If \(a\), \(b\), and \(c\) are any numbers, then $$a + (b + c)
= (a + b) + c. $$
Using P1, you can derive the same rules for all higher cardinality
sums of numbers. Here are all the expansions of cardinality 4:
$$((a + b) + c) + d =\\
(a + (b + c)) + d =\\
a + ((b + c) + d) =\\
a + (b + (c + d)) =\\
(a + b) + (c + d)$$
The most important thing to take from the above is that we can
reapply this procedure to show that any cardinality of sums can be
substituted down to property \(a + (b + c) = (a + b) + c\). For
instance the 4 cardinality sum of \((d + (e + f) + g)\) is really
just \(d = a\), \((e + f) = b\) and \(g = c\).
For any cardinality-N of sums, it would seem to rely on
having the cardinality-(N-1) of sums defined in order to have a
pattern that you can substitute into?
What is this grouping thing (the parentheses)? How do I
know that I can freely add and remove parentheses around numbers?
Where was grouping defined?
Why is it that grouping only makes sense at cardinality 3
and greater sums?
If \(a\) is any number, then $$ a + 0 = 0 + a = a. $$
For every number \(a\), there is a number \(-a\) such that
$$a + (-a) = (-a) + a = 0. $$
It is then stated that if \(a + x = a\), then \(x = 0\). The proof,
$$
a + x = a\\
(-a) + (a + x) = (-a) + a;\\
(-a) + (a + x) = 0;\\
((-a) + a) + x = 0;\\
0 + x = 0;\\
x = 0.
$$
I would like to be able to have some sort of mouseover for
each line of the proof. So that I could mouseover a line and it
would give me a little message like "P2 was applied" or "(-a) was
added to both side". How to do that?
I would like to have all the equals signs aligned vertically
in the above proof. How do I say that with Latex?
If \(a\) and \(b\) are any numbers, then
$$ a + b = b + a. $$
If \(a\), \(b\), and \(c\) are any numbers, then
$$ a \cdot (b \cdot c) = (a \cdot b) \cdot c.$$
If \(a\) is any number, then
$$ a \cdot 1 = 1 \cdot a = a\\
Moreover; 1 \neq 0$$
I do not understand why I do need to specify \(Moreover; 1 \neq 0\)
for multiplication, but I do not specify that \(Moreover; 0 \neq 1\)
in P2? What if P2 was a system that only had one number, namely 1?
For every number \(a \neq 0\), there is \(a\) number \(a^{-1}\)
such that
$$a \cdot a^{-1} = a^{-1} \cdot a = 1.$$
If \(a\) and \(b\) are any numbers, then
$$ a \cdot b = b \cdot a $$
There are several interesting things to note here:
\(0 \cdot 0^{-1} = 1\) is undefined!
because of the \(a \neq 0\) in P7.
It is interesting that people say that it is undefined. I
am not sure if being undefined is the same thing as not
existing?
\(a \over b\) is just a different way of writing \(a \cdot
b^{-1}\)
If \(a \cdot b = a \cdot c\), it does not necessarily follow
that \(b = c\).
$$
a \cdot b = a \cdot c and a \neq 0,\\
a^{-1} \cdot (a \cdot b) = a^{-1} \cdot (a \cdot c);\\
(a^{-1} \cdot a) \cdot b) = (a^{-1} \cdot a) \cdot c;\\
1 \dot b = 1 \cdot c);\\
b = c;
$$
If \(a \cdot b = 0 \) then either \(a = 0\) or \(b = 0\).
The above point is commonly used to find the solutions of
equations of the form \((x - 1)(x - 2) = 0.\)
$$
a \cdot b = 0 and a \neq 0,\\
a^{-1} \cdot (a \cdot b) = 0;\\
(a^{-1} \cdot a) \cdot b) = 0;\\
1 \dot b = 0;\\
b = 0;
$$
It would be really neat if I could have a sort of inline link
where I could click on the statement of 3 or 4 above, which states
both the given and the conclusion of a proof, and that would cause a
row based proof of the statement to expand out. Can I write
that?
If \(a\), \(b\), \(c\) are any numbers, then
$$ a \cdot (b + c) = a \cdot b + a \cdot c. $$
$$
a - b = b - a,\\
(a - b) + b = (b - a) + b = b + (b - a);\\
a = b + b - a;\\
a + a = (b + b - a) + a = b + b.\\
a \cdot (1 + 1) = b * (1 + 1),
a = b.
$$
$$
a \cdot 0 + a \cdot 0 =\\
a \cdot (0 + 0) =\\
a \cdot 0;\\
\\
(a \cdot 0) + (a \cdot 0) - (a \cdot 0) = (a \cdot 0) - (a \cdot 0),\\
(a \cdot 0) = 0
$$
This proof really amazes me. You start by proving that the
summation of Z is equal to Z. And then you use this fact to take
your original premise and zero its left equality. That is crazy.
First we assert that \((-a) \cdot b = - (a \cdot b)\)
$$
(-a) \cdot b + (a \cdot b) = \\
[(-a) + a] \cdot b = \\
0 \cdot b = \\
0.
$$
By adding \(-(a \cdot b)\) to both sides,
$$
(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot (-b) + (-a) \cdot b \\
(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot [(-b) + b] \\
(-a) \cdot (-b) + [-(a \cdot b)] = (-a) \cdot 0 \\
(-a) \cdot (-b) + [-(a \cdot b)] = 0
$$
Finally, adding \((a \cdot b)\) to both sides,
$$
(-a) \cdot (-b) = a \cdot b.
$$
So evidently, this is enough to "force upon us" that the product
of product of two negative numbers is positive. Hmm, I am not
quite seeing it. I can see that the steps all make logical
sense, but I am failing to actually grasp P1-P9 force the
product of two negative numbers to be positive.
P9 is kind of a big deal. It is basically the justification for
almost all algebraic manipulations
Most factorization is really just a repeated use of P9
Pretty neat example of how P9 is actually used whenever you multiply
Arabic numerals.
The numbers a satisfying \(a > 0\) are called positive, while
the numbers a satisfying \(a < 0\) are called negative.
Positivity can be defined in terms of \(<\), it is possible to reverse the procedure to mean that \(a < b\) means \(b - a\) is positive.
\(P\) is defined as being the set of all positive numbers. Note in
particular that \(a\) is in \(P\) iff \(a > 0\).
For every number \(a\), one and only one of the following holds:
(i) \(a = 0\)
(ii) \(a\) is in the collection \(P\),
(iii) \(-a\) is in the collection \(P\)
If \(a\) and \(b\) are in \(P\), then \(a + b\) is in \(P\).
If \(a\) and \(b\) are in \(P\), then \(a \cdot b\) is in \(P\).
There is a great deal here about inequalities, followed by
a rather elaborate proof showing that \(|a + b| \leq |a| + |b|.\)
