DLA Examples

Section 16.5 Examples

In this section.

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.

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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.

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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,

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

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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.

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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,

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

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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.

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Subsection 16.5.2 Verifying axioms: the space of functions

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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.

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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.

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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.

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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,

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

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with justifications

definition of the sum of $f$ and $g+h\text{;}$
definition of the sum of $g$ and $h\text{;}$
definition of the sum of $f+g$ and $h\text{;}$ and
definition of the sum of $f$ and $g\text{.}$

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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.

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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,

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

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as desired, with justifications

definition of the sum of $f$ and $-f\text{;}$
definition of the negative of $f\text{;}$
algebra of numbers;
algebra of numbers; and
definition of the zero function.

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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.

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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,

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

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with justifications

definition of scalar multiplication of $f$ by $k+m\text{;}$
definition of the sum of $kf$ and $mf\text{;}$ and
definition of scalar multiplication of $f$ by $k$ and by $m\text{.}$

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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.