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18.2.1. Age Group and Pigeonhole Principle Application
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Today, we’ll start by discussing subsequences! Who can tell me what a strictly increasing sequence looks like?
Is it like when each number is larger than the previous one?
Exactly! An increasing subsequence will look like (a1, a2, ..., an) where a1 < a2 < ... < an. How about a strictly decreasing sequence?
It should have each number smaller than the one before it!
Correct! An example would be (b1, b2, ..., bm) where b1 > b2 > ... > bm. Remember, subsequences can skip some elements!
Wait, how do we find these subsequences in a set of numbers?
Great question! We’ll see how the pigeonhole principle helps us assure the existence of such subsequences regardless of the chosen numbers.
So, each number in a sequence can have its own increasing or decreasing length?
Exactly! That's essential for our proof!
To summarize, we learned about increasing and decreasing subsequences. Next, we’ll apply the pigeonhole principle for our proof.
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Now let’s dive into how we can demonstrate the existence of these subsequences using the pigeonhole principle. What do we mean by that?
Is that where we have more objects than containers?
Exactly! If you have more pigeons than holes, at least one hole must contain more than one pigeon. Here, we’re dealing with lengths of subsequences!
So we assign lengths to each number?
Right! We denote Li as the length of the longest increasing subsequence starting at ai. If we assume every length is ≤ n, we can explore pairs (Li, Di) as potential values.
This sounds complex! How do we derive a contradiction?
Good query! If we assume that all lengths are less than or equal to n, we'd end up with more pairs than possible values. Hence, we have to have the same lengths for two different starting points.
And that means we can find one subsequence that’s longer!
Exactly! Summarizing today, we’ve seen how the pigeonhole principle helps us find increasing or decreasing subsequences. Now, let’s look at an illustrative example using an age group.
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For our final segment, let’s see how we can apply our findings to ages of individuals. What limitations did we set for a group of 9 people?
Ages from 18 to 58!
Correct! So if we pick any 9 people, how many possible non-empty subsets can be formed?
511! Since there are 2^9 subsets minus the empty set.
Exactly! Now, what about the sum of these ages?
The minimum is 18 and the maximum is 522 if everyone is 58!
Perfect! Now, applying the pigeonhole principle, we have 511 subsets as our pigeons and a limited range of sums as our holes. What does that guarantee?
That there must be at least two groups of people whose age sums are equal!
Exactly! And even if any groups share members, we can modify them to ensure they’re disjoint. Great job today everyone! To summarize, we’ve seen the principles of increasing and decreasing subsequences with an impactful real-world application.
Overview
Short Summary
This section explores how to apply the pigeonhole principle to demonstrate the existence of increasing or decreasing subsequences within any selected group of distinct real numbers.
Medium Summary
In this section, we discuss the application of the pigeonhole principle in proving that for any selection of distinct real numbers, there will always exist a subsequence that is either strictly increasing or strictly decreasing. It includes examples and explores specific cases, such as the grouping of ages within given limits, and highlights the significance of both increasing and decreasing subsequences.
Detailed Summary
In this section, we begin with a statement that for any sequence comprised of (n + 1) distinct real numbers, there always exists a subsequence of length (n + 1) that is either strictly increasing or strictly decreasing. Concepts of subsequences are clarified, highlighting that these are not required to be consecutive. The proof employs the pigeonhole principle and contradiction, leading to the conclusion that some elements must share properties that force a subsequence of the required nature to exist. An illustrative example is provided using an age group of people, demonstrating the method of forming disjoint groups to arrive at equal sums of ages. This highlights the essence of the pigeonhole principle in real-world applications.
Reference YouTube Videos
Audio Book
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Create a free accountIn this question we want to show the following: you are arbitrarily picking 9 people in the age group of 18 to 58. That means the minimum age is 18 and the maximum age is 58.
Detailed Explanation
This chunk introduces the context of the problem, which involves choosing 9 individuals from a specified age range. The key point is to understand that each person's age must fall between 18 and 58, inclusive. This sets the boundaries for our age sums when we create groups later.
Examples & Analogies
Imagine you are hosting a birthday party for your friends and they are all aged between 18 and 58. You need to consider everyone's age when planning games or activities – just like we will when forming age groups.
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Create a free accountWe want to prove that irrespective of what exactly are their ages, as long as they are in the range 18 to 58, it is always possible to choose 2 disjoint groups of people out of these 9 people whose sum of ages is the same.
Detailed Explanation
This statement asserts a fundamental idea: no matter how the ages are distributed, two separate groups of the selected individuals can always be formed with equal total ages. This is done using the pigeonhole principle, which states that if you have more items than containers, at least one container must hold more than one item.
Examples & Analogies
Think of a classroom with students of different ages, each age being a container that can hold scores or sums of ages. If you try to create groups from your class marks, you'll often find that two groups yield the same total, just as with ages in this scenario.
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Create a free accountIf I have 9 people, then the number of non-empty groups that I can form out of those 9 people is 511. If I consider the minimum sum of ages possible in a group, it could be 18. The maximum possible sum can occur when all 9 people are 58, resulting in a maximum sum of 522.
Detailed Explanation
The number of subsets (groups) that can be formed from 9 people is calculated as 2^9 - 1 = 511. This accounts for all possible combinations except the empty set. The understanding of minimum and maximum ages helps compute the range of possible sums of ages, which is key in applying the pigeonhole principle.
Examples & Analogies
Imagine you have a box with 9 different fruits. You can pick any combination of these fruits to make different fruit salads, which represents the subsets. The variety in the fruit ages (18 to 58) helps us predict the taste of the salads (sum of ages), just like we are predicting the age sums here.
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Create a free accountMy pigeons are the various possible non-empty sets of people that I can form out of this group of 9 people. My holes are the range of possible sums. Since there are more subsets (511) than possible sums of ages (505), by pigeonhole principle, there must exist at least one pair of groups with equal sums.
Detailed Explanation
Here we define our 'pigeons' as the subsets of people and 'holes' as the sums of ages. With 511 potential groups and only 505 potential distinct age sums, some groups must share the same sum by virtue of the pigeonhole principle. This guarantees that two groups will exist with equal age totals.
Examples & Analogies
It’s like having 511 students trying to fit into only 505 lockers. Since there are more students than lockers, at least two students will have to share a locker, which is like our groups sharing the same age sum.
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Create a free accountBut my question wants me to show that the groups should be disjoint. If the sets are not the same and have some common members, I can remove the common people from both sets; the same amount will be removed from both groups, thus still yielding two groups with equal sums.
Detailed Explanation
This part of the explanation emphasizes how to ensure the groups remain disjoint even after finding two with equal sums. If there are common members in the two groups, removing them won't affect the equality of the sum. This shows that we can still create two completely separate groups that add up to the same number.
Examples & Analogies
Imagine two overlapping teams for a group project. If both teams initially included some members working together, you can easily assign different tasks to different people, ensuring that both teams are distinct while still achieving the same goal (the equal sum of contributions).
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Key concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
- Subsequence:
A selection of elements from a sequence where order is preserved.
- Pigeonhole Principle:
A combinatorial argument that ensures that if there are more items than containers, at least one container must hold more than one item.
- Strictly Increasing/Subsequence:
A sequence with each term larger than the previous term; crucial for proving the existence of specific subsequences.
- Strictly Decreasing/Subsequence:
A sequence where each term is less than the previous; also essential for the proof.
Examples
Memory aids
Imagine a classroom of distinct students, each standing in a line. Some are taller, some shorter. If there are more students than heights available, there must be a repeat in a height category!
To remember increasing subsequences, think 'Ladder': Each rung gets higher, never lower.
Flash Cards
Glossary
Strictly Increasing Sequence
A sequence where each element is larger than the one preceding it.
Strictly Decreasing Sequence
A sequence where each element is smaller than the one preceding it.
Subsequence
A sequence derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
Pigeonhole Principle
A principle stating that if more items are put into fewer containers than there are items, at least one container must hold more than one item.
Contradiction
A logical inconsistency, where a statement contradicts another; often used in proofs.