Instructor’s Lecture Notes

Cognition & Learning

What is Algebra?

The word “algebra” comes from Muhammad ibn Musa al-Khwarizmi’s early 9^th century treatise titled al-Kitāb al-Mukhtaṣar fī Ḥisāb al-Jabr wal-Muqābalah (The Concise Book of Calculation by Restoration and Balancing), but this etymology doesn’t really answer the titular question: what is algebra?

There is no algebra without first arithmetic, the study of numbers and of adding and subtracting and multiplying and dividing and sometimes applying exponents or logarithms to those numbers. Arithmetic expressions always resolve to a number. Algebra is just arithmetic where we allow for indeterminates, symbols that behave as if they were numbers yet to be determined or declared. Sometimes the indeterminates are called variables, just a difference of connotation, whether we think of the indeterminate as varying vs being a static unknown quantity.

In a sense, arithmetic is algebra’s shadow, cast upon evaluating the indeterminates at some value and resolving the arithmetic operations. Remembering this gives us a way to check our algebraic calculations.

When in doubt, check your algebraic calculations with literal numbers.

Mathematical Topics

Arithmetic

Initially there is only addition, denoted with the \(+\) sign. Subtraction, denoted with the \(-\) sign, is introduced as an “undo” operation to addition, and bears us negative numbers. Multiplication \(\times\) is defined as iterated addition: \(5\times 7\) is \(5\) added to itself \(7\) times. The operation of multiplication comes up so often that the × sign is often dropped, and the two factors are written next to each other, wrapped in parenthesis if they are both numbers — \({5\times 7 = (5)(7)}\) — or not if at least one is a variable/indeterminate — \({5\times y = 5y.}\) Then multiplication is given its own “undo” operation, division \(\div,\) from which fractional numbers are born.

Addition and multiplication are commutative operations whereas subtraction and division are not. \[ a+b = b+a \qquad ab = ba \qquad\qquad a-b \neq b-a \qquad a\div b \neq b \div a \]

Addition and multiplication are also associative operations whereas subtraction and division are not. \[ (a+b)+c = a+(b+c) \qquad (ab)c = a(bc)\] \[(a-b)-c \neq a-(b-c) \qquad (a \div b) \div c \neq a \div (b \div c) \] Parenthesis are not an operation; they are only used to assign precedence of operations, to group operations together to write arithmetic clearly. The division sign \(\div\) is never used in practice due to its tendency to misread without careful use of parenthesis. Instead the fraction notation \(\frac{a}{b}\) is used to denote the division \(a \div b.\) The inequality involving the division sign \(\div\) above can be rewritten more clearly without parenthesis. \[ \frac{a/b}{c} \neq \frac{a}{b/c} \]

The distributive law relates the operations of addition and multiplication. \[ a(b+c) = ab + ac \] The inequality involving the subtraction above becomes obvious once the first negative sign on the right-hand side is distributed into the difference. \[ a-b-c \neq a -b +c \] That negative sign is a \(-1\) being distributed.

The word “binomial” is just a fancy word for a sum of two things. The expression \(a+b\) is a binomial. Similarly \(a+b+c\) is a trinomial, and so on. We can multiply binomials by applying the distributive law with diligence. \[ (a+b)(c+d) = (a+b)c + (a+b)d = ac+bc+ad+bd \] This is usually memoized as the acronym “FOIL” for “First, Outer, Inner, Last” to account for the four products that result, but it’s more prudent to understand that the final result consists of the products of all pairs of terms, one coming from the first factor and the other coming from the second factor. This understanding more readily generalizes. \[ ({\color{darkgoldenrod}a}+{\color{chocolate}b}+{\color{tomato}c}) ({\color{orchid}d}+{\color{teal}e}+{\color{olivedrab}f}) = {\color{darkgoldenrod}a}{\color{orchid}d} +{\color{darkgoldenrod}a}{\color{teal}e} +{\color{darkgoldenrod}a}{\color{olivedrab}f} +{\color{chocolate}b}{\color{orchid}d} +{\color{chocolate}b}{\color{teal}e} +{\color{chocolate}b}{\color{olivedrab}f} +{\color{tomato}c}{\color{orchid}d} +{\color{tomato}c}{\color{teal}e} +{\color{tomato}c}{\color{olivedrab}f} \]

The standard algorithm for multiplying multi-digit numbers is simply an efficient application of the distributive law.

Fractions

A fraction is a number expressed as the result of division operation: \( \frac{a}{b} = a \div b .\) The rational numbers are all the fractional numbers that can be expressed as fractions \(\frac{a}{b}\) for whole numbers \(a\) and \(b.\) A decimal expression for a number is simply a sum of fractions all having powers of \(10\) as denominators. Many rational numbers have decimal expressions that don’t terminate, that continue indefinitely, and so can never be written down completely; in this case any written decimal expression can only be an approximation to the fraction. \[\begin{align*} \frac{1}{7} = 1 \div 7 &= 0.1428571428571428571428571428571428571428\dots \\&\approx 0.142857 \\&\approx \frac{0}{1} +\frac{1}{10} +\frac{4}{100} +\frac{2}{1000} +\frac{8}{10000} +\frac{5}{100000} +\frac{7}{1000000} \end{align*}\]

Think of the denominators of fractions as units. They’re “denominations”. When multiplying fractions the units are also multiplied. \[\begin{align*} \tfrac{2}{5}\times\tfrac{3}{7} = (2\text{ fifths})(3\text{ sevenths}) = 6\text{ thirty-fifths} = \tfrac{6}{35} \end{align*}\]

When dividing fractions it’s helpful to remember that two division operations “undo” each other and turn into multiplication. \[\begin{align*} \frac{\tfrac{2}{5}}{\tfrac{3}{7}} = \frac{2 \div 5}{3 \div 7} = (2 \div 5) \div (3 \div 7) &= \bigl((2 \div 5) \div 3\bigl) \times 7 \\&= (2 \times 7) \div (5 \times 3) = \frac{2 \times 7}{5 \times 3} = \frac{14}{15} \end{align*}\] This idea is usually memoized as the phrase “flip and multiply”. \[\begin{align*} \frac{\tfrac{2}{5}}{\tfrac{3}{7}} = \tfrac{2}{5}\div\tfrac{3}{7} = \tfrac{2}{5}\times\tfrac{7}{3} = \frac{14}{15} \end{align*}\]

When adding fractions the denominators may first have to be converted into a common set of units. \[\begin{align*} \tfrac{1}{2}+\tfrac{1}{3} &= 1\text{ half} + 1\text{ third} \\&= (1\text{ half})\tfrac{3}{3} + (1\text{ third})\tfrac{2}{2} \\&= (1\text{ half})(3\text{ thirds}) + (1\text{ third})(2\text{ halves}) \\&= 3\text{ sixths} + 2\text{ sixths} = 5\text{ sixths} = \tfrac{5}{6} \end{align*}\] \[\begin{align*} \tfrac{2}{5}+\tfrac{3}{7} &= 2\text{ fifths} + 3\text{ sevenths} \\&= (2\text{ fifths})\tfrac{7}{7} + (3\text{ sevenths})\tfrac{5}{5} \\&= (2\text{ fifths})(7\text{ sevenths}) + (3\text{ sevenths})(5\text{ fifths}) \\&= 14\text{ thirty-fifths} + 15\text{ thirty-fifths} = 29\text{ thirty-fifths} = \tfrac{29}{35} \end{align*}\]

Common factors (not summands) in a fraction’s numerator and denominator can be cancelled out, reducing the fraction to one that is equivalent. \[ \frac{210}{1155} =\frac{2\times 3 \times 5 \times 7}{3 \times 5 \times 7 \times 11} =\frac{2\times \cancel{3} \times \cancel{5} \times \cancel{7}}{\cancel{3} \times \cancel{5} \times \cancel{7} \times 11} =\frac{2}{11} \]

None of this is any different in the case that the numerator or denominator of the fraction contains variables (indeterminates). \[ \frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd} \qquad \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \qquad \frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc} \qquad \frac{a\cancel{c}}{b\cancel{c}} = \frac{a}{b} \] A few notational quirks though when mixing numbers (scalars) with variables. When not be explicitly divided a number is “in the numerator” by default, and \(-1 = \tfrac{1}{-1}.\) \[ -\tfrac{2}{3}x = \tfrac{-2}{3}x = \tfrac{2}{-3}x = -\tfrac{2x}{3} = -2\tfrac{x}{3} = (-2)\tfrac{1}{3}x = 2\tfrac{1}{3}(-x) = 2\tfrac{-1}{3}x = -(x)\tfrac{1}{3}(2) \]

Exponents & Radicals

Just like multiplications is iterated addition, exponentiation is iterated multiplication: \(5^7\) is \(5\) multiplied by itself \(7\) times. Exponentiation is not commutative. Radical operations, square roots \(\sqrt{\phantom{0}},\) etc, are the results of exponentiating to a fractional power.

Basic rules of exponentiation: \[ x^a x^b = x^{a+b} \qquad (xy)^a = x^a y^a \qquad \bigl(x^a\bigr)^b = x^{ab} \] Conventions of exponentiation: \[ x^{0} = 1 \qquad \qquad x^{-a} = \frac{1}{x^a} \qquad \qquad x^{1/a} = \sqrt[a]{x} \] Corollaries from the rules and conventions: \[ \left(\frac{x}{y}\right)^a = \frac{x^a}{y^a} \qquad \qquad x^{a} = \frac{1}{x^{-a}} \qquad \qquad x^{a} = \sqrt[1/a]{x} \]

What about \( x^a + x^b ?\) Nothing … except you could factor some copies of \(x.\) we could write \(x^a + x^b\) as \(x^a \bigl(1 + x^{b-a}\bigr)\) or as \(x^b \bigl(x^{a-b}+1\bigr).\)

What about \( (x+y)^a ?\) Nothing simple. If \(a\) is a whole number you could multiply it out. \[\begin{align*} (x\!+\!y)^2 &= (x\!+\!y)(x\!+\!y) = x^2 \!+\! 2xy \!+\! y^2 \\ (x\!+\!y)^3 &= (x\!+\!y)(x\!+\!y)(x\!+\!y) = (x\!+\!y)\bigl(x^2 \!+\! 2xy \!+\! y^2\bigr) = x^3 \!+\! 3x^2y \!+\! 3xy^2 \!+\! y^3 \\ (x\!+\!y)^4 &= (x\!+\!y)(x\!+\!y)(x\!+\!y)(x\!+\!y) = (x\!+\!y)\bigl(x^3 \!+\! 3x^2y \!+\! 3xy^2 \!+\! y^3\bigr) = \dotsb \\ (x\!+\!y)^5 &= \dotsb \end{align*}\] At the very least, it’s critical to never forget that \((x+y)^2 \neq x^2+y^2.\)

What about \(\sqrt[a]{x+y} ?\) Similarly nothing simple unless we can manage to write \(x+y\) as an \(a^{\text{th}}\) power. At the very least, it’s critical to never forget that \(\sqrt[a]{x+y} \neq \sqrt[a]{x}+\sqrt[a]{y}.\)

Polynomial Expressions

A polynomial in \(x\) is an expression built exclusively by applying the operations of addition, subtraction, multiplication, and scaling to \(x;\) no division-by-\(x\) nor radicals nor exponentials nor logarithms nor trigonometry. Such a polynomial in “standard form” looks like \[ a_n x^n + a_{n-1}x^{n-1} + \dotsb + a_2x^2 + a_1x + a_0\,. \] The highest power of \(x\) is \(n\) so we say the polynomial has degree \(n.\) The \(a_i\) are called the coefficients. Specifically \(a_n\) is called the leading coefficient since its on the leading term of highest power, and \(a_0\) is called the constant term.

Polynomials of low degree have their own dedicated names.

Even if an expression doesn’t look exactly like the above template of a polynomial it may still be algebraically manipulated into that form. \[\begin{align*} x(3-x)^2-7(x-1) \;\;&=\;\; x\bigl(9-6x+x^2\bigr)-7(x-1) \\\;\;&=\;\; 9x-6x^2+x^3-7x+7 \\\;\;&=\;\; x^3-6x^2+2x+7 \end{align*}\]

A fundamental question about polynomials is what numbers \(x\) would zero-out the polynomial? Regarding a polynomial expression as a function, this amounts to asking which inputs \(x\) give an output of \(0\)? Such numbers \(x\) are called zeros or roots of the polynomial. Determining a polynomial’s roots typically amounts to factoring the polynomial into linear terms. Since \[ 2x^3 + 3x^2 - 32x + 15 = \bigl(2x-1\bigr) \bigl(x-3\bigr) \bigl(x+5\bigr) \] \[\begin{align*} &\qquad 2x^3 + 3x^2 - 32x + 15 = 0 \\\qquad\implies&\qquad \bigl(2x-1\bigr) \bigl(x-3\bigr) \bigl(x+5\bigr) = 0 \\\qquad\implies&\qquad \bigl(2x-1\bigr) = 0 \quad\text{ or }\quad \bigl(x-3\bigr) = 0 \quad\text{ or }\quad \bigl(x+5\bigr) = 0 \\\qquad\implies&\qquad x = \tfrac{1}{2} \quad\text{ or }\quad x=3 \quad\text{ or }\quad x=-5 \end{align*}\]

Factoring polynomials is hard in general, and not always possible. A quadratic might be factored by guessing: \[ x^2+4x-21 = \bigl(x-\_\_\bigr)\bigl(x+\_\_\bigr) \] What two numbers add to \(4\) and multiply to \(-21?\) If the roots cannot be guessed in this manner, there is a formula, popularly called the quadratic formula for the roots of roots of a quadratic polynomial \(ax^2+bx+c:\) \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \] Evidently from this formula, if \(b^2-4ac \lt 0\) then the quadratic has no real roots. This number \(b^2-4ac\) is called the discriminant of the polynomial. Similar formulas exist for cubic and quartic polynomials, however those are rather unwieldy. Famously, there are no such formulas for polynomials of degree five or higher. Here are some guiding facts for determining the roots of polynomials in general.

• The Fundamental Theorem of Algebra – A polynomial of degree \(n\) has \(n\) complex roots. I.e. it has at most \(n\) real roots.
• A polynomial of odd degree has at least one real root.
• Rational Roots Theorem – The rational roots of a polynomial \( a_n x^n + \dotsb + a_0 \) must have the form \(p/q\) where \(p\) divides \(a_0\) and \(q\) divides \(a_n.\)
• Descartes’ Rule of Signs – The number of the positive real roots of a polynomial will be at most the number of times the sign \(+/-\) of the coefficients change reading the polynomial in standard form \( a_n x^n + \dotsb + a_0 \) from left to right. The difference in the number of sign-changes and roots must be even. The transformation \(x \mapsto -x\) tells us similar information about the number of negative roots.
• The Polynomial Remainder Theorem – If \(r\) is a root of a polynomial \(P\) then the linear term \((x-r)\) divides \(P,\) which means there is some quotient polynomial \(Q\) of degree one-less than \(P\) such that \(P = Q(x-r).\) The quotient \(Q\) can be calculated by performing polynomial long division of \(P\) by \((x-r),\) resulting in no remainder.
• The Fundamental Theorem of Algebra (Corollary) – A polynomial of degree \(n\) can be factored completely into a product of linear factors over the complex numbers. Over the real numbers it can be factored into a product of linear and irreducible quadratic factors.
• A polynomial having “quadratic form” can be factored like a quadratic. \[ ax^{2n} + bx^{n} + c = 0 \quad\implies\quad x = \sqrt[n]{\frac{-b \pm \sqrt{b^2-4ac}}{2a}} \]

Rational Expressions

A rational expression in \(x\) is a polynomial expression that additionally allows for division by \(x.\) Every rational expression can be written as a quotient of two polynomials. \[ \frac{3}{x} + x^2 - \frac{x-1}{5} \;\;=\;\; \frac{15}{5x} + \frac{5x^3}{5x} - \frac{x^2-x}{5x} \;\;=\;\; \frac{5x^3-x^2+x+15}{5x} \]

In general a rational expression may be in terms of multiple variables. \[ \frac{3z}{xy+x}+\frac{x-1}{xz} \;\;=\;\; \frac{3z(z)}{x(y+1)z}+\frac{(x-1)(y+1)}{x(y+1)z} \;\;=\;\; \frac{3z^2+xy+x-y-1}{x(y+1)z} \]

Like with rational numbers, any factors common to the numerator and denominator may be “cancelled out.” \[\begin{align*} \frac{2x^3+3x^2-32x+15}{x^2+4x-21} \;\;&=\;\; \frac{\bigl(2x-1\bigr)\cancel{\bigl(x-3\bigr)}\bigl(x+5\bigr)}{\cancel{(x-3)}(x+7)} \\[1em]\;\;&=\;\; \frac{\bigl(2x-1\bigr)\bigl(x+5\bigr)}{x+7} \;\;=\;\; \frac{2x^2+9x-5}{x+7} \end{align*}\]

A rational expression is proper if the degree of the numerator is less than the degree of the denominator. Every rational expression can be written as the sum of a polynomial and a proper rational expression. In general, doing this requires polynomial long division, but sometimes this sum can more easily be determined. \[ \frac{x+1}{x} = 1 + \frac{1}{x} \qquad\qquad \frac{x}{x+1} = \frac{x {\color{chocolate}\;+\; 1-1}}{x+1} = 1 - \frac{1}{x+1} \] But sometimes the long division is necessary. \[\begin{align*} &\phantom{\;\;=\;\;} \frac{x^2-5x+7}{x+3} \;\;=\;\; \frac{x^2 {\color{chocolate} \;+\; 3x - 3x}-5x+7}{x+3} \;\;=\;\; \frac{x\bigl(x {\color{chocolate} \;+\; 3}\bigr) + \bigl({\color{chocolate}- 3x}-5x+7\bigr)}{x+3} \\[1em]&\;\;=\;\; x + \frac{-8x+7}{x+3} \;\;=\;\; x + \frac{-8x {\color{orchid}\;-\;24+24} +7}{x+3} \;\;=\;\; x + \frac{-8\bigl(x {\color{orchid}\;+\;3}\bigr)+\bigl({\color{orchid}24} +7\bigr)}{x+3} \\[1em]&\;\;=\;\; x - 8 + \frac{31}{x+3} \end{align*}\]

Conjugation & Rationalization

Any fractional expression containing radicals can be rewritten various different equivalent ways. While no equivalent expression is better than any other, sometimes one is more useful in a given situation. The important skill is to fluidly move between them. Doing so, mechanically speaking, always comes down to multiplying by \(\color{chocolate}{1}\) in some clever form. \[ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} {\color{chocolate}\Biggl(\frac{\sqrt{2}}{\sqrt{2}}\Biggr)} = \frac{\sqrt{2}}{2} \]

Some folks become irrationally upset by irrational numbers appearing in the denominator of a fraction, like the \(\sqrt{2}\) initially in the previous example. The phrase they use for fixing this, for removing the source of their anguish by performing the algebra done in that example, is “rationalizing the denominator”. This can always be done per some clever choice of a factor of \(\color{chocolate}{1}\). \[ \frac{2}{3+\sqrt{5}} = \frac{2}{3+\sqrt{5}} {\color{chocolate}\Biggl(\frac{3-\sqrt{5}}{3-\sqrt{5}}\Biggr)} = \frac{2\bigl(3-\sqrt{5}\bigr)}{(3+\sqrt{5})(3-\sqrt{5})} = \frac{6-2\sqrt{5}}{9-5} = \frac{3-\sqrt{5}}{2} \] In this example \(\color{chocolate}{1}\) was chosen to have a numerator and denominator of \(3-\sqrt{5}\) based on the original denominator of \(3+\sqrt{5}.\) To ascribe terminology, \(3-\sqrt{5}\) is a conjugate of \(3+\sqrt{5}.\) In general the numbers \(a+b\sqrt{c}\) and \(a-b\sqrt{c}\) form a conjugate pair, named so because their product contains no square root, no irrational component. \[ \bigl(a+b\sqrt{c}\bigr) \bigl(a-b\sqrt{c}\bigr) = a^2 = b^2c \]

This trick of “multiplying by a conjugate” can sometimes lead to simpler ways of writing an expression. \[ \frac{\sqrt{x}-4}{x-16} = \frac{\sqrt{x}-4}{x-16} {\color{chocolate}\Biggl(\frac{\sqrt{x}+4}{\sqrt{x}+4}\Biggr)} = \frac{x-16}{(x-16)\bigl(\sqrt{x}+4\bigr)} = \frac{1}{\sqrt{x}+4} \]