## Saturday, January 4, 2014

### Basic Properties of Numbers (Ch 1)

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:

1. $$0 \cdot 0^{-1} = 1$$ 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?

2. $$a \over b$$ is just a different way of writing $$a \cdot b^{-1}$$

3. If $$a \cdot b = a \cdot c$$, it does not necessarily follow
that $$b = c$$.

$$a \cdot b = a \cdot c and a \neq 0,\\ a^{-1} \cdot (a \cdot b) = a^{-1} \cdot (a \cdot c);\\ (a^{-1} \cdot a) \cdot b) = (a^{-1} \cdot a) \cdot c;\\ 1 \dot b = 1 \cdot c);\\ b = c;$$

4. 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
$$a \cdot (b + c) = a \cdot b + a \cdot c.$$

When does $$a - b = b - a$$?
$$a - b = b - a,\\ (a - b) + b = (b - a) + b = b + (b - a);\\ a = b + b - a;\\ a + a = (b + b - a) + a = b + b.\\ a \cdot (1 + 1) = b * (1 + 1), a = b.$$

Prove that $$a \cdot 0 = 0$$?
$$a \cdot 0 + a \cdot 0 =\\ a \cdot (0 + 0) =\\ a \cdot 0;\\ \\ (a \cdot 0) + (a \cdot 0) - (a \cdot 0) = (a \cdot 0) - (a \cdot 0),\\ (a \cdot 0) = 0$$

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 $$|a + b| \leq |a| + |b|.$$