Linear Algebra **Problem 7: Vector Space Verification****Objective:** Demonstrate why the set $\mathbb{Z}$ of integers is NOT a vector space.To prove that $\mathbb{Z}$ is not a vector space, we need to check whether it satisfies the axioms required for a vector space over a field, typically the field of real numbers $\mathbb{R}$.### Key Vector Space Axioms:1. **Closure under addition:** For any two integers $a, b \in \mathbb{Z}$, the sum $a + b$ should also be in $\mathbb{Z}$, which holds true.2. **Existence of additive identity:** There exists an element $0 \in \mathbb{Z}$ such that $a + 0 = a$ for all $a \in \mathbb{Z}$, which also holds true.3. **Existence of additive inverses:** For each $a \in \mathbb{Z}$, there exists an element $-a \in \mathbb{Z}$ such that $a + (-a) = 0$, which holds true.4. **Closure under scalar multiplication:** For a scalar $c \in \mathbb{R}$ and an integer $a \in \mathbb{Z}$, the product $c \cdot a$ must also be in $\mathbb{Z}$. This axiom fails because, for example, if $c = 0.5$ (a real number) and $a = 2$, then $c \cdot a = 1$, which is not an integer.5. **Other axioms** regarding associativity, distributivity, etc., are irrelevant if closure under scalar multiplication fails.### Conclusion:Since the set of integers $\mathbb{Z}$ fails to satisfy closure under scalar multiplication, it does not meet all the criteria to be a vector space. Thus, $\mathbb{Z}$ is NOT a vector space.

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Answered: 7. Show that the set Z of integers is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Linear Algebra

**Problem 7: Vector Space Verification**

**Objective:** Demonstrate why the set $\mathbb{Z}$ of integers is NOT a vector space.

To prove that $\mathbb{Z}$ is not a vector space, we need to check whether it satisfies the axioms required for a vector space over a field, typically the field of real numbers $\mathbb{R}$.

### Key Vector Space Axioms:
1. **Closure under addition:** For any two integers $a, b \in \mathbb{Z}$, the sum $a + b$ should also be in $\mathbb{Z}$, which holds true.
2. **Existence of additive identity:** There exists an element $0 \in \mathbb{Z}$ such that $a + 0 = a$ for all $a \in \mathbb{Z}$, which also holds true.
3. **Existence of additive inverses:** For each $a \in \mathbb{Z}$, there exists an element $-a \in \mathbb{Z}$ such that $a + (-a) = 0$, which holds true.
4. **Closure under scalar multiplication:** For a scalar $c \in \mathbb{R}$ and an integer $a \in \mathbb{Z}$, the product $c \cdot a$ must also be in $\mathbb{Z}$. This axiom fails because, for example, if $c = 0.5$ (a real number) and $a = 2$, then $c \cdot a = 1$, which is not an integer.
5. **Other axioms** regarding associativity, distributivity, etc., are irrelevant if closure under scalar multiplication fails.

### Conclusion:
Since the set of integers $\mathbb{Z}$ fails to satisfy closure under scalar multiplication, it does not meet all the criteria to be a vector space. Thus, $\mathbb{Z}$ is NOT a vector space.

Branch of mathematics concerned with mathematical structures that are closed under operations like addition and scalar multiplication. It is the study of linear combinations, vector spaces, lines and planes, and some mappings that are used to perform linear transformations. Linear algebra also includes vectors, matrices, and linear functions. It has many applications from mathematical physics to modern algebra and coding theory.

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