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that there exist a, b,erd EA Such thant a tb = ctd. la,b) # l,d) and %3D

that there exist a, b,erd EA Such thant a tb = ctd. la,b) # l,d) and %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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this is the solution for this question. Actually I do not understand at all. Please explain it to me. How and why do we need define function in this way？ Whats the meaning of A x A to solve this problem？ How can we relate this question to pigeonhole principle.
Sure, here is the transcription of the handwritten text that appears in the image: --- **Problem** 1. Suppose that \( A = \{1, 2, \ldots, 12\} \) with \(|A| \geq 25\). Show that there exist \( a, b, c, d \in A \) such that \((a, b) \neq (c, d)\) and \( a + b = c + d \). **Solution** 1. Define \( f : A \times A \rightarrow \{2, 3, \ldots, 24\} \) by \( f(a, b) = a + b \) for \( a, b \in A \). \( f \) is well-defined because \( a, b \in A \). The sum \( 2 = 1+1 \leq a + b \leq 12 + 12 = 24 \). \(|A \times A| = 12^2 = 144 \geq 25\), \( f \) is not injective by the pigeonhole principle. Therefore, there exist \( a, b, c, d \in A \) such that \( a \neq c \) or \( b \neq d \) and \( f(a, b) = f(c, d) \), i.e., \( a + b = c + d \). --- This transcription is suitable for an educational website detailing a solution that uses the pigeonhole principle to show the existence of distinct pairs in a set that sum to the same value.
Transcribed Image Text:Sure, here is the transcription of the handwritten text that appears in the image: --- **Problem** 1. Suppose that \( A = \{1, 2, \ldots, 12\} \) with \(|A| \geq 25\). Show that there exist \( a, b, c, d \in A \) such that \((a, b) \neq (c, d)\) and \( a + b = c + d \). **Solution** 1. Define \( f : A \times A \rightarrow \{2, 3, \ldots, 24\} \) by \( f(a, b) = a + b \) for \( a, b \in A \). \( f \) is well-defined because \( a, b \in A \). The sum \( 2 = 1+1 \leq a + b \leq 12 + 12 = 24 \). \(|A \times A| = 12^2 = 144 \geq 25\), \( f \) is not injective by the pigeonhole principle. Therefore, there exist \( a, b, c, d \in A \) such that \( a \neq c \) or \( b \neq d \) and \( f(a, b) = f(c, d) \), i.e., \( a + b = c + d \). --- This transcription is suitable for an educational website detailing a solution that uses the pigeonhole principle to show the existence of distinct pairs in a set that sum to the same value.
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