2.1: Mathematical Fundamentals

Last updated
Save as PDF

Page ID: 211563

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Commutative Property of Addition

Changing the order of addends does not change their sum
\[A+B=B+A \\ 3+2=5 \;\; while, \;\; 2+3=5 \]

note, subtraction is not commutative

\[A-B \neq B-A \\ 3-2=1\;\; while, \;\; 2-3=-1 \]

Commutative Property of Multiplication

\[AB=BA\]

Note division not commutative
\[ \frac{A}{B} \neq\frac{B}{A} \\ \frac{3}{2}=1.5,\;\;while\;\;\frac{2}{3}=0.67\]

Associative Property of Addition.

\[(A+B)+C=A+(B+C) \\ \text{left side: } (2+3)+4 = 5+4=9\\and\\ \text{right side: } 2+(3+4)=2+7=9\]

Associative Property of Multiplication.

\[2(3x4)=(2x3)4 \\ \text{left side: }2(12)=24 \\and \\ \text{right side: }(6)4=24\]

Distributive Property of multiplication .

\[A(B+C)=AB+AC \\ \text{left side: }2(3+4) =2(7)=14 \\ \text{right side: }2(3) +2(4) = 6+8=14\]

note the reverse process allows you to factor out a value if you want to solve for it, so in the next equation, say you want to solve for T

\[m_1c_1T+m_2C_2T=Q\\T(m_1c_1+m_2c_2)=Q\] and you can now solve for T

\[T=\frac{Q}{m_1c_1+m_2c_2}\]

Exponentiation

Exponentiation is the repetition of multiplication

\[2^n \text{ is 2 times 2 repeated n times } \\ 2^4=2(2)(2)(2)=16\]

Roots of a number

The root of a number to the power of n is the number that when multiplied n times equals the original number. The root of a number is actually taking the number to the power of the reciprocal of the exponent: the third root of 64 is 64 to the power of the reciprocal of three.

\[\sqrt[3]{64} =64^{ \frac{1}{3}} \text{ means what number when multiplied by itself 3 times equals 64}\\ \sqrt[3]{64} =4 \text{ because 4(4)(4) = 64}\]

Order of Operations: PEMDAS

PEMDAS is an acronym standing for Parenthesis Exponents Multiplication Division Addition Subtraction, and used to remember the order of performing operations for calculations involving multiple arithmetic operations. The PEMDAS convention that is used in both math and computer science is to perform operations in the following order

P first compete all calculations in parenthesis
E second complete all calculations involving exponents (and roots - which are essentially exponents)
MD third perform all calculations involving multiplication and division, typically going left to write (these are of priority, just as roots and exponents are)
AS last addition and subtraction, typically going left to write (these are of equivalent priority)

Consider the following operation

\[2(3+4)\]

If you incorrectly performed operations left to right you would multiple 2 time 3 and come up with 6 and then add 4 giving the incorrect value of 10.

the proper sequence is to add the 3+4 to get 7 and then multiply by two giving the correct solution of 14.

That is \[\begin{align*} 2(3+4) & \; \neq \; 2(3) +4 \\ \; \\ 2(7) & \; \neq \; 6+4 \\ \; \\ 14 & \; \neq \; 10 \end{align*}\]

The correct answer for \(2(3+4)\) is 14.