lilith's maths!

Because it has:

0: the additive unit

1: the multiplicative unit

2: the smallest prime number

π: a delicious food

e: amount of mathematics discovered by Euler in 3 units of math

To start, we should focus on the exponent on e, or the phrase

π + i²π

From the definition of i, or i = √(-1), it follows that i² = -1. Now, we can simplify our exponent to

π - π

or zero. This means our equation is now as follows:

1 - e⁰

All numbers, including irrational numbers, are equal to 1 when having an exponent equal to 0. The irrational constant e, which can be approximated to 2.718281828, is no different. Our equation is now

1 - 1

chemists: don't think they know enough to be an authority in other fields

biologists: aren't even sure they know enough to be an authority in their own firld

Okay so let's try to do it with as little proof by contradiction as possible. The set of arithmetic progressions (meaning sets of the form { an + b : n ∈ Z } for integers a,b) forms a basis for a topology on Z. Fine. All progressions with the same common difference (meaning they are 'cosets' of one another) partition Z, so each of these basic open sets is also closed. In particular, Z is totally disconnected. In fact, Z is also Hausdorff, as you can always find two cosets that contain either point.

As a first-countable space, the topological properties of Z are fully determined by the limiting behaviour of sequences. When does a sequence (aᵢ) converge to n? That's if the difference n - aᵢ is eventually divisible by every number as i increases. So the topology has a certain metric quality, where you can measure the distance between two points as inversely proportional to the number of factors of their difference. This doesn't actually satisfy the triangle inequality, though (the sum of two very composite differences might be prime, for example).

Anyway, it's also clear that any open set that contains at least one point contains an arithmetic progression, and therefore must be infinite. Together with Hausdorffness this shows that no point is isolated. Now say you have any finite set of prime numbers. The combined set of all their multiples is a finite union of arithmetic progressions, hence a closed set. Its complement is open and contains 1 and so must be infinite, and therefore there exists a number not divisible by any of these primes, QED.

So no proof by contradiction necessary, really. It's actually a rather constructive proof. So if I have a finite set of prime numbers, how does Harry Furstenberg find us a new prime? Well, for a single prime p we get that its set of multiples is a basic open set, and its cosets are too. So pick the coset containing 1 and you have numbers not divisible by p. In essence, for the proof it suffices that p+1 is not divisible by p, hence it is divisible by another prime number.

Now for two primes p₁, p₂, their sets of multiples are basic open sets, but because their cosets are too we find that their complements

U₁ = { n ∈ Z : n is not divisible by p₁ }, U₂ = { n ∈ Z : n is not divisible by p₂ },

are open as well. Now the next step is that the union of the multiples is a finite union of closed sets, hence closed. This corresponds to the fact that the intersection U₁ ∩ U₂ is open. In fact, all we need for the proof is that U₁ ∩ U₂ contains at least one basic open set.

Why is this? I sort of glossed over this, but the fact that the arithmetic progressions form a basis for a topology is a somewhat non-trivial fact. It follows from the fact that the intersection of two arithmetic progressions { an + b } and { cn + d } is either empty or another arithmetic progression. If they intersect at the number k, then adding any multiple of the least common multiple lcm(a,b) lands in the intersection again, and if the difference between m and k is not divisible by lcm(a,b) (so not by both a and b), then m cannot lie in the intersection. So

{ an + b } ∩ { cn + d } = { lcm(a,b) ⋅ n + k }.

As someone who has been in university for around 5 years already, my advice as a STEM student, would be to not allow men to pressure you out of your field. The statics won't be on your side, it will usually be disheartening to see. As someone who goes to a school where STEM is the biggest thing, you will still face some misogynistic stares no matter where you go. That's why you don't say anything, you prove your worth by out shinning them. Even if you fail one test, look at everything from a larger scale. You're already completing more than you know. You keep going to prove that women are just as capable.

As a woman in STEM it is hard to be taken seriously by male peers. Some won't be outwardly misogynistic but they'll say things that you know aren't for the faint of heart. "Do you need help?" "It's okay if you don't get it. No one expects you to." This is your fuel, this is your fire. Complete things with an open mind. Some nights will rough you up more than you know. Other nights will be so exhausting you won't want to get up. The curriculum will get hard to the point it has your second guessing your major. Yet you don't quit. You can't quit. Not when so many women out there would kill to have your spot. That's why you keep going, to prove your worth in a society who doesn't see it. To beat the odds, the statistics, and be the greatest anyone has seen.

This sounds silly but it's how I keep myself from flaking out of my doctorates ahh!!! And I was thinking maybe someone else will understand this small mantra I say to myself!!!! A lot of this is my personal experience and such but still!!!! I have always been determined academically and I know a lot of girls and women here can understand!!!