MaryAnn said:
I am reading a chapter section on Ordered Field that starts off with the additive and multiplicative properties:
To my untrained eyes, they do not mean anything at all. Could somebody therefore give an intuitive significance of the two properties, perhaps with examples - please. Are they about closure?
Thank you for your time and gracious helps. ~MA
Hey MaryAnn,
Yes, they are about closure. And they include the substitution-property of the equality-relation although I'm not exactly sure why. I haven't seen that before. Usually the substitution-property is part of the separate definition of the
equality-relation and can then just be used everywhere, including as it is used here in these properties about closure.
We're starting with a set that may contain literally anything, could be symbols that resemble numbers, could be apples and pears.
Then we add the required structure to it to be able to call it an Ordered Field.
First thing to add the operations
addition and
multiplication.
Typically we
define for instance that $1+1=2$ and that $2\times 2=4$.
If the set contains apples and pears, we have to define what the result is if we add an apple and a pear together.
That is, if we want to call it an Ordered Field.
Closure means that it doesn't suffice to just say that $1+1=2$.
Instead we have to define addition for every combination of 2 elements that are in the set.
And the result must be in the set as well.
Suppose we have the 2-element set $\{0, 1\}$.
What does it take to be able to call it and Ordered Field?
Well, first off we need to define addition, and ensure that it is closed under addition.
One way to define addition is with an addition table.
For instance:
$$\begin{array}{c|cc}+&0&1 \\
\hline
0 & 1 & 1 \\
1 & 0 & 0
\end{array}$$
There you go, now for any $x$ and $y$ in the set, the addition $x+y$ is defined and is included in the set.
As for the equality-substitution-property, the only way that it can fail is if $0=1$.
But then our set would only contain 1 element, which contradicts that we started with a 2-element set.
Therefore the equality-substitution-property is also satisfied.
Next we define multiplication, say with the table:
$$\begin{array}{c|cc}\times &0&1 \\
\hline
0 & 0 & 1 \\
1 & 1 & 1
\end{array}$$
Now for any $x$ and $y$ in the set, the multiplication $x\times y$ is defined and is included in the set.
And the equality-substitution-property is satisified for the same reason as before.
