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(=) Assume I + a = I + b. .. a El + a :: a El + b : a = i + b for some i EI .. i = a – b .. a – bE I (E) Assume a - bEI Assume x E I + a .: x = i + a for some i EI . x = i + a + (-b+ b) : x = (i + a – b) + b .. x E I + b .. I + a CI + b Assume x E I + b .: x = i + b for some i E I :: x = i +b + (-a + b) .x = (i + b – a) + a .. x EI + a .. I +b CI+a .: I + a = 1 + b

(=) Assume I + a = I + b. .. a El + a :: a El + b : a = i + b for some i EI .. i = a – b .. a – bE I (E) Assume a - bEI Assume x E I + a .: x = i + a for some i EI . x = i + a + (-b+ b) : x = (i + a – b) + b .. x E I + b .. I + a CI + b Assume x E I + b .: x = i + b for some i E I :: x = i +b + (-a + b) .x = (i + b – a) + a .. x EI + a .. I +b CI+a .: I + a = 1 + b

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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Justify each claim of the proof
Certainly! Here's a transcription and explanation of the content: --- **Proof of Set Equivalence** **Forward Direction (⇒):** Assume \( I + a = I + b \). 1. \( a \in I + a \) 2. Therefore, \( a \in I + b \) 3. Hence, \( a = i + b \) for some \( i \in I \) 4. Let \( i = a - b \) 5. So, \( a - b \in I \) **Backward Direction (⇐):** Assume \( a - b \in I \). 1. Assume \( x \in I + a \) 1. Thus, \( x = i + a \) for some \( i \in I \) 2. \( x = i + a + (-b + b) \) 3. Therefore, \( x = (i + a - b) + b \) 4. Hence, \( x \in I + b \) 5. Conclusion: \( I + a \subseteq I + b \) 2. Assume \( x \in I + b \) 1. Thus, \( x = i + b \) for some \( i \in I \) 2. \( x = i + b + (-a + a) \) 3. Therefore, \( x = (i + b - a) + a \) 4. Hence, \( x \in I + a \) 5. Conclusion: \( I + b \subseteq I + a \) Therefore, \( I + a = I + b \). --- This text demonstrates a proof involving set theory and algebra, showing the equivalence of \( I + a \) and \( I + b \) under certain assumptions. The proof is divided into two parts: proving the forward direction and the backward direction. Each direction uses set membership and algebraic manipulation to establish the equivalence.
Transcribed Image Text:Certainly! Here's a transcription and explanation of the content: --- **Proof of Set Equivalence** **Forward Direction (⇒):** Assume \( I + a = I + b \). 1. \( a \in I + a \) 2. Therefore, \( a \in I + b \) 3. Hence, \( a = i + b \) for some \( i \in I \) 4. Let \( i = a - b \) 5. So, \( a - b \in I \) **Backward Direction (⇐):** Assume \( a - b \in I \). 1. Assume \( x \in I + a \) 1. Thus, \( x = i + a \) for some \( i \in I \) 2. \( x = i + a + (-b + b) \) 3. Therefore, \( x = (i + a - b) + b \) 4. Hence, \( x \in I + b \) 5. Conclusion: \( I + a \subseteq I + b \) 2. Assume \( x \in I + b \) 1. Thus, \( x = i + b \) for some \( i \in I \) 2. \( x = i + b + (-a + a) \) 3. Therefore, \( x = (i + b - a) + a \) 4. Hence, \( x \in I + a \) 5. Conclusion: \( I + b \subseteq I + a \) Therefore, \( I + a = I + b \). --- This text demonstrates a proof involving set theory and algebra, showing the equivalence of \( I + a \) and \( I + b \) under certain assumptions. The proof is divided into two parts: proving the forward direction and the backward direction. Each direction uses set membership and algebraic manipulation to establish the equivalence.
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