Addition & Multiplication

This study sheet is called Addition & Multiplication because once you understand the rules for addition and multiplication, you will be able to do all the mathematics you encounter in school.

True, there are functions to learn, procedures to master, and calculus to understand, but the methods needed to solve equations and manipulate Mathematical Expressions (called MEs for the rest of the study sheet) are outlined in this study sheet.

Mathematical Expressions & Symbols

MEs (or Algebraic Expressions) are mathematical representations of There are rules (called axioms or properties) which govern how to manipulate MEs and change their structure, much like the rules for moving things around in a game or in a sport you play.

Two differences between mathematical axioms and sports' rules are:

The axioms are based on fundamental mathematical truths which we have discovered after millennia of study, thought, and wonder.

Understanding the structure of the universe through mathematics is one of mankind's greatest achievements. You can learn in the next few moments what it took human kind a long time to formulate.

Examples of some MEs: $$ \begin{gather} 16\\ 10^{\ 3}\\ \frac{1}{\sqrt{2\pi}}\int^t_0e^{-x^2/2}\\ (ax^2 + bx + c)\\ f(x)\\ (2+2)\\ \frac{\sqrt{n}(\bar{x}-\mu^*)}{\sigma} \end{gather} $$ It is important to understand that the axioms apply to all MEs, not just simple expressions like \(2+2\).

Symbols are used in MEs to represent individual quantities (which can be further MEs) in the ME. Examples: $$0,1,2,3,4,5,\Box,\triangle,6,7,8,9,A,B,C...,a,b,c,\hslash,...\alpha,\beta,\gamma,\pi, \xi, \Psi, \eta, \text{etc.}$$ It doesn't matter what symbols we use. The symbols are just used to represent a mathematical entity we can think of, much like words are used to express verbally an object or concept. We could just as correctly write the quadratic equation $$(ax^2 + bx + c)$$ as $$(\heartsuit \cdot \Delta^2 + \lozenge \cdot \Delta + \xi)$$ although it would take some getting used to!

Capital letters will be the symbols used for the rest of this study sheet to represent MEs.

The Operations\(\ +\ -\ \times\ \div\)

The binary operators are addition, subtraction, multiplication, and division. Binary means they work on (operate on) and join two MEs.

It is very important to realize that once you join two MEs with a binary operator, you can only manipulate them and change their structure by rules of the axioms.

The symbols used for the binary operators of addition and subtraction (\(+, -\)) are the same symbols used to represent the sign of a value, positive or negative (\(\texttt{+, -}\)) of a ME (these are unary operators, as they operate on one object to indicate or change the sign of that object).

Multiplication of two MEs can be indicated many different ways, eg. $$ \begin{gather} M \times P,\\ M*P,\\ M\cdot P,\\ MP,\\ M(P),\\ (M)(P) \end{gather} $$ Division can be represented different ways also, which will be discussed below.

Grouping, Blocks, and Parenthesis

Whenever the binary operations are applied to MEs, those two MEs are joined and can not be unjoined except by the axioms. When these axioms are applied, it can be shown that $$ \require{cancel} \frac{A\times B}{A}=\frac{\cancel{A}\times B}{\cancel{A}}=B\tag{eqn 1} $$ is correct, but $$ \frac{A+B}{A}=\frac{\cancel{A}+B}{\cancel{A}}\tag{eqn 2} $$ is incorrect.

Students frequently make the mistake shown in (eqn 2).

You should train yourself to see blocks of symbols in MEs.

One way to do this is to put parenthesis around all binary operators. Therefore, \(2\times 2\) should be \((2\times 2)\). As an example, there are many way that the quadratic equation can be written with parenthesis. One way is $$ ((a\cdot x)\cdot x) + ((b\cdot x) + c) $$ Depending on how you want to change the structure of the above quadratic equation, you can see it written in blocks like $$B_1\cdot B_2 + B_3 + B_4$$ where $$B_1=a\cdot x,\ B_2=x,\ B_3=b\cdot x, \text{ and } B_4=c$$ Or, maybe you should see it as just two blocks (depending on how you want to move the symbols around) like $$C_1+C_2$$ where $$C_1=[(a\cdot x\cdot x) + (b\cdot x)], \text{ and } C_2=c$$ The point to emphasize is that:

the symbols in MEs are moved around in blocks, and the only way those blocks are moved are by the rules of the axioms
From these axioms, you can get (eqn 1), but not (eqn 2).

Equality

The Equal sign \((=)\) is not an operator like \(\ +\ -\ \times\ \text{ or }\div\). It is a statement of the equality of the MEs that are on each side of the equal sign. When you are given a statement of equality, you want to use the axioms to change the MEs to extract additional information from the mathematics while preserving the equality you were initially given.

How multiplication with negatives works

When like signs are multiplied together, they yield a positive (\(\texttt{+}\)). $$ (\texttt{+}P_1)\times(\texttt{+}P_2)=\texttt{+}(P_1\times P_2) $$$$ (\texttt{-}N_1)\times(\texttt{-}N_2)=\texttt{+}(N_1\times N_2) $$
Examples
$$ \begin{gather} (\texttt{+}2)\times(\texttt{+}5)=\texttt{+}10\\ (\texttt{-}1)\times(\texttt{-}1)=\texttt{+}1\\ (\texttt{-}2)\times(\texttt{-}2)=\texttt{+}4\\ \frac{\texttt{-}3}{\texttt{-}4}=\texttt{+}\frac{3}{4} \end{gather} $$ When opposite signs are multiplied together, they yield a negative (\(\texttt{-}\)). $$(\texttt{+}P)\times(\texttt{-}N)=\texttt{-}(P\times N)\\$$
Example
$$ \begin{gather} (\texttt{-}2)\times(\texttt{+}3)=\texttt{-}6\\ \end{gather} $$ $$ \begin{gather} \frac{\texttt{-}2}{3}=\frac{2}{\texttt{-}{3}}=\texttt{-}\frac{2}{3} \end{gather} $$

The Axioms of Algebra

The axioms apply to addition & multiplication. Subtraction & division can be changed to addition & multiplication as follows.

Division by zero (\(0\)) is undefined.

Changing Subtraction to Addition

For any subtractions, change $$A=B-C$$ to $$A=B+(\texttt{-}C) \tag{StoA}$$
Examples:
$$6-7 = 6+(\texttt{-}7) = \texttt{-}1$$ $$8-(-15) = 8+(\texttt{+}15)=23$$

Changing Division to Multiplication

For any divisions, change $$\require{enclose}D = F \div G \ \ \ \ \ \text{or}\ \ \ \ \ D=G\enclose{longdiv}{F\phantom{.}}$$ to $$D=F\times \frac{1}{G}\tag{DtoM}$$
Example:
$$D = 2 \div 10 \ \ \ \ \ \text{or}\ \ \ \ \ D=10\enclose{longdiv}{2\phantom{.}}$$ changing to multiplication then gives $$D=2\times\frac{1}{10}= 0.2$$

Additive Equality

If \(A=B\), then \(A+C=B+C \tag{ADD-EQ}\) Given an equality \(A=B\), the equality is preserved when an ME, \(C\), is added to each side of the equality.

Multiplicative Equality

If \(A=B\), then \(A\times C=B\times C \tag{MULT-EQ}\) Given an equality \(A=B\), the equality is preserved when an ME, \(C\), is multiplied to each side of the equality.

The Commutative Axiom

$$H+J=J+H\tag{COMM-ADD}$$ $$K\times L = L\times K\tag{COMM-MULT}$$ The order in which MEs are added does not affect the result.
The order in which MEs are multiplied does not affect the result.
Examples:
$$\begin{align*} 4+5 & = 5+4 \qquad & 8\times 6 &= 6\times 8 \\ 9 & =9 \qquad & 48 & = 48 \end{align*}$$
Note:
Now you can see why subtraction and division are rewritten as addition and multiplication. $$H-J\neq J-H$$ $$K\div L \neq L\div K$$ but of course $$H+(\texttt{-}J)=(\texttt{-}J)+H$$ $$K*\frac{1}{L}= \frac{1}{L}*K$$

The Associative Axiom

$$(M+N)+P=M+(N+P)\tag{ASSOC-ADD}$$ $$(Q\times R)\times T=Q\times(R\times T)\tag{ASSOC-MULT}$$ The order in which groups of MEs are added does not affect the result.
The order in which groups of MEs are multiplied does not affect the result.
Examples:
$$\begin{align} (35 +20) + 12 & = 35+(20+12) \qquad & (3\times 5)\times 7 &= 3\times(5\times 7) \\ (55)+12 & =35+(32) \qquad & (15)\times 7 & = 3\times(35) \\ 67 & =67 \qquad & 105 &= 105 \end{align}$$

The Distributive Axiom

$$\begin{align} U\times(W+X)&=(U\times W)+(U\times X) = UW + UX\tag{DIST}\\ (Z\times Y)+(Z\times A)&=Z\times(Y+A) \end{align}$$ When a ME multiplies a sum, it doesn't matter whether the sum is evaluated first, and then multiplied, or, whether the multiplication is distributed through to the individual summands, and then that result is summed.
If a common factor is present in two summands, then that common factor can be removed from both summands, and multiplied to the sum of the remaining factors.
Examples:
$$ \begin{align} 6*(10+5) & = (6*10)+(6*5) \\ 6*(15) & =(60)+(30) \\ 90 & =90 \end{align} $$ $$63=54+9=(9*6)+(9*1) = 9*(6+1) = 9*(7) = 63$$

The Additive Identity\(\ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \text{(zero)}\)

$$A+0=A\tag{ADD-ID}$$ Whenever \(0\) is added in an equation, the equality of the equation is preserved. Always.
Example:
The equation \(x^2-6x-7=y\) can be solved to find the x-axis intercepts (values of \(x\) which result in \(y\) being equal to \(0\)). Adding \(0\) changes the structure of the equation so the square can be completed. $$ \begin{align} x^2-6x-7\ \ \ \ \ \ \ \ \ &=y \\ \newline x^2-6x-7\ \ \ \ \ \ \ \ \ &=0 \tag{set y=0}\\ x^2-6x\phantom{+((9-9)} &=7 \tag{use ADD-EQ}\\ x^2-6x +\ \ \ \ 0\ \ \ \ \ \ &=7 \tag{add 0}\\ x^2-6x +(9-9) &=7 \tag{0=9\(-\)9}\\ (x^2-6x +9)-9\ &=7\ \ \checkmark\tag{use ASSOC-ADD}\\ (x^2-6x +9)\phantom{-9}\ \ &=16\\ (x-3)^2 & =16 \tag{square completed!}\\ (x-3)\ \ & =\pm 4 \\ x & =3 \pm 4 \\ x & =-1, 7 \end{align} $$ Therefore values of \(x\) = \(-1\) & \(7\) will result in \(y = 0\) for the equation \(y=x^2-6x-7\).

The Additive Inverse

$$A+(\texttt{-}A)=0\tag{ADD-INV}$$

The Multiplicative Identity\(\ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \text{(one)}\)

$$B\times 1 = B\tag{MULT-ID}$$ Whenever any number or term in an equation is multiplied by \(1\), the equality of the equation is preserved. Always.
Examples:
  1. Fractions
    To add 2 fractions, the denominators must be the same. So to add \(\frac{1}{3}+\frac{2}{5}\), multiply each fraction by \(1\) to change the denominators (and of course the numerators will change, too). $$ \begin{align*} &\frac{1}{3} &&+\frac{2}{5} &&=\ ?\\ &\frac{1}{3}*\ 1 &&+\frac{2}{5}*\ 1 &&=\ ?\\ &\frac{1}{3}*\frac{5}{5} &&+\frac{2}{5}*\frac{3}{3} &&=\ ?\\ &\ \ \ \frac{5}{15} &&+\ \ \ \frac{6}{15} &&=\frac{11}{15} \end{align*} $$
  2. Unit Conversions
    All unit conversions use the Multiplicative Identity. For instance, to convert from miles per hour \((mi/hr)\) to feet per second \((ft/s)\), use $$ \begin{align*} 3600 s & = 1 hr\\ \nonumber \newline \text{so}\ \ \ \ \ \ \ \ \ \ \ \ \ 1 & = \frac{1hr}{3600s}\tag{eqn. 1}\\ \nonumber \newline \text{use also}\ \ \ \ \ \ \ \ \ \ \ 5280ft & = 1mi\\ \nonumber \newline \text{so}\ \ \ \ \ \ \ \ \ \ \ \ \ 1 & = \frac{5280ft}{1mi}\tag{eqn. 2} \end{align*} $$ Now, to convert \(60mi/hr\) to \(ft/s\), just multiply by the \(1\)'s (ones) above (eqn. 1 & eqn. 2): $$\begin{align} \frac{60mi}{hr}\phantom{*\frac{hr}{3600s}*\frac{5280ft}{mi}}\ \ & =\ ?\frac{ft}{s}\\ \newline \frac{60mi}{hr}\ \ *1\ \ \ \ \ \ \ \ *1\ \ \ \ \ \ \ \ \ \ & =\ ?\frac{ft}{s}\\ \newline \frac{60mi}{hr}*\frac{hr}{3600s}*\frac{5280ft}{mi} & =\ ?\frac{ft}{s}\\ \newline \require{cancel}\frac{60\cancel{mi}}{\cancel{hr}}*\frac{\cancel{hr}}{3600s}*\frac{5280ft}{\cancel{mi}} & =\ ?\frac{ft}{s}\\ 60*\frac{1}{3600s}*5280ft & =\ 88\frac{ft}{s}\\ \text{so}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 60\frac{mi}{hr} & = 88\frac{ft}{s} \end{align}$$

The Multiplicative Inverse

$$B\times \frac{1}{B} = 1\tag{MULT-INV}$$