Appendix A
Introduction
This appendix to the elementary arithmetic course provides proofs, additional explanations and derivations of some of the statements, theorems and rules used in the course.
Why a negative times a negative is a positive
$$\begin{array}{rclr}-1\times 0& =& 0& \\ -1\times \left(1+-1\right)& =& 0& \\ \left(-1\times 1\right)+\left(-1\times -1\right)& =& 0& \\ -1+\left(-1\times -1\right)& =& 0& \\ -1& =& -\left(-1\times -1\right)& \\ 1& =& -1\times -1& \square \end{array}$$
In the second line, 0 in the left side of the equation is replaced by (1+ −1). The distributive rule applied to the second line result in the third line. Subtracting (−1 × −1) from both sides of the equation in the fourth line result in the fifth line. Finally, negating both sides result in what was to be demonstrated: −1 × −1 = 1.
Proof that the set of prime numbers is infinite
We use a proof by contradiction. The set of prime numbers is either finite or infinite. Let's assume it is finite and try to show that this assumption leads to a contradiction, which proves that it must be infinite.
So, let the finite number of primes all be collected in set $P=\left\{{p}_{1},{p}_{2},{p}_{3},\mathrm{\dots},{p}_{n}\right\}$ and let number $N={p}_{1}\times {p}_{2}\times {p}_{3}\times \mathrm{\dots}\times {p}_{n}+1$.
Now, N is either prime, in which case we have a new prime number other than the original ones, which contradicts our assumption, or N is composite, in which case $N={q}_{1}\times {q}_{2}\times {q}_{3}\times \mathrm{\dots}\times {q}_{m}$, where ${q}_{1},{q}_{2},{q}_{3},\mathrm{\dots},{q}_{m}$ are N's prime factors. But no q can be a prime number from the original set P, because dividing N by any original prime or product of original primes leaves a remainder of 1 (N = p_{1} × p_{2} × p_{3} × ... × p_{n} + 1). Thus ${q}_{1},{q}_{2},{q}_{3},\mathrm{\dots},{q}_{m}$ are new primes, other than the original ones, which also contradicts our start assumption. So, the set of prime numbers cannot be a finite; it must be infinite ☐.
Why the divisibility rules work
Divisibility by 2 or 5
$$4578=\left(4\times 1000\right)+\left(5\times 100\right)+\left(7\times 10\right)+8$$
Powers of 10 i.e. 10, 100, 1000, etc. are always divisible by 2. So, if and only if the last digit, the ones digit, is divisible by 2, the whole number is divisible by 2. Divisibility by 5 works the same.
Divisibility by 3
$$\begin{array}{l}4578=\left(4\times \left(999+1\right)\right)+\left(5\times \left(99+1\right)\right)+\left(7\times \left(9+1\right)\right)+8\\ \phantom{4578}=\left(4\times 999\right)+4+\left(5\times 99\right)+5+\left(7\times 9\right)+7+8\\ \phantom{4578}=\left(4\times 999\right)+\left(5\times 99\right)+\left(7\times 9\right)+4+5+7+8\end{array}$$
9, 99, 999, etc. are always divisible by 9 and thus also by 3. So, if and only if the addition of all digits is divisible by 3, the whole number is divisible by 3.
Divisibility by 7
$$4578=\left(457\times 10\right)+8$$
If 4578 is divisible by 7, then 2 × 4578 is as well, and vice versa:
$$\begin{array}{l}2\times 4578=457\times 20+\left(2\times 8\right)\\ \phantom{2\times 4578}=457\times \left(21-1\right)+\left(2\times 8\right)\\ \phantom{2\times 4578}=\left(457\times 21\right)-457+\left(2\times 8\right)\end{array}$$
(457 × 21) is divisible by 7 (because 21 is divisible by 7). If $-457+\left(2\times 8\right)$ is divisible by 7, then $457-\left(2\times 8\right)$ is divisible by 7 and 4578 is divisible by 7 (and vice versa).
So, a number is divisible by 7 if and only if the subtraction of 2 times the last digit from the number formed by the rest of the digits is divisible by 7.
Divisibility by 11
$$\begin{array}{l}4578=457\times \left(11-1\right)+8\\ \phantom{4578}=\left(457\times 11\right)-457+8\end{array}$$
If 4578 is divisible by 11, then −457 + 8 is divisible by 11, then 457 − 8 is divisible by 11 (and vice versa).
So, a number is divisible by 11 if and only if the subtraction of the last digit from the number formed by the rest of the digits is divisible by 11.
Or:
$$\begin{array}{l}4578=\left(4\times \left(1001-1\right)\right)+\left(5\times \left(99+1\right)\right)+\left(7\times \left(11-1\right)\right)+8\\ \phantom{4578}=\left(4\times 91\times 11\right)-4+\left(5\times 9\times 11\right)+5+\left(7\times 11\right)-7+8\\ \phantom{4578}=\left(4\times 91\times 11\right)+\left(5\times 9\times 11\right)+\left(7\times 11\right)-4+5-7+8\end{array}$$
If and only if the alternating subtraction and addition of all digits is divisible by 11, the whole number is divisible by 11.
Divisibility by 13
$$\begin{array}{l}9\times 4578=457\times 90+\left(9\times 8\right)\\ \phantom{9\times 4578}=457\times \left(91-1\right)+\left(9\times 8\right)\\ \phantom{9\times 4578}=\left(457\times 91\right)-457+\left(9\times 8\right)\\ \phantom{9\times 4578}=\left(457\times 7\times 13\right)-457+\left(9\times 8\right)\end{array}$$
If and only if the subtraction of 9 times the last digit from the number formed by the rest of the digits is divisible by 13, the whole number is divisible by 13.
Derivation of some of the arithmetic rules for fractions
$$\frac{0}{a}=0\times \frac{1}{a}=0$$
$$\frac{a}{b}=1\times \frac{a}{b}=\frac{c}{c}\times \frac{a}{b}=\frac{a\times c}{b\times c}$$
$$\begin{array}{l}\frac{a}{b}\pm \frac{c}{b}=a\times \frac{1}{b}\pm c\times \frac{1}{b}\\ \phantom{\frac{a}{b}\pm \frac{c}{b}}=\frac{1}{b}\times \left(a\pm c\right)\\ \phantom{\frac{a}{b}\pm \frac{c}{b}}=\frac{a\pm c}{b}\end{array}$$
$$\begin{array}{l}\frac{a}{b}\pm \frac{c}{d}=\frac{a\times d}{b\times d}\pm \frac{b\times c}{b\times d}\\ \phantom{\frac{a}{b}\pm \frac{c}{d}}=\frac{\left(a\times d\right)\pm \left(c\times b\right)}{b\times d}\end{array}$$
$$\begin{array}{l}\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{\frac{a}{b}\times \frac{d}{c}}{\frac{c}{d}\times \frac{d}{c}}=\frac{\frac{a}{b}\times \frac{d}{c}}{1}\\ \phantom{\frac{\frac{a}{b}}{\frac{c}{d}}}=\frac{a}{b}\times \frac{d}{c}\end{array}$$
$$\begin{array}{l}-\frac{a}{b}=-1\times \frac{a}{b}=\frac{-1\times a}{b}\\ \phantom{-\frac{a}{b}}=\frac{-a}{b}\\ \phantom{-\frac{a}{b}}=\frac{-1}{-1}\times \frac{-a}{b}=\frac{-1\times -a}{-1\times b}\\ \phantom{-\frac{a}{b}}=\frac{a}{-b}\end{array}$$
Why does the Euclidean algorithm work?
Let:
- $\mathrm{gcd}\left(a=px,b=py\right)=p$, a > b, b > 0,
- t be any positive integer,
- q_{k} be a quotient and r_{k} be a remainder:
$$a-tb=px-tpy=p\left(x-ty\right)$$
Thus,
$$\mathrm{gcd}\left(a,b\right)=\mathrm{gcd}\left(a,b,a-tb\right)$$
$$\frac{a}{b}={q}_{0}+\frac{{r}_{0}}{b}=\frac{{q}_{0}b}{b}+\frac{{r}_{0}}{b}$$
$$\begin{array}{rcl}\frac{a}{b}& =& \frac{{r}_{0}+{q}_{0}b}{b}\\ a& =& {r}_{0}+{q}_{0}b\\ a-{q}_{0}b& =& {r}_{0}\end{array}$$
Thus,
$$\begin{array}{lr}\mathrm{gcd}\left(a,b\right)=\mathrm{gcd}\left(b,a-{q}_{0}b\right)& \\ \phantom{\mathrm{gcd}\left(a,b\right)}=\mathrm{gcd}\left(b,{r}_{0}\right)& b>{r}_{0}\end{array}$$
$$\begin{array}{lr}\mathrm{gcd}\left(a,b\right)=\mathrm{gcd}\left(b,{r}_{0}\right)& b>{r}_{0}\\ \mathrm{gcd}\left(b,{r}_{0}\right)=\mathrm{gcd}\left({r}_{0},{r}_{1}\right)& {r}_{0}>{r}_{1}\\ \mathrm{gcd}\left({r}_{0},{r}_{1}\right)=\mathrm{gcd}\left({r}_{1},{r}_{2}\right)& {r}_{1}>{r}_{2}\\ \mathrm{\dots}& \\ \mathrm{gcd}\left({r}_{k},{r}_{k-1}\right)=\mathrm{gcd}\left({r}_{n},{r}_{n-1}\right)& {r}_{n-1}>{r}_{n}\end{array}$$
The remainder r_{k} (always an integer) is always smaller than the remainder in the step before. So, the remainder must finally be: ${r}_{n}=0$. Thus we can write:
$$\begin{array}{lr}\mathrm{gcd}\left(a,b\right)=\mathrm{gcd}\left({r}_{k},{r}_{k-1}\right)& \\ \phantom{\mathrm{gcd}\left(a,b\right)}=\mathrm{gcd}\left(0,{r}_{n-1}\right)& \\ \phantom{\mathrm{gcd}\left(a,b\right)}={r}_{n-1}& \u2610\end{array}$$