Mathematics, 24.06.2020 21:01, sophoyak
Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text.
(a) To show that a set is not a vector space, it is sufficient to show that just one axiom is not satisfied.
1. True, for a set with two operations to be a vector space all 10 axioms must be satisfied. Therefore, if just one of the axioms fails, then this set cannot be a vector space.
2. False, for a set with two operations to be a vector space a majority of 10 axioms must be satisfied. Therefore, if just one of the axioms fails, then this set can still be a vector space.
3. False, for a set with two operations to be a vector space at least one of the 10 axioms must be satisfied. Therefore, if just one of the axioms fails, then this set can still be a vector space.
4. False, for a set with two operations to be a vector space at least 9 of the 10 axioms must be satisfied. Therefore, if just one of the axioms fails, then this set can still be a vector space.
5. False, for a set with two operations to be a vector space at least 8 of the 10 axioms must be satisfied. Therefore, if just one of the axioms fails, then this set can still be a vector space.
(b) The set of all first-degree polynomials with the standard operations is a vector space.
1. True. The set is a vector space because all 10 axioms are satisfied.
2. False. The set is not a vector space because it is not closed under addition.
3. False. The set is not a vector space because it does not have the commutative property of addition.
4. False. The set is not a vector space because it does not have the associative property of addition.
5. False. The set is not a vector space because a scalar identity does not exist.
(c) The set of all pairs of real numbers of the form (0, y), with the standard operations on R^2, is a vector space.
1. True. The set is a vector space because all 10 axioms are satisfied.
2. False. The set is not a vector space because it is not closed under addition.
3. False. The set is not a vector space because an additive inverse does not exist.
4. False. The set is not a vector space because it is not closed under scalar multiplication.
5. False. The set is not a vector space because a scalar identity does not exist.
Answers: 3
Mathematics, 21.06.2019 18:00, woebrooke11
Me, prove a quadrilateral with vertices g(1,-1), h(5,1), i(4,3) and j(0,1) is a rectangle using the parallelogram method and a rectangle method.
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Mathematics, 21.06.2019 19:30, sk9600930
Sundar used linear combination to solve the system of equations shown. he did so by multiplying the first equation by 5 and the second equation by another number to eliminate the y-terms. what number did sundar multiply the second equation by? 2x+9y=41 3x+5y=36
Answers: 1
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