So the Caltech Y offered a few trips on Saturday to celebrate the end of rotation and the beginning
of the rest of our lives doing ** sets,** and I went on the kayaking trip, which was a blast.
It was sea kayaking, which I thought meant being out on open ocean with waves and sharks and everything,
but in reality we were at Long Beach, CA, which is this kind of harbor town with a few canals.
I guess you might call it a lagoon. Anyway, most everyone there had boats, or jet skis, and probably family money.

There weren’t any waves.

I was paddling along and this giant boat starts to pass me. It’s the kind with three tiers, the driver’s at the top and people can sunbathe on the second level, with, I don’t know, maybe a kitchenette on the lowest bit. The guy who was driving gave me a little wave, which I returned, but I thought about how an actual conversation between me and him would probably have gone…

**Him:** Hey man.

**Me:** Hey, so, I don’t know you at all and I don’t have much to go on to make this judgement,
but based on the facts that you’re driving a three-tier boat called *Rowdy* and smoking an
e-cigarette, we probably wouldn’t like each other too much, so maybe we should just quit
while we’re ahead.

**Him:** Oh yeah you’re probably right. Bye.

**Me:** Bye.

Also, my TAs are amazing.

My physics TA is an old man who literally wrote the textbook, so you already know he’s one of the coolest human beings on the planet and also probably the literal most qualified person to teach this physics class. The other day he brought up these squids that propel themselves through the water by pushing water through their bodies, and we talked about that for a little while.

In math, we have this Chinese dude who is so very funny. He gives so much
sass to anyone who asks him a question, and he also showed me this ridiculous
proof that is like my new favorite thing. I’m putting it in spoiler tags
because of how many of you will be *so very bored.*

## Math

So this is an alternate way to prove that the sum of the first n consecutive numbers is (n + 1) choose 2.

First you construct a set called A and fill it with all the numbers up to n + 1. So A = {1, 2, 3, … n + 1}.

Then you make another set, called C. C is a subset of A x A, so it’s made up of ordered pairs (x, y), where x and y are both chosen from A, but there’s another condition: y > x. This means C is made up of all the ways to choose two distinct elements from A, and so the number of elements in C is equal to (n + 1) choose 2.

But we’re not done making sets! Next we need a set called C

_{j}, where j is a member of A. We define C_{j}to be a subset of C, made of all of the elements of C where y = j. For example, C_{4}would be = {(1, 4), (2, 4), (3, 4)}. You can see that the number of elements in C_{j}is just equal to j - 1.Futher, you can see that all of the C

_{j}s are completely different: none of them could contain any shared elements. This means we can add all of them up to get the entirety of C without any overcounting! So the number of elements in each set C_{j}from j = 1 all the way up to j = n + 1 add up to be the number of elements in big C! And since the number of elements in each C_{j}are just the counting numbers 1, 2, 3, all the way up to n, we’ve proved that the sum of all those numbers is just (n + 1) choose 2!Turns out, you can also generalize this to the sum of n choose k being equal to (n + 1) choose (k + 1), which is also super cool.

But there’s only like three people who might still be reading this because it’s not actually super cool in general.

Anyway, that has been #math.

If you wanted to hear about something else… Leave a comment? I know people think every web page should have comments now, but I think that tends to turn out pretty horribly.

This is not an open forum, not even for people I would want to hear from.

This is my place, and nobody but me should have the right to say whatever they want here.

Plus it’s faster as a static site and comments would be nuts to implement.

If you do want to reach me I suggest Twitter or a telephone.