English is not my native language so I was a bit confused by "200 times less", which I (wrongly) imagined to mean starting amount (x) minus 200x, getting to -199x, which didn't make sense. Math in speech is a tricky thing.
You're adding speed ("going faster").
You're adding a 200% of speed, which is twice the nominal speed (the 100%). Given the nominal speed is 1x and you're adding 2x, you end with triple the magnitude of the original nominal speed.
People will rail about them using it wrong but it's pretty useless when you have to basically guess whether people subscribe to your definition of "right" before you can understand something.
I'm using the exact same terminology, so splitting hairs on phrasing isn't going to work for me.
After some clever engineering, it now runs 50% faster. Its speed is now 100% (baseline) + 50% (improvement) = 150% of 1 m/s (original speed) = 1.5 * 1 m/s = 1.5 m/s
The budget option runs 20% slower than the original model. Its speed is 100% (baseline) - 20% (derating) = 80% * 1 m/s (original speed) = 0.8 m/s.
There’s not really any particular advantage to saying 0.005 of the amount (or 1 – 0.995 of the amount) vs. 200 times less. Personally I find it significantly less clear (though not really any more or less “impressive”), because doing mental decimal arithmetic takes some extra effort and leaves more room for confusion. That is, it is easier to reason about multiplying or dividing some quantity by 200 vs. multiplying or dividing by (1 – 0.995).
But the two numbers are reciprocals; this is grade-school rational arithmetic, not some kind of trick.
Before that equations were written out in words!
Latin: Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.
Translation: It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
Fermat's Little Theorem (much more useful in practice):
French:Tout nombre premier mesure infailliblement une des puissances − 1 de quelque progression que ce soit, et l'exposant de la dite puissance est sous-multiple du nombre premier donné − 1; et, après qu'on a trouvé la première puissance qui satisfait à la question, toutes celles dont les exposants sont multiples de l'exposant de la première satisfont tout de même à la question.
Translation: Every prime number [p] divides necessarily one of the powers minus one of any [geometric] progression [x, x², x³, ... ] [that is, there exists a such that p divides xª – 1], and the exponent of this power [a] divides the given prime minus one [divides p – 1]. After one has found the first power [a] that satisfies the question, all those whose exponents are multiples of the exponent of the first one satisfy similarly the question [that is, all multiples of the first a have the same property].
From an early/first English translation of Euclid.[1]
The book is written in Latin and contains diagrams and text to describe each lemma and law. Geometric proof seems to feature heavily!
From wikipedia: https://en.wikipedia.org/wiki/Philosophi%C3%A6_Naturalis_Pri...