarily talking realism here since often it’s still pretty far from realistic, but more like, “someone has to be running spies in this fantasy kingdom, and we’ve seen the whole royal court, so which background character is it? How does that change these three major interactions?” Or “real life historical nobility did in fact have some things to do that were like jobs, how does this human disaster cope with running an estate?” Or “there’s no reason for a sci-fi robot detective to know how to whitewater kayak, where’d she learn?” Or “if this guy is serving the emperor directly he has to be way high up in the space empire servant hierarchy, why is he doing this menial task for someone else? What’s his motive? Does he perhaps have the secret space telepathy?”

The beauty of it is that you could place that exact same dialogue into any time and place in human history and it would make just as much sense, you could have it as a conversation between sailors in a 1600s ship, a group of bronze-age warriors cleaning their weapons, stone age hunters walking home from a hunt, or a bunch of 21st century redditors on the internet, and it would hit just as splendidly.

There was this conversation about if you could have sex with any specific person from any time through human history, which one would you choose? Famous legendary lovers were brought up, queens, emperors and such were popular, might as well fuck Rasputin just to see what the hype was all about, the kind of answers one would expect.

Then some guy said he would choose his own late wife. He went on to talk about how brutal it is, that you might find the love of your life, but there's no promise that this life couldn't be suddenly cut short. That his body aches for hers and how there isn't much he wouldn't do for one more opportunity to make love to her one last time.

Now, I'd like you to imagine a pause in the conversation. A solemn, contemplative moment of silence, with several of the people present nodding in agreement that this is indeed the finest answer - it wasn't a competition, but there's no doing better than that.

And then someone breaks this solemn silence and says:

From the second equation we get that cheeseburger plus fries-squared is five. Subtracting cheeseburger, which is six, from both sides, we get that fries-squared is negative-one. Math fans will know that there are two solutions to this; either fries are the "imaginary unit" 𝒾 or they are its negative, -𝒾. We'll do the rest of the problem with 𝒾, keeping in mind that at the end we should also take the complex conjugates as solutions.

Finally, we have that cup to the power of fries, minus cup, equals three. Replacing fries with 𝒾, and moving a cup to the other side, we get that cup-to-the-𝒾 is equal to cup-plus-three.

Now, the weird part about this is the cup-to-the-i. The problem with this is that complex exponentiation is technically not a thing. That is to say, there is no one function which is mathematically equal to "input-to-the-power-of-𝒾". In fact, there are infinitely many such functions.

Fortunately, due to reasons that take about six pages to explain (trust me I've done it), there is one particular function that many people have agreed is "the most reasonable one". This is not a mathematical notion, but a human preference. Seeing as this question was presumably written by a human, I am comfortable with using this function.

So, what function is this? Well, given a complex number r∠θ written in polar form (if you don't know what that means don't worry), where -π < θ ≤ π, then (r∠θ)^𝒾 = e^(-θ)∠ln(r).

Applying this to our problem a value r∠θ will be a possible solution for cup if e^(-θ)∠ln(r) = r∠θ + 3. Splitting this into real and imaginary parts, we get two equations: e^(-θ) cos(ln(r)) = r cos(θ) + 3 and e^(-θ) sin(ln(r)) = r sin(θ). We can graph these equations on Desmos:

let the value of the glass = z, for z ∈ ℂ:

z^(±i) = 3+z

let z = r·e^(iθ) for r,θ ∈ ℝ, -π < θ ≤ π

∴ z = r·e^i(θ+2πn) for all n ∈ ℤ

∴ (r·e^i(θ+2πn))^(±i) = 3+r·e^i(θ+2πn)

distribute powers (apologies for the use of ∓):

r^(±i)·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

convert to the same base:

e^i(±ln(r))·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

split into real and imaginary components:

re: cos(±ln(r))·e^(∓(θ+2πn)) = 3+r·cos(θ)

im: sin(±ln(r))·e^(∓(θ+2πn)) = r·sin(θ)