Math Problem Statement
Exam 2.pdf
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show a counter example for 1a
Solution
To find a counterexample for part 1a, let’s examine why this structure fails to be a vector space. We need to check if this modified version of (with the non-standard addition defined as ) fails any of the vector space axioms.
One of the critical axioms of a vector space is that addition must be commutative:
With the given definition of addition, however, we can find a counterexample:
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Select two vectors in , say:
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Compute using the given definition:
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Compute using the same addition rule:
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Observe the result:
Since , this addition operation is not commutative, which violates the axioms required for a vector space. Therefore, this serves as a counterexample to show that the given set with this addition is not a vector space.
Do you have any questions or would you like more details? Here are some related questions to expand on this topic:
- What other vector space axioms could be checked for failure in this structure?
- Could an alternative definition of addition make this a vector space?
- How would scalar multiplication be affected if a different addition rule is chosen?
- Can you provide examples of common vector spaces that satisfy all vector space properties?
- What is the impact of non-standard operations on the structure of mathematical objects like vector spaces?
Tip: When verifying if a set and operations form a vector space, always begin by checking the axioms of vector addition and scalar multiplication, as these often reveal whether the structure can indeed be classified as a vector space.
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Math Problem Analysis
Mathematical Concepts
Vector Spaces
Counterexamples
Non-standard Operations
Formulas
Addition operation for \(\mathbb{R}^2\): \(\begin{bmatrix} a \\ b \end{bmatrix} + \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} a - c \\ b - d \end{bmatrix}\)
Theorems
Vector Space Axioms
Commutative Property of Addition
Suitable Grade Level
Undergraduate Level
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