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Section 16.5 Examples

Subsection 16.5.1 Verifying axioms: the space of positive numbers

Here we will continue our “weird” example from the end of Subsection 16.4.2, and verify some of the other axioms for vectors in that space.

Let's start with Axiom A 2. Here we would like to verify that the vector equality \(\uvec{v}+\uvec{w} = \uvec{w}+\uvec{v}\) is always true when the vectors are positive numbers and vector addition is defined to be ordinary multiplication, as defined in (\(\dagger\)) in Subsection 16.4.2.

For this space, vectors are positive numbers, so we should take \(\uvec{v} = (a)\) and \(\uvec{w} = (b)\) for arbitrary, unspecified positive numbers \(a\) and \(b\) (where again we use brackets to distinguish between numbers that are vectors and numbers that are scalars). Then,

\begin{align*} \text{LHS} \amp= \uvec{v} + \uvec{w} \amp \text{RHS} \amp= \uvec{w} + \uvec{v}\\ \amp= (a) \oplus (b) \amp \amp= (b) \oplus (a)\\ \amp= (ab), \amp \amp= (ba). \end{align*}

Now, we know that ordinary multiplication of numbers can be performed in either order, so \(ba=ab\text{,}\) and thus \(\text{LHS}=\text{RHS}\) as desired.

We will leave the other addition axioms up to you, but let's verify one of the scalar multiplication axioms. Consider Axiom S 2. We need to verify that \(k(\uvec{v}+\uvec{w}) = k\uvec{v}+k\uvec{w}\) is always true for all scalars \(k\) and all vectors \(\uvec{v}\) and \(\uvec{w}\text{,}\) where scalar multiplication is defined as in (\(\dagger\dagger\)) in Subsection 16.4.2. When considering the left- and right-hand sides of this vector equality, we need to be sure to pay attention to the order of operations on each side. Again, take \(\uvec{v} = (a)\) and \(\uvec{w} = (b)\) for arbitrary, unspecified positive numbers \(a\) and \(b\text{.}\) Then,

\begin{align*} \text{LHS} \amp= k(\uvec{v} + \uvec{w}) \amp \text{RHS} \amp= k\uvec{v} + k\uvec{w}\\ \amp= k\odot\bbrac{(a) \oplus (b)} \amp \amp= \bbrac{k\odot(a)} \oplus \bbrac{k\odot(b)}\\ \amp= k\odot(ab) \amp \amp= \left(a^k\right) \oplus \left(b^k\right)\\ \amp= \left((ab)^k\right), \amp \amp= \left(a^k b^k\right). \end{align*}

We can now see that \(\text{LHS}=\text{RHS}\) as desired because of the exponent law \((ab)^k = a^k b^k\) from the algebra of ordinary numbers.

Subsection 16.5.2 Verifying axioms: the space of functions

Here we will verify some of the axioms for vectors in the space \(F(D)\text{.}\) We will be verifying equality of functions, so we need to make sure we know what it means for two functions to be equal.

Definition 16.5.1. Equality of functions.

Two functions are equal when the input-output processes they represent always produce the same outputs. That is, functions \(f\) and \(g\) are equal if \(f(x)=g(x)\) for all possible input \(x\)-values.

Let's start with Axiom A 3. Here we would like to verify that the vector equality \(\uvec{u} + (\uvec{v}+\uvec{w}) = (\uvec{u} + \uvec{v}) + \uvec{w}\) is always true when the vectors are functions and addition is defined in \(F(D)\) as in (\(\star\)) in Subsection 16.4.2.

Let's take \(\uvec{u} = f\text{,}\) \(\uvec{v} = g\text{,}\) and \(\uvec{w} = h\text{,}\) where \(f,g,h\) are arbitrary, unspecified functions that are all defined on the domain \(D\text{.}\) From Definition 16.5.1 above, we see that we need to verify that the sum functions \(f+(g+h)\) and \((f+g)+h\) always produce the same output when fed the same input. So suppose \(x\) is an input value in the domain \(D\text{.}\) Then,

\begin{align*} \text{LHS} \amp= \bbrac{f+(g+h)}(x)\\ \amp= f(x) + (g+h)(x) \amp \amp\text{(i)}\\ \amp= f(x) + \bbrac{g(x) + h(x)}, \amp \amp\text{(ii)}\\ \\ \text{RHS} \amp= \bbrac{(f+g)+h}(x)\\ \amp= (f+g)(x) + h(x) \amp \amp\text{(iii)}\\ \amp= \bbrac{f(x)+g(x)} + h(x), \amp \amp\text{(iv)} \end{align*}

with justifications

  1. definition of the sum of \(f\) and \(g+h\text{;}\)
  2. definition of the sum of \(g\) and \(h\text{;}\)
  3. definition of the sum of \(f+g\) and \(h\text{;}\) and
  4. definition of the sum of \(f\) and \(g\text{.}\)

Now, \(f(x),g(x),h(x)\) are just numbers — they are the output \(y\)-values produced by the functions from the input value \(x\) — and we know that we can group numbers with brackets in any combination when adding. So \(\text{LHS}=\text{RHS}\) as desired.

Now let's verify Axiom A 5, using the definition \((-f)(x) = -f(x)\) from Subsection 16.4.2 (where we have also defined \(\zerovec(x) = 0\)). We must verify that the sum function \(f+(-f)\) is the same as the zero function \(\zerovec\text{,}\) which means we must verify that these functions always have the same outputs. So suppose \(x\) is an input value in the domain \(D\text{.}\) Then,

\begin{align*} \text{LHS} \amp= \bbrac{f+(-f)}(x)\\ \amp= f(x) + (-f)(x) \amp \amp\text{(i)}\\ \amp= f(x) + \bbrac{-f(x)} \amp \amp\text{(ii)}\\ \amp= f(x) - f(x) \amp \amp\text{(iii)}\\ \amp= 0 \amp \amp\text{(iv)}\\ \amp= \zerovec(x) \amp \amp\text{(v)}\\ \amp= \text{RHS}, \end{align*}

as desired, with justifications

  1. definition of the sum of \(f\) and \(-f\text{;}\)
  2. definition of the negative of \(f\text{;}\)
  3. algebra of numbers;
  4. algebra of numbers; and
  5. definition of the zero function.

As a last example, let's verify Axiom S 3 in this space. We need to verify that \((k+m)\uvec{v} = k\uvec{v}+m\uvec{v}\) is always true for all scalars \(k\) and \(m\) and all vectors \(\uvec{v}\) when the vectors are functions and the vector operations are defined in \(F(D)\) as in (\(\star\)) and (\(\star\star\)) in Subsection 16.4.2. And when considering the left- and right-hand sides of this vector equality, we need to be sure to pay attention to the order of operations on each side.

Again, take \(\uvec{v} = f\) for arbitrary, unspecified function \(f\text{.}\) Then we actually need to verify that the function \((k+m)f\) always produces the same outputs as the function \(kf+mf\text{.}\) So suppose \(x\) is an input value in the domain \(D\text{.}\) Then,

\begin{align*} \text{LHS} \amp= \bbrac{(k+m)f}(x)\\ \amp= (k+m)f(x), \amp \amp\text{(i)}\\ \\ \text{RHS} \amp= (kf+mf)(x)\\ \amp= (kf)(x) + (mf)(x) \amp \amp\text{(ii)}\\ \amp= kf(x) + mf(x), \amp \amp\text{(iii)} \end{align*}

with justifications

  1. definition of scalar multiplication of \(f\) by \(k+m\text{;}\)
  2. definition of the sum of \(kf\) and \(mf\text{;}\) and
  3. definition of scalar multiplication of \(f\) by \(k\) and by \(m\text{.}\)

Again, \(f(x)\) is just a number, and \(k,m\) are also numbers, and we know that we can distribute the multiplication of \(f(x)\) across the sum of \(k\) and \(m\) in the expression \((k+m)f(x)\text{.}\) Therefore, \(\text{LHS}=\text{RHS}\) as desired.