Month: September 2022

Hello again,

So originally I had intended to write this blog about math proofs and how annoying they are. However, I happened across a much more fun anecdote to share this week, so that’s what everyone gets to hear.

I am a big fan of nighttime walks. Much of my day is spent cooped up in my room either reading or working on my computer, so it’s nice to take some time to go out and stroll around. Daylight is work time and I prefer the quiet anyway, so going at night is an obvious choice for me. I think the dark is more comfortable, and the stars being out is very relaxing.

In my mind, one of the things that makes Penn State special is the arboretum. Not many universities have access to built-in garden space we do, and I think that we’re quite lucky for it. It’s really beautiful during the day, all of the plants and flowers and such, but my favorite time to visit is actually at night. It’s the only place I’ve found on campus that doesn’t have any lights. When dusk falls and the sun dips below the horizon you can see the entire night sky light up with stars.

I sorta wish I had taken a picture by now, but hindsight is 20/20 and I didn’t expect to write a blog about this. One of the coolest fun facts about campus I can share is actually about the stars at the arboretum; when you walk toward the garden there is a big open clearing with a nice fountain at the other side, and if you look up as you walk forward you can see the big dipper pasted perfectly in the sky directly about the fountain. It is legitimately picturesque.

Very fun, good times at the arboretum, very cool. I got a little distracted and rambly, but the point is that’s where I go when I walk at night.

Tonight started out like most of my walks do: I walk all the way across campus to get to the arboretum, take a right when I walk in, pass the koi ponds, and take a short break to listen to music at the little cabin at the back. Relaxing vibe, having a good time, everything is good. Eventually, I get up and leave. As I walk out of the cabin, on a whim, I take a little gravel train instead of the main concrete path. Walking along, bopping to the Wallows, thinking about writing my blog when I get back, I look down to the ground. Vaguely, out of the corner of my eye, lit only by the moon, I see a small white ball move toward the bush.

SHIT. FUCK. SHIT. It’s a skunk. I fucked up. Its piss is all over me.

Once I had run a sufficient distance away, I texted my roommate and asked him to bring me clothes and shower stuff so that I didn’t track the smell into the room.

I don’t have tomato juice, but after washing for 20 minutes, taking my clothes down to the laundry, washing for another 10, and half a bottle of febreeze, I think I got the smell out. My roommate gave me the all-clear, but I suppose if I didn’t get it all out then you all will be the first to know.

This whole blog post is really just a public service announcement to tell you that there are in fact skunks in the arboretum.

Hello again,

SO, I was reasonably sickly this week and have accomplished very little beyond making sure I get enough sleep and food to live. As a consequence of my laziness, I didn’t come up with another fun little math problem to share this week. Fortunately, I’m taking a couple math classes this semester to draw from if I ever have an off week, so today I’m gonna draw from that pool and share a cool proof I wrote for homework.

To understand any of what I’m about to say, you’re first gonna need at least a fuzzy idea of what congruence classes are, so that’s where I’ll start:

An integer, a, is said to be congruent to another integer, b, if it can be written in the form a ⦀ b mod c [1].

For those who don’t know, the mod operator is like division, but it only returns the remainder of the two numbers. For example 4 mod 3 = 1 and 4 mod 5 = 4. Pretty simple, but as the congruence suggests, this can quickly become much more complicated. Let’s see some examples of numbers that are congruent to each other:

9 ⦀ 1 mod 8
25 ⦀ 9 mod 8
7 ⦀ -1 mod 8
8 ⦀ 0 mod 8

Very cool. Now we have this, thing…, but it’s pretty unclear what it means and, more importantly, what we’re supposed to do with it. I honestly had a hard time seeing what problem could be solved with this algebraic structure until I found this:

Prove that 8n+7 cannot be written as the sum of 3 squares.

In math, the easiest way to prove that a given statement is true is by proving that if it were false then nothing would make sense and everything would be bad [2]. This process is called proof by contradiction, and here’s mine for this problem:

Suppose 8n+7 = a^2 + b^2 + c^2 for some non negative integers n, a, b, and c.

If this were true, then we could write 8n = a^2 + b^2 + c^2 – 7 by subtracting 7 from both sides.

Because n is a positive integer, we know that if we divide both sides by 8 we will have an integer on both sides and that means that (a^2 + b^2 + c^2 – 7) is divisible by 8.

This means that (a^2 + b^2 + c^2 – 7) ⦀ 0 mod 8 because 8 divides evenly with no remainder.

Given this congruence, we can now say (a^2 + b^2 + c^2) ⦀ 7 mod 8 [3]
The second congruence is very important because now we have worked the problem to a point where it’s almosssst clear that a contradiction is appearing. All we have left to do is prove that the sum of the squares of three numbers cannot be congruent to 7 mod 8, or, in more clear terms, the squares of 3 numbers cannot sum to 7. This is very easy because there are only so many numbers that can sum to 7, specifically (0,0,7), (1,1,5), and (2,2,3).

From here, you might be inclined to say that clearly none of the potential sums are all squared numbers, so we’ve found this contradiction and can leave. However, these aren’t normal squares, they’re congruence classes that represent positive integers less than 8, which means its possible that there is a square greater than 8 that is congruent to n mod 8 so that n is one of the numbers that can potentially sum to 7 [4].

To prove that no number squared can sum with any other two squares to equal 7, we need to construct a table.

We’re almost done at this point. Because any positive integer can be represented as congruent to the integers 0-7 mod 8, their squares can also be represented by what is seen in the table above [5].

Because the congruences of the squares are all either 0, 1, or 4, it is not possible to take these numbers and sum them to 7. This finally indicates that it is not possible to represent 8n+7 as the sum of 3 squares and ends our proof.

Because it is not possible to find 3 integers a, b, and c such that the sum of their squares is congruent to 7 mod 8, we have found a contradiction with our supposition at the very beginning of our proof and demonstrated that it must be false.

This process ended up being a little more difficult to explain than I expected, which is really my bad cause I had already done the work so I knew what it took to achieve the understanding. Anyways, I’ll do something that can be represented physically so that it’s easier to visualize next week. I hope you found this cool, have a nice week until I write my next blog

[1] – The triple bar is supposed to be horizontal, Docs is uncooperative.

[2] – Fun example you’ve probably seen before: https://www.youtube.com/watch?v=hI9CaQD7P6I

[3] – I know I didn’t fully explain the algebraic nature of adding, subtracting, and multiplying congruence classes, but I really can’t write another 1300-word blog this week, so my source here is just gonna have to trust me bro.

[4] – I can tell this sentence is probably a little difficult to digest, but it’s too late to turn back now, so we keep marching forward : ).

[5] – trust me bro

Hello : ),

I don’t have a very clear idea of what I’d like to write my blog about yet, so I’ve decided to share a lil personal anecdote for this assignment and see how that feels.

On my bus ride back from Philidelphia over the labor day weekend, I started thinking about paper and how it’s folded. I began thinking about ‘hamburger’ and ‘hotdog’ style folds from elementary school (folding a paper short ways or long ways so that it looks more like either a hamburger or hotdog bun respectively) and how I had always thought those names were silly, but then I started thinking about math and I guess the two got jumbled together. It’s unclear exactly how I reached the problem statement, but basically, I spent the rest of the bus ride trying to reason this out:

If I take two sheets of paper, start by folding one hamburger ways and the other hotdog ways, and then fold them both the same number of times, alternating between hotdog and hamburger each fold, will there ever be a point where the total length of folds on the hamburger paper exceeds that of the hotdog paper.

This is an exceptionally goofy thing to be thinking about, and it is probably difficult to understand what I was trying to solve if I just explain it with words, so I will take a moment and explain what I mean visually.

We’ll start with two regular pieces of paper, 8.5 by 11 units in width and length each. If we fold the first piece of paper hotdog ways, we will have created a fold that is 11 units long

And if we fold the second hamburger ways, we will have created a fold that is 8.5 units long.

These two folds are obviously not the same length, so we can then take the first paper and fold it hamburger style this time, getting us a new fold of 4.25 units, for a total fold length of 15.25 units.

Looking back to the second paper, we will fold it hotdog style on its second fold, getting a new fold of 5.5 units, for a total length of 14 units.

The papers’ total fold lengths are still clearly not the same, but, noticeably, the total lengths of the folds have gotten closer. From having a difference of (11-8.5) = 2.5 to 15.25-14 = 1.25, the two total fold lengths appear to be approaching one another. The question then stands, does this pattern continue? Will our total fold lengths ever be equal?

Once I had gotten back to my dorm, I was able to sit down with a piece of paper and start writing out formulas to really model the problem in full. A few things were immediately apparent to me:

This problem would be modeled by a summation series
Each time the paper was folded, it would shrink the length of the other side by half, leaving each subsequent fold of either hamburger or hotdog to be half of what the previous fold of the same type was
This would need to be modeled by two series for each paper, one for the hamburger folds and one for the hotdogs, which would then be summed together to give the total fold length

It’s always good to start off with a series write-out because it can help visualize what the summation will look like.

L = length of the paper
W = width of the paper
f = number of folds (used in the summation)

Total fold length of the paper originally folded hotdog style =
L + W/2 + L/2 + W/4 + L/4 + W/8…

Total fold length of the paper originally folded hamburger style =
W + L/2 + W/2 + L/4 + W/4 + L/8…

If I graph this in desmos where the x-axis represents a given number of folds and the y-axis represents the total fold length after that many folds then we get a graph like this:

-blue line is the hotdog style original fold
-red line is the hamburger style original fold

With this write-out and graph, I can clearly see that I want to create two summations for each folded paper that each handle the lengths or the widths and add up a number of terms equal to half the total number of folds where each term is half the previous. The final equation will look something like [1]

Hotdog Paper after 6 folds: (L + L/2 + L/4…) + (W/2 + W/4 + W/8…) =

Hamburger Paper after 6 folds: (W + W/2 + W/4…) + (L/2 + L/4 + L/8…) =

After only a little work we have a clear understanding of what this series will look like. The only issue with what we have now is that it doesn’t account for what happens where then is an odd number of folds. When this occurs, n/2 is decimal instead of a whole number, so we need to round one of our term’s n/2 up and the other down so that only one new term is added for each fold that occurs.

The easiest way to round numbers up or down is to use a ceiling and a floor function. Ceiling takes a decimal and rounds up to the nearest whole number; floor does the same but rounding down. Adding these two functions to the original equations for each paper gives us the final answer[2]:

The graph of these summations with a lil desmos magic is very cool:

-green line is the difference between the total fold lengths after each fold

With this graph, we can draw a few observations, make conclusions, and answer the original question. First, we can see that both the graphs do converge to a single total fold length as you approach an infinite number of folds. Secondly, we can see that these graphs never cross or even touch, meaning that the total fold length of the hotdog paper is always greater than the hamburger. Third, we can see that the two total fold lengths that the graphs converge to are actually the same distance apart as the two original fold lengths; this is actually an interesting conclusion and not something that I would’ve guessed at the beginning. Finally, it is worth noting that if you look at the green line graph, you can see that the difference between the two lengths bounces up and down until they converge. In fact, the second fold is where the difference between the two fold lengths is the least, so the original observation that I got so worked up over on the bus was actually just a cruel bait that meant nothing.

In reflection, this project was very silly, but I did have a good amount of fun with it. I suppose the moral of the story can be that you should always check more than two examples of something before you assume there’s a pattern. It’s a little hard for me as the person writing this to tell if the reasoning I’m trying to convey makes any sense at all written down, so I hope that you got what I was saying for most of this. If you didn’t understand a word because of my poor explanation of the subject, then I hope you at least appreciated the cool graphics and equations that went along with it.

Authors Notes:

[0] I’m way over word count, but I still didn’t fully explain the specifics of my equations, so I’ve included this bit to better elaborate on some of the stuff I just assert.

[1] The first summation starts at 0 because I want it to have a term where the original fold length is divided by only 1, where the first term of the second fold should be divided by 2.

[2] The reason why I put the floor and ceiling functions where they are instead of swapping them is that I want the first summation to count up only when the number of folds is even and the second to count up only when the number of folds is odd. The even/odd notation only works if you think about the first fold as a sort of zeroth fold and the second one as the first on the graph, which, just…, this is already way too long, so give me the benefit of the doubt on this one, go look at the graphs are something.