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Prove that there is only one vector that satisfies the condition of the additive identity in Axiom 3 .

Short Answer

Expert verified
The theorem states that there is only one additive identity in a vector space. When attempting to assume there are two (0 and 0'), it can be deduced that 0 equals 0', meaning they are not distinct. Thus, there can only be one additive identity in a vector space.

Step by step solution

01

Theorem Statement and Assumptions

Theorem: All vector spaces have one and only one additive identity vector, denoted as 0. Assume to the contrary that there are two additive identity vectors in some vector space, denoted as 0 and 0'.
02

Deduction

To prove that 0 and 0' are not distinct, examine the sum 0 + 0'. According to the properties of 0, any vector added to 0 is the vector itself. Therefore, 0 + 0' equals 0'. However, also by the properties of 0', 0 + 0' equals 0.
03

Conclusion

Based on the deduction, 0 equals 0'. Therefore, by contradiction there can only be one additive identity in a vector space, so the initial assumption of having two distinct additive identities is incorrect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Additive Identity
In the study of vector spaces, the **additive identity** is a crucial concept. It represents a unique element within the space that, when added to any vector in that space, results in the same vector. This element is often denoted by the symbol 0. For example, in the vector space of all two-dimensional vectors, the additive identity is the vector \((0, 0)\). The essential property of the additive identity, 0, can be defined mathematically as follows: for any vector \(\mathbf{v}\) in a vector space, \(\mathbf{v} + 0 = \mathbf{v}\). This demonstrates its role as a neutral element in addition.
It is important to understand that the additive identity must exist in every vector space. Without it, the fundamental operations and properties we rely on would not hold true. This identity helps in establishing core theorems and lemmas crucial to the study of vector spaces.
Axiom 3
Axiom 3 is a foundational rule in the context of vector spaces that speaks specifically about the existence of an additive identity. This axiom is part of a set of axioms required for a structure to be called a vector space. Specifically, it states that there exists a vector, denoted as 0, such that for any vector \(\mathbf{v}\) in the vector space, adding 0 to \(\mathbf{v}\) leaves \(\mathbf{v}\) unchanged. In mathematical terms, \(\mathbf{v} + 0 = \mathbf{v}\) for any vector \(\mathbf{v}\).
Axiom 3 reinforces the idea of reversibility or neutrality in addition, asserting that there is an insinuated but necessary element (0) which assures that the sum of the two retains the integrity of the original vector. This axiom is crucial as it ensures the consistency and predictability of operations within a vector space.
Theorem Proof
The **proof of a theorem** often involves demonstrating that certain properties or conditions hold true under given assumptions. In this case, the theorem is about the uniqueness of the additive identity in a vector space. We start by making an assumption for the sake of argument. Let's assume there are two different additive identities denoted as 0 and 0'.
The proof involves examining the consequences of this assumption. By definition, the additive identity property states that any vector added to an additive identity results in the vector itself. Thus, for two potential identities: when we add 0 to 0', the result should be 0', and similarly, when we add 0 to 0', the result should also be 0. These simultaneous equalities lead us logically to conclude that 0 must equal 0', confirming that our initial assumption is incorrect. This concludes the proof, supporting the theorem that there can be only one unique additive identity in any vector space.
Contradiction Method
The **contradiction method** is a powerful proof technique used to demonstrate the truth of a statement by initially assuming the opposite. In the realm of mathematics, it's a strategy employed to show that a supposition, different from what you are trying to prove, leads to a contradiction.
For the additive identity theorem, we initially assumed there exist two distinct additive identities in a vector space. If this were true, both would satisfy the conditions of being an additive identity, but this leads to contradictory outcomes (i.e., 0 + 0' = 0 and 0 + 0' = 0').
This inconsistency confirms our initial assumption is false or flawed, thereby affirming that there can only be one additive identity. Proof by contradiction is effective because it clearly illustrates that different assumptions may disrupt established truths, which in turn supports the correctness of the proposition being proven.

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Most popular questions from this chapter

a. What is the smallest subset of \(\mathbb{R}\) that contains \(\frac{1}{2}\) and is closed under addition? b. What is the smallest subset of \(R\) that contains \(\frac{1}{2}\) and is closed under multiplication?

Determine whether we obtain a vector space from the following subset of \(\mathbb{R}^{2}\) with the standard operations: $$ S=\left\\{\left(v_{1}, v_{2}\right) \in \mathbb{R}^{2} \mid v_{1} \text { and } v_{2} \text { are integers }\right\\} . $$ Here the identities stated in the eight axioms are not particularly in doubt. The real question is whether the definitions apply in as much generality as they should. The definition of a vector space requires that when we add together any pair of vectors or multiply any vector by any real number, the result is an element of the set. In the terminology of Section 1.1, the set must be closed. under addition and scalar multiplication.

Prove that \(\frac{1}{2} \mathbf{v}+\frac{1}{2} \mathbf{v}=\mathbf{v}\)

Write the plane in \(\mathbb{R}^{3}\) that is tangent to the unit sphere at the point \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) in the form \(\\{r \mathbf{v}+s \mathbf{w}+\mathbf{x} \mid r, s \in \mathbb{R}\\}\). (Suggestion: To find direction vectors, use the fact that the plane intersects the three coordinate axes at points that are equidistant from the origin.)

Is it possible to add two sets? For sets of real numbers, we might define addition of sets in terms of addition of the elements in the sets. Let us introduce the following meaning to the symbol \(\oplus\) for adding a set \(A\) of real numbers to another set \(B\) of real numbers: \(A \oplus B=\\{a+b \mid a \in A\) and \(b \in B\\}\). a. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,10\\}\). b. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,6\\}\). c. List the elements in the set \(\\{1,2,3\\} \oplus \varnothing\). d. If a set \(A\) contams \(m\) real numbers and a set \(B\) contains \(n\) real numbers, can you predict the number of elements in \(A \oplus B\) ? If you run into difficulties, can you determine the minimum and naximum numbers of elements possible in \(A \oplus B\) ? e. Does the commutative law hold for this new addition? That is, does \(A \oplus\) \(B=B \oplus A\) ? f. Reformulate other laws of real-number addition in terms of this new addition of sets. Which of your formulas are true? Can you prove them or provide counterexamples? g. What about laws that combine set addition with union and intersection? For example, does \((A \cup B) \oplus C=(A \oplus C) \cup(B \oplus C)\) ? h. Is there any hope of extending the other operations of arithmetic to sets of real numbers? What about algebra? Limits? Power series?

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