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 \(|a + b| \leq |a| + |b|.\)

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