18.4. Counting Solutions to Equations

Here we want to show the following that you take any sequence of n + 1 distinct real numbers. They are arbitrary real numbers; may be positive, negative in any order you take them. The only condition is that they have to be distinct. Then the claim is that irrespective of the n + 1 real numbers that you have in your sequence you always have a subsequence of length n + 1 which is either strictly increasing or strictly decreasing.

In this chunk, we introduce a claim regarding sequences of distinct real numbers. Given any sequence of n + 1 distinct numbers, we state that there will always be a subsequence of these numbers that is either strictly increasing or strictly decreasing. A strictly increasing sequence means each number is less than the next, while a strictly decreasing sequence means each number is greater than the next.

Think of a group of friends standing in a line, each holding a different height marker. Whether they stand in increasing height order or decreasing height order, despite any arbitrary arrangement, you will always find a way to choose certain heights that either increase or decrease.

A subsequence means here that the values may not be consecutive. That means I am allowed to miss a few numbers. In the sense, say I take a sequence 1, 3, 0, -5, 2, 8 and so on. Then I can choose to pick 1 and then exclude 3 and 0 and -5. This is a subsequence. In the same way I can pick a subsequence saying 3, 2, and 8 that means I skip 0, I skip -5.

This chunk focuses on the concept of subsequences. A subsequence can be formed from a sequence without needing to select consecutive elements. For instance, from the sequence 1, 3, 0, -5, 2, 8, one could create a subsequence by choosing some numbers (like 1, 3) while skipping others (like 0 and -5). This flexibility in number choice is crucial for understanding the greater claim about subsequences being either increasing or decreasing.

Imagine a treasure hunt where you can pick and choose clues from a list without needing them to be next to each other. You might solve clues 1, 3, and later 5 without needing to solve clue 2, making a ‘subsequence’ of your clue-solving journey.

So let me first define two values. I define L as the length of the longest increasing subsequence starting at a_i. One might be of length 1, a trivial increasing subsequence starting at a_i is the value a_i itself. But I might be having a subsequence of strict length 2 starting at a_i. In the same way, I define D as the length of the longest decreasing subsequence starting at a_i.

In this part, we define two variables: L, representing the length of the longest increasing subsequence starting from a specific number in the sequence (let's call it a_i), and D, which represents the length of the longest decreasing subsequence from a_i. The trivial subsequence of one element is evident, and we could potentially have longer subsequences. This will help us analyze the properties of our original sequence.

Consider a line of children who can grow taller or shorter as they stand in line. One can measure the longest stretch of heights going up (increasing) or down (decreasing). If you start at any child, you can measure how many, at maximum, line up behind them in those patterns.

The question basically asks me to show that there always exists some a_i such that either L is greater than or equal to n + 1 or D is n + 1. I have to show the existence of one such number a_i in this subsequence. I prove that by assuming a contradiction...

In this chunk, we assume for contradiction that for every element a_i in our sequence, the lengths L and D (the lengths of the longest increasing and decreasing subsequences) are both bounded by n. This implies a limit to how many increasing or decreasing sequences we can find. Using this assumption, we leverage the pigeonhole principle to show that if the sequences are limited (pigeons) while the actual numbers vary (holes), a contradiction arises.

Imagine everyone in a classroom of varying heights trying to form groups based on increasing or decreasing heights. If we assume it's impossible to form a larger group than a set limit, yet we have more students than spots in gradual height groups, eventually, some students would need to overlap in their heights, contradicting the initial assumption.

What I can say is the following. I know that there is a decreasing subsequence starting at a_i and its length is D. My claim is if you take that subsequence and put an a_i at the beginning then that now gives me a new decreasing subsequence starting at a_j.

In the final thoughts, we conclude that our assumption leads to contradictory results. Regardless of assuming either of the sequences' lengths to be bounded by n, we ultimately discover that at least one element must exist where either L exceeds n or D does, confirming subsequences of required lengths always exist.

It's like building a tower using building blocks. If you claim that you can't stack blocks taller than a certain limit but find you have a deeper selection of blocks than that limit, you will eventually find ways to build a taller structure—an undeniable contradiction to your original claim.

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