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 $c \le d$

Euclid's Division Lemma

Theorem 2:  Euclid's Division Lemma

Given two integers $a$ and $b$ with $b \ne 0$, there exist unique integers $q$ and $r$ such that:

• $a = bq+r$
• $0 \le r < |b|$

Note:  Where $|b|$ is the absolute value of $b$.

Proof:

(1)  Case 1:  $a \ge 0$

(2)  Let $S$ be the set of all integers $a - bk$ where $k$ is an integer and $a - bk \ge 0$

(3)  $S$ is non-empty since either $a = a - b\times0 \ge 0$ which is in $S$

(4)  By the Well-Ordering Principle, there exists an integer $q$ such that $a - bq$ is the least element in $S$

(5)  Let $r = a - bq$.  $0 \le r$ by definition of $S$.  Further $r < b$ since if $r \ge b$, then it is not the least element since $r - b = a - bq - b = a - b(q+1) \ge 0$ would be less than $r$ and an element in $S$.

(6)  $r,q$ must be unique.  Assume that there exists $r', q'$ such that $a = qb + r = q'b + r'$  and $|r - r'| > 0$ and $|q - q'| > 0$.  Then, $b > |r - r'| = b|q - q'| > b$ which is impossible.

(7)  Case 2:  $a < 0$  In this case, let $a' = -a$.  It now follows that $a' \ge 0$ and there exists unique $r,q$.  It now follows that $a = -(bq +r) = b(-q) -r$  If $r > 0$, then $a = b(-q+1) + (b-r)$ with $0 < b-r < b$

Well-Ordering Principle

Theorem 1:  Well-Ordering Princple

Every non-empty set of non-negative integers contains a least element

Proof:

(1)  Assume that there exists a non-empty set $S$ of positive integers that does not contain a least element.

(2)  Let $P(n)$ be true if $n$ is an element of $S$

(3)  $P(0)$ is false.  If true, then $0$ would be the least element for $S$

(4)  Assume that for all $i \le n, P(i)$ is false. 

(5)  It follows that $P(n+1)$ is false.  If not, then $n+1$ would be the least element for $S$.

(6)  But, then by the Principle of Mathematical Induction, there is no positive integer in $S$

(7)  This contradicts our assumption in (1) that $S$ is non-empty.  Therefore, we can reject our assumption in step (1).

Note:  This is an example of a proof by contradiction.

Principle of Mathematical Induction

Axiom 1:  Principle of Mathematical Induction

A proof by induction consists of 2 steps:

(1)  Base Case:  Proves that the statement is true for one case independent of all other cases.

(2)  Inductive Case: Proves that if the statement holds for any case $n=k$, then it must also hold for $n=k+1$

The principle can be understood with the analogy of a ladder.  if we can climb on the bottom rung of a ladder and for each rung, we can climb to the next, then it follows that we can climb as high as we like up to the top of the ladder. 

This principle is very important in the establishment of advanced mathematical arguments.