Download Solutions

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