01 - Axioms for Arithmetic in ℝ

Axioms for Arithmetic in $\mathbb{R}$

Sum: Unique real number $a+b$ for every $a,b\in \mathbb{R}$
Product: Unique real number $a\cdot b$ for every $a,b\in \mathbb{R}$
Additive Inverse / Negative: Unique real number $-a$ for $a\in \mathbb{R}$
Multiplicative Inverse / Reciprocal: Unique real number $\frac{1}{a}$ for $a\in \mathbb{R}$ with $a\ne 0$
Zero / Additive Inverse: Special element $0\in \mathbb{R}$
One / Multiplicative Identity: Special element $1\in \mathbb{R}$

For $a,b,c\in \mathbb{R}$ we also have
$+$ is commutative:

$$a+b=b+a$$

$+$ is associative:

$$a+(b+c)=(a+b)+c$$

Additive Identity:

$$a+0=a$$

Additive Inverse:

$$a+(-a)=0$$

$\cdot$ is commutative:

$$a\cdot b=b\cdot a$$

$\cdot$ is associative:

$$a\cdot (b\cdot c)=(a\cdot b)\cdot c$$

Multiplicative Identity:

$$a\cdot 1=a$$

Multiplicative Inverse:

$$\text{ if }a\ne 0\text{ then}a\cdot \frac{1}{a}=1$$

$\cdot$ distributes over $+$:

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

The first 4 are addition and the rest are for multiplication

To Avoid Total Collapse

$$0\ne 1$$

Properties of arithmetic in $\mathbb{R}$

Let $a,b,c,x,y$ be real numbers

1. Uniqueness of $0$: If $a+x=a$ for all $a$ then $x=0$

2. Cancellation for $+$: If $a+x=a+y$ then $x=y$

3. $-0=0$

4. $-(-a)=a$

5. $-(a+b)=(-a)+(-b)$

6. Uniqueness of $1$: If $a\cdot x=a$ for all $a\ne 0$ then $x=1$

7. Cancellation of $\cdot$: If $a\ne 0$ and $a\cdot x=a\cdot y$ then $x=y$

8. If $a\ne 0$ then $\frac{1}{\frac{1}{a}}=a$

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

10. $a\cdot 0=0$

11. $a\cdot (-b)=-(a\cdot b)$
    In particular, $(-1)\cdot a=-a$

12. $(-1)\cdot (-1)=1$

13. If $a\cdot b=0$ then $a=0$ or $b=0$
    If $a\ne 0$ and $b\ne 0$ then $\frac{1}{a\cdot b}=\frac{1}{a}\cdot \frac{1}{b}$

Proof

1. Suppose that $a+x=a$ for all $a$ then

$$\begin{array}{rl}x& =x+0(\text{additive identity})\\ & =0+x(+\text{ is commutative})\\ & =0(\text{by hypothesis, with }a=0)\end{array}$$

2. Suppose that $a+x=a+y$ then

$$\begin{array}{rl}y& =y+0(\text{additive identity})\\ & =y+(a+(-a))(\text{additive inverses})\\ & =(y+a)+(-a)(+\text{ is associative})\\ & =(a+y)+(-a)(+\text{ is commutative})\\ & =(a+x)+(-a)(\text{hypothesis})\\ & =(x+a)+(-a)(+\text{is commutative})\\ & =x+(a+(-a))(+\text{ is associative})\\ & =x+0(\text{additive inverse})\\ & =x(\text{additive identity})\end{array}$$

3. We have that

$$\begin{array}{r}0+0=0(\text{additive identity})\\ 0+(-0)=0(\text{additive inverses})\end{array}$$

Hence

$$0+0=0+(-0)\text{ hence }0=-0(\text{cancellation for }+(2))$$

4. We have

$$\begin{array}{rl}(-a)+a& =a+(-a)(+\text{ is commutative})\\ & =0(\text{additive inverses})\end{array}$$

and $(-a)+(-(-a))=0(\text{additive inverses})$
so $(-a)+a=(-a)+(-(-a))$,
so $a=(-(-a))(\text{cancellation for }+(2))$

5. TODO onwards… (also refer to lecture notes for some of them)