Solution to the Depressed Cubic Equation

Theorem 9:  Depressed Cubic Equation

If $x^3 + bx = c$, then:

$$x = \sqrt[3]{\frac{c}{2} + \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}} + \sqrt[3]{\frac{c}{2} - \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}}$$

Proof:

(1)  $x^3 = -bx + c$

(2)  There exists real numbers $u,v$ such that $x = u + v$

(3)  $(u+v)^3 = u^3 + 3u^2v + 3uv^2 + v^3 = 3uv(u+v) + (u^3 + v^3)$

(4)  Since $b = -3uv$, it follows that $v = \dfrac{-b}{3u}$

(5)  $c = u^3 + v^3 = u^3 + \dfrac{-b^3}{27u^3}$

(6)  It follows that:

$$u^6  - u^3c  + \dfrac{-b^3}{27}= 0$$

(7)  Using the solution for the quadratic equation, we have:

$$u^3 = \frac{c \pm \sqrt{c^2 + \frac{4b^3}{27}}}{2} = \frac{c}{2} \pm \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(8)  From the symmetry of the equation, we can assume that:

$$u^3 = \frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

$$v^3 = c - u^3 = c - \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3} = \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(9)  $x = u + v = \sqrt[3]{\frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}} + \sqrt[3]{\frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}}$

Solution to the Quadratic Equation

Theorem 8:  If $ax^2 + bx + c=0$, then:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Proof:

(1)  $x^2 + \dfrac{bx}{a} + \dfrac{c}{a} = 0$

(2)  $x^2 + \dfrac{bx}{a} = -\dfrac{c}{a}$

(3)  $x^2 + \dfrac{bx}{a} + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(4)  $\left(x + \dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

(5)  $x + \dfrac{b}{2a} = \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2}$

(6) $x = -\dfrac{b}{2a} \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2} = \dfrac{-b \pm  \sqrt{b^2 - 4ac}}{2a}$

The Square Root of 2 is irrational

Theorem 7: The Square Root of $2$ cannot be represented by the ratio of two integers

Proof:

(1)  Assume that the squre root of $2$ could be represented by a ratio of two integers with $a$ and $b$ the reduced form.

$$\frac{a}{b} = \sqrt{2}$$ 

(2)  Squaring both sides:

$$a^2 = 2b^2$$

(3)  Since $a$ must be even, there exists $c$ such that $a=2c$ and:

$$a^2 = (2c)^2 = 4c^2 = 2b^2$$

(4)  It follows that $b$ must be even since:

$$2c^2 = b^2$$

(5)  Then there exists $d$ such that $b=2d$ and:

$$\sqrt{2} = \frac{2c}{2d} = \frac{c}{d}$$

(6)  But then we have a contradiction since $a$ and $b$ can be reduced to $c$ and $d$.

Infinitude of Prime Numbers

Theorem 6:  Infinitude of Prime Numbers

There are an infinite number of prime numbers.

Proof:

(1)  Assume that there is a a finite number of primes. 

(2)  Let $P$ be the product of all primes.

(3)  Let $q = P+1$

(4)  By the Fundamental Theorem of Arithmetic, either $q$ is prime or $q$ is a product of primes.

(5)  If $q$ is prime, then $q | (P + 1) - P = 1$ which s impossible.

(6)  If $r$ is a prime that divides $q$, then we have the same problem since $r | (P + 1) - P = 1$

Fundamental Theorem of Arithmetic

Theorem 5:  The Fundamental Theorem of Arithmetic

Every integer greater than $1$ is either prime or can be represented as the unique product of powers of prime numbers.

Proof:

(1)  First, prove existence.  It is true for the base case where $n=2$ since $2$ is a prime.

(2)  Assume that it is true for all $i$ with $i \le n$

(3)  For $n+1$, if $n+1$ is prime, then it is true.  If $n+1$ is not prime, then there exists integers $a,b$ such that $n+1 = ab$ with $a \le n$ and $b \le n$.  Since $a \le n$ and $b \le n$, they can be represented by a prime or a product of primes with $a=p_1p_2\dots{p_n}$ and $b=q_1q_2\dots{q_n}$

(4)  Then $n+1 = ab = p_1p_2\dots{p_n}q_1q_2\dots{q_n}$ which means that $n+1$ can be represented as a product of powers of prime numbers.

(5)  Assume that there is an integer that has two different product of powers of prime numbers.

(6)  Let $y$ be the least integer where this is true so that $y = p_1p_2\dots{p_m} = q_1q_2\dots{q_n}$ with $p_1\dots{p_m}$  and $q_1\dots{q_n}$.

(7)  From Euclid's Generalized Lemma, $p_1$ divides $q_i$ where $1 \le i \le n$.  But, then $\frac{y}{p_1} = p_2\dots{p_m} = \frac{q_1q_2\dots{q_n}}{p_1}$ which is smaller than $n$ and contradicts our assumption in step(6).

Euclid's Lemma

Theorem 4:  Euclid's Lemma

If a prime $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ divides $a$ or $p$ divides $b$

Proof:

(1)  Assume that $p | ab$ but $p \nmid a$

(2)  From Bezout's Identity, there exists $r,s$ such that $rp + sa = 1$

(3)  Multiply both sides by $b$ to get:  $rpb + sab = b$

(4)  Since $p | ab$, it follows that $p | b$

Corollary 4.1:  Euclid's Generalized Lemma

If a prime $p$ divides $a_1a_2\dots{a_m}$, then $p$ divides $a_i$ where $1 \le i \le m$

Proof:

(1)  Base Case:  The base case is Euclid's Lemma.  If $p | a_1a_2$, then $p | a_1$ or $p | a_2$

(2)  Assume that up to $n$ if $p | a_1a_2\dots{a_n}$, then $p|a_i$ with $1 \le i \le n$

(3)  Let $b = a_1a_2\dots{a_n}$.  

(4)  Inductive Case:  From Euclid's Lemma, if $p | ba_{n+1}$, then $p | b$ or $p|a_{n+1}$.

Bezout's Identity

Theorem 3:  Bezout's Identity

Let $a$ and $b$ be nonzero integers with greatest common divisor $d$.  Then, there exists integers $x$ and $y$ such that $ax + by = d$.

Proof:

(1)  For integers $a,b$ let $S$ be a set that includes all $ax+by$ where $ax + by > 0$

(2)  The set is non-empty since it contains either $a$ or $-a$ where $y=0$ and $x=1$ or $x=-1$

(3)  By the Well-Ordering Principle, $S$ has a minimal element $as + bt$ and let $d = as+bt$

(4)  Using Euclid's Divsion Lemma, there exists $q_1, q_2, r_1, r_2$ such that:

• $a = q_1d + r_1$
• $b = q_2d + r_2$

(5)  It follows that:

• $r_1 = a - q_1d = a - q_1(as + bt) = a(1 - q_1s)  + b(-q_1t)$ which must equal $0$ since $r_1 < d$ but $d$ is the least element in $S$.  The same argument applies to $r_2$ so that $r_2 = 0$

(6)  Then, $d$ divides both $a$ and $b$.

(7)  Let $c$ be any common divisor of $a$ and $b$ such that $a =cu$ and $b = cv$

(8)  $d = as + bt = cus + cvt = c(us + vt)$ so that it follows that $c$ divides $d$ and further that $c \le d$