What Are the Zero and Unit Elements in R^2 for Proving Vector Space Axioms?

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In R^2, the zero element is represented as the vector (0, 0), which satisfies the condition for vector addition, where any vector added to (0, 0) remains unchanged. The unit element, however, is not represented as the vector (1, 1); instead, it refers to the scalar 1 used in scalar multiplication, where 1 multiplied by any vector yields the same vector. To prove that R^2 is a vector space, one must demonstrate that all eight vector space axioms hold true. The discussion emphasizes that understanding the distinction between vector and scalar elements is crucial for proving vector space properties. Therefore, confirming the axioms is sufficient to establish R^2 as a vector space.
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Let R^2 be a set containing all possible rows: (a b)

when using the 8 axioms to prove whether (a,b) is indeed a vector space, i have to show that there is a zero element and a unit element.

Is the zero element 0? or is it in matrix form such that W = (0 0) and W is contained in R^2?

Is the unit element 1? or is it in matrix form such that F=(1 1) and F is contained in R^2?

If I showed the 8 axioms are true, then does that show that R^2 is indeed a vector space?
 
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Your right about (0 0) and (1 1).
And yes just show the 8 axioms hold.
 
Last edited:
ircdan said:
Your right about (0 0) and (1 1).
And yes just show the 8 axioms hold.
No, he's not right about (1, 1).

indigojoker, you shouldn't even have to think about that. The "zero element" acts like 0: x+ 0= 0 in the VECTOR addition. If you are adding vectors the 0 has to be a vector: (0, 0). On the other hand, scalar multiplication involves multiplying a scalar by a vector: in "1v= v", the "1" is a number, not a vector.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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