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Solutions to Tutorial 2 Simon Rose March 21, 2013 We will only provide solutions to problems 3, 5 from the lecture notes (“Examples (c), (f)”) due to our not having covered the vector space FX (R; R). Problem 3: We are considering (R≥0 , +, ·). Let us run through the list of axioms to see which fails first. (a) The operation + is associative. This is true, since addition in this case is simply that from the real numbers, which we know to be associative. (b) There is a zero vector. This is also true, since 0 ∈ R≥0 , and this is the usual zero element for addition. (c) There exist additive inverses. This fails: recall that we want, for each x ∈ R≥0 some element y ∈ R≥0 such that x + y = 0. However, these would be the usual additive inverses, i.e. the negative real numbers! Specifically, consider (for example) 1 ∈ R≥0 . Then the only number which satisfies 1+y =0 is y = −1, but −1 ∈ / R≥0 . Thus since at least one additive inverse does not exist (that’s all we need!) this can not be a vector space. Note that (almost) all of the other axioms are true! The only place this fails otherwise is that it is not true that scalar multiplication is well defined: recall that this is a function · : R × R≥0 → R≥0 However, (−1) · x < 0 if x ≥ 0, and so this does not lie in the target space of this function. Problem 5: We are looking at the space ([−1, 1], +, ·). Let us run through the list. (a) Associativity of addition: True* (b) Existence of zero: True (c) Existence of additive inverses: True (d) Addition is commutative: True* 1 (e) Multiplication is associative: True* (f) Multiplication is distributive (version 1): True* (g) Multiplication is distributive (version 2): True* (h) Multiplication by 1 does nothing: True. So is this a vector space? It seems to satisfy all of the conditions... The answer is no! All of the conditions marked * are not really true, since they don’t even make sense! Similar to the previous problem, neither addition nor multiplication are well defined at all: Consider 1 + 1 =? ∈ [−1, 1] Obviously this lies outside of that set, so addition isn’t even an operation on this set. So despite all of the axioms being “true”, this is not a vector space. A similar argument, it should be noted, can be made regarding multiplication. 2