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Is the collection of irrational numbers R\ Qa field? Why or why not?

Is the collection of irrational numbers R\ Qa field? Why or why not?

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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**Is the collection of irrational numbers \(\mathbb{R} \setminus \mathbb{Q}\) a field? Why or why not?** In order for a set to be considered a field, it must satisfy several properties under two operations: addition and multiplication. It must be closed under these operations, have an additive identity (0) and a multiplicative identity (1), have additive inverses for every element, and every nonzero element must have a multiplicative inverse. The collection of irrational numbers \(\mathbb{R} \setminus \mathbb{Q}\) does not form a field because it fails to satisfy the closure property under addition and multiplication. For example, the sum or product of two irrational numbers can be rational: - \( \sqrt{2} \cdot \sqrt{2} = 2 \), which is a rational number. - A specific example where an addition results in a rational number is \( (\sqrt{2} - \frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} = \sqrt{2} \), which should remain irrational, but does not showcase closure directly without specific examples resulting in rationals naturally. Hence, \(\mathbb{R} \setminus \mathbb{Q}\) does not satisfy the field properties, primarily due to lack of closure under these operations.
Transcribed Image Text:**Is the collection of irrational numbers \(\mathbb{R} \setminus \mathbb{Q}\) a field? Why or why not?** In order for a set to be considered a field, it must satisfy several properties under two operations: addition and multiplication. It must be closed under these operations, have an additive identity (0) and a multiplicative identity (1), have additive inverses for every element, and every nonzero element must have a multiplicative inverse. The collection of irrational numbers \(\mathbb{R} \setminus \mathbb{Q}\) does not form a field because it fails to satisfy the closure property under addition and multiplication. For example, the sum or product of two irrational numbers can be rational: - \( \sqrt{2} \cdot \sqrt{2} = 2 \), which is a rational number. - A specific example where an addition results in a rational number is \( (\sqrt{2} - \frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} = \sqrt{2} \), which should remain irrational, but does not showcase closure directly without specific examples resulting in rationals naturally. Hence, \(\mathbb{R} \setminus \mathbb{Q}\) does not satisfy the field properties, primarily due to lack of closure under these operations.
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