Order of operations

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


When we calculate with numbers, there is a certain order of operations in which we establish the order in which we perform mathematical operations.


When we interpret a mathematical expression consisting of additions, subtractions, multiplications, divisions and exponentiations of real numbers, the following rules apply:

  • addition and subtraction are carried out from left to right, in the order in which these operations appear in the expression
  • multiplications and divisions are carried out from left to right, in the order in which these operations appear in the expression
  • multiplication and division take precedence over addition and subtraction
  • exponentiation takes precedence over multiplication and division
  • exponentiation is carried out from right to left, i.e. in the reverse order in which these operations appear in the expression

An easy way to remember the order of operations is:

Please Excuse MDear Aunt Sally!

We first work on everything inside the parentheses, then continue on exponents (including square roots), then continue on multiplication and division, and finally addition and subtraction.

You can change the normal order of operations using parentheses. Everything between parentheses always has priority.


Let’s practice!

Calculate \(7.5 \cdot 5+5 \cdot 8.5\).

Multiplications take precedence over additions. You can rewrite the exercise with brackets around the multiplications. This results in:

\(\begin{array}{rcl} 7.5 \cdot 5+5 \cdot 8.5 &=& (7.5 \cdot 5)+(5 \cdot 8.5) \\ &=& 37.5 +42.5 \\ &=& 80 \end{array}\)

Answer:

\(80\)

Roots of integers

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


If \(a\) is a non-negative real number, then there is exactly one non-negative real number \(b\), in such a way that \(b^2 = a\). This number is indicated by \(\sqrt{a}\) and is called the root of \(\sqrt{a}\).

Some roots are well known: \(\sqrt{1} = 1\), \(\sqrt{4} = 2\) and \(\sqrt{9} = 3\).


Rules of calculation for roots

Let \(a\) and \(b\) be non-negative numbers. The following rules apply:

  • \(\left(\sqrt{a}\right)^2 =a= \sqrt{a^2}\)
  • \(\sqrt{a}\cdot\sqrt{b} =\sqrt{a\cdot b}\)

With these rules, you can write a product with roots as a product with no more than one root. This for example helps to write down each product of integers and roots of integers in a unique way, the standard form.


Standard form for roots of integers

Every positive integer \(a\) can be written as \(b^2\cdot{c}\), in which \(b\) and \(c\) are integers and \(c\) a product of prime numbers that each appear exactly once. The standard form for \(\sqrt{a}\) is \(b\cdot\sqrt{c}\).


Let’s practice!

Write \(-4\sqrt{7}\cdot -5\sqrt{3}\) in standard notation.

This can be seen as follows:

\(\begin{array}{rcl}-4\sqrt{7}\cdot -5\sqrt{3}&=&-4 \cdot -5 \sqrt{7\cdot 3}\\ &=&20\sqrt{21}\end{array}\)

Answer:

\(-4\sqrt{7}\cdot -5\sqrt{3}={20\sqrt{21}}\)

Decimal numbers

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


The picture of the real numbers as points on a line is geometrically appealing. On the other hand, an algebraic definition is not easy. Before we look into calculating with real numbers, we elaborate on the following special rational numbers.


digit is one of the numbers \(0, 1, 2, 3, 4, 5, 6, 7, 8, 9.\)

A number of the form \(\frac{a}{10^n}\), wherein\(a\) and \(n\) are integers with \(n\ge0\) is called a decimal number. This number is often written as \(a\) (possibly complemented with zeros from the left side) with a decimal point before the \(n\)-th digit from the right.

The digits behind the decimal point are called the decimals.


Example

\(\frac{3}{10}= .3 =0.3\) \(\frac{1}{2}=\frac{5}{10} = 0.5\)
\(\frac{192}{10^0}= 192\) \(\frac{31 51 49 2}{10^6}=3.141592\)
\(\frac{-12}{10^7}=-0.00 00 012\) \(\frac{-1234567890}{10^{5}}= -12345.6789\)

Addition, subtraction and multiplication of decimal numbers results in another decimal number.

Division of a decimal number by \(2\) or \(5\) as well.

But in general, dividing a decimal number by another decimal number does not result in a decimal number.


Let’s practice!

The division of the decimal number \(542.107\) by \(50\) is another decimal number. Which one?

Division by \(50\) is the same as multiplying by \(2\) and dividing by \(100\).

  • Multiplying \(542.107\) by \(2\) results in the decimal number \(1084.214\).
  • Division of \(1084.214\) by \(100\) corresponds with the digits moving \(2\) places to the left of the decimal point. (If there are not enough decimal places you can of course add zeros) Hence, the answer is \(10.84214\)

Answer:

\(10.84214\)

Simplifying fractions

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


The following definition will help get a unique expression for every rational number:

When both numerator  and denominator can be divided by the same integer (except \(1\)), we can simplify the fraction by doing so. This is called simplifying fractions.

We speak about fractions that can’t be simplified if

  • The greatest common divisor (gcd) of both numerator and denominator is equal to \(1\),
  • the denominator is positive, and
  • the denominator is equal to \(1\) if the numerator is equal to \(0\)..

Let’s practice!

Simplify the following fraction as much as possible.

\(\frac{18}{-135}\)

You can divide the numerator and denominator by \(9\). This gives:

\(\frac{18}{-135}=-\frac{18}{135} =-\frac{9 \cdot 2}{9 \cdot 15}=-\frac{2}{15}\)

You can also simplify in steps:

\(\begin{array}{rcl}\frac{18}{-135}&=&\frac{3 \cdot 6}{3 \cdot -45}\\ &=&\frac{6}{-45} \\ &=&\frac{3 \cdot 2}{3 \cdot -15}\\ &=&\frac{2}{-15}\\ &=&\displaystyle -{{2}\over{15}} \\ &&\phantom{xx}\color{blue}{\text{minus sign moved}}\end{array}\)

Answer:

\(\frac{18}{-135}=-{{2}\over{15}}\)

The number \(9\) is, in fact, the greatest common denominator of \(18\) and \(-135\).

Fractions

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


When you divide an integer by another integer not equal to \(0\), you either get an integer, or a rational number. We can calculate with rational numbers the same way we do with integers.

Rational numbers are real numbers who can be written as a fraction of two integers. In other words, a rational number is the result of dividing an integer (the numerator) by another integer (the denominator). We write a fraction with both the numerator and the denominator, divided by a horizontal line or forward slash.


There are different ways to write a rational number: \(\frac{2}{3}\) and \(\frac{4}{6}\) are the same rational number. We therefore write \(\frac{2}{3}=\frac{4}{6}\).


Characteristics of fractions

The following is true for fractions of any real number:

  1. When you multiply both numerator and denominator by the same integer (except \(0\)), the value does not change.
  2. For any two fractions, if you multiply the numerator of the first with the denominator of the second, and it’s the same value as multiplying the denominator of the first with the numerator of the second, both fractions have the same value.

A rational number with denominator \(1\) is an integer. For example, \(\frac{9}{-1}=\frac{-9}{1}=-9\). Every integer is therefore also rational.


Let’s practice!

Enter the missing numbers on the red and blue dots:

\(\frac{24}{30}=\frac{16}{\color{blue}{\text{…}}}=\frac{\color{red}{\text{…}}}{50}\)

The value of a fraction does not change if you multiply the numerator and denominator by the same number. First we calculate the blue dots:

\(\begin{array}{rclrcl} \frac{24}{30}=\frac{16}{\color{blue}{\text{…}}} &&&\phantom{xx}\color{blue}{\text{numerator multiplied by }\frac{16}{24}={{2}\over{3}} \text{}} \\ \frac{24}{30}=\frac{16}{20}&&&\phantom{xx}\color{blue}{\text{multiply the denominator by } {{2}\over{3}}} \\ \end{array}\)

Next, we calculate the red dots:

\(\begin{array}{rclrcl} \frac{24}{30}=\frac{\color{red}{\text{…}}}{50} &&&\phantom{xx}\color{blue}{\text{denominator multiplied by }\frac{50}{30}= {{5}\over{3}} \text{}} \\ \frac{24}{30}=\frac{40}{50}&&&\phantom{xx}\color{blue}{\text{multiply the numerator by } {{5}\over{3}}} \\ \end{array}\)

This gives \(\color{blue}{\text{…}}= 20\) and \(\color{red}{\text{…}}= 40\).

You can solve this exercise by simplifying \(\frac{24}{30}\) into \(\frac{4}{5}\).

Answer:

\(\color{blue}{\text{…}}= 20\)
\(\color{red}{\text{…}}= 40\)

Calculating with integers

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


When several operations must be performed one after another, the notation will have to represent this. For this purpose brackets are used. If a sub-expression is placed between brackets, then the part between the brackets has to be calculated first. For example:

\(\begin{array}{rcl}3\times (6+9) &=& 3\times 15 = 45\\ &\text{ and }&\\ (3\times 6) + 9 &=& 18+9 = 27\tiny.\end {array}\)

Because too many brackets could complicate the reading of the expression, we leave them out if the rules of precedence allow us to do so without changing the meaning of the expression.


Order of operations with numbers

  • In the case of repeated addition and repeated multiplication the order does not matter.
  • In the case of subtraction and repeated exponentiation the order does matter.
  • If there are no brackets at the repetition of subtraction, we interpret the expression as if there are brackets from left to right.
  • If there are no brackets at the repetition of exponentiation, we interpret the expression as if there are brackets from left to right.

Sum and product

An expression which consists of an addition of two or more sub-expressions is called a sum. The sub-expressions are called terms or summands.


Order of operations

The order of operations is:

  1. Calculate what is inside the brackets
  2. Exponentiation (in case of more operations in a row: from left to right)
  3. Multiplication
  4. Addition and subtraction (in case of more operations without brackets: from left to right)

Let’s practice!

Calculate \(-18\cdot 68 + 9\) and \(-18\cdot (68 + 9)\).

\(-18\cdot 68 + 9=-1224+9=-1215\)

\(-18\cdot (68 + 9) = -18\cdot 77 = -1386\)

 


Let’s try one with exponentiation!

\({{2}^{3}}^{2}={2}^{\displaystyle\left({3}^{2}\right)} = {2}^{9} = 512\tiny.\)

Answer:

\(512\)

The notion of integer

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


The counting of objects usually starts with \(1\). From here on you can reach all other natural numbers by counting up, or, in other words, by continuing to add \(1\) sufficiently often. For example,

\(\begin{array}{rcl}2 &=&1+1\\ 3 &=&1+1+1=2+1\\ 7 &= & 6+1\\ 2016 &= & 2015+1\\ \end {array}\)


The natural numbers are \( 1, 2, \ldots \) These numbers are greater than zero, that is, positive. The notation \(12\gt0\) expresses that \(12\) is bigger than \(11\).

If we count backwards from \(0\), then the negative integers arise: \( -1, -2, -3, \ldots\) The notation \(-10\lt0\) expresses that \(-10\) is negative. With \(-9\lt -8\) we indicate that \(-9\) is smaller than \(-8\).

To describe the set of all natural numbers and \(0\), we use the term non-negative integers. The notation \(10\ge0\) expresses that \(10\) is non-negative. With \(7\ge5\) we indicate that \(7\) is greater than \(5\).

The non-positive integers are \(0, -1, -2, \ldots \) The notation \(-3\le0\) expresses that \(-3\) is non-positive. By \(-9\le -8\) we indicate that \(-9\) is less than or equal to \(-8\).

The negative integers and the non-negative integers taken together are all integers. They arranged as follows

\(\cdots\lt-5\lt-4\lt-3\lt-2\lt-1\lt0\lt 1\lt2\lt3\lt4\lt5\lt\cdots\)

For negative number we can say that it has sign \(–\). A positive number has sign \(+\).


Addition, subtraction, multiplication, and exponentiation are called operations, sometimes arithmetic operations, because they create new integers from old ones.


Addition

If we add seven times \(1\) to \(5\), then the result is the number \(12\). Because seven times \(1\) is indicated by \(7\), we can write this activity briefly as \( 7 + 5 \). The equality \(7+5=12\) indicates the result of this operation. The addition of two numbers can be explained that way.


Subtraction

For counting backwards, we can use the minus sign. If we count backwards \(6\) starting from \(8\), we arrive at \(2\). This is denoted as \(8-6=2\). We can also use the negative number \(-6\) to describe this process: with \(8+ (-6)\) we mean the number you obtain from \(8\) by counting backwards \(6\). Hence \(8+(-6)=8-6=2\). This is convenient if we do not have enough natural numbers to make the counting backwards succeed: \(6 – 8 = -2\). After all, at \( 6 – 6\) we encounter \(0\), and from there we have to subtract \(2\) times \(1\) again, so that we end on the negative integer \(-2\). We can use the same convention when subtracting a negative integer \(-14\): which is equivalent to adding \(14\). Hence \(5-(-14) = 5+14=19\). The subtraction of two integers can be explained this way.


Multiplication

The equality \(3 \cdot 1 =3\) expresses that adding \(3\) times \(1\) to \(0\) yields the result \(3\). If instead of \(1\) we take for example \(4\), then \(3 \cdot 4\) expresses that we have to add \(4\) three times to \(0\). This result is \(12\), because:

\(3\times 4 = \underbrace{1+1+1+1}_{\text{one time}} + \underbrace{1+1+1+1}_{\text{two times}} + \underbrace{1+1+1+1}_{\text{three times}}=12\tiny.\)

The multiplication of integers can be explained this way.


Exponentiation

Finally, exponentiation does for multiplication what multiplication does for adding: \(3^5\) stands for \(3\cdot 3\cdot 3\cdot 3\cdot 3\), which is the same as \(243\).

In \(3^5\) , \(3\) is the base and \(5\) is the exponent.

Finding a common denominator

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


Addition and subtraction of fractions with variables is just like regular addition and subtraction of fractions:

  • If two or more fractions in a sum or difference have the same denominator, the denominator of the final result is the same, and the numerator is the sum or the difference of the original numerators.
  • If the fractions have different denominators, they must be brought under one denominator. This can be achieved by taking the product of the denominators of the two fractions as the new denominator and taking the product of the old numerator and the denominator of the other fraction as the new numerator for each fraction.

The corresponding formulas for the case of fractions \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c \) and \(d\) are polynomials, are not different from the formulas for rational numbers:

\(\begin{array}{lrcl}\text{same denominator}:\ \ &\frac{a}{b}+\frac{c}{b} &=& \frac{a+b}{b}\\ \text{in general: }&\frac{a}{b}+\frac{c}{d} &=& \frac{a\cdot d+b\cdot c}{b\cdot d} \end {array}\)


Let’s practice!

Put over a common denominator and expand all brackets from the expression \(\frac{x}{x-10}-\frac{1}{x+10}\):

\(\begin{array}{rcl}\displaystyle\frac{x}{x-10}-\frac{1}{x+10}&=&\displaystyle\frac{x(x+10)}{(x-10)\cdot(x+10)}-\frac{x-10}{(x-10)\cdot(x+10)}\\&&\phantom{ssppaacce}\color{blue}{\text{put over common denominators}}\\&=&\displaystyle\frac{x(x+10)+x-10}{(x-10)\cdot(x+10)}\\&&\phantom{ssppaacce}\color{blue}{\text{fractions with common denominators added}}\\&=&\displaystyle\frac{x^2+10x – x-10}{x^2-10^2}\\&&\phantom{ssppaacce}\color{blue}{\text{brackets expanded}}\\&=&\displaystyle\frac{x^2+9\cdot x+10}{x^2-100}\\&&\phantom{ssppaacce}\color{blue}{\text{simplified}}\end{array}\)

Answer:

\(\frac{x^2+9\cdot x+10}{x^2-100}\)


Now, let’s take a look at the reverse process:

The expanding of a fraciton as a sum of fractions with denominators which divide the original denominator is called partial fraction decomposition.


Let’s practice!

Split \({{p^2+1}\over{p}}\) in fractions with only one term in the numerator.

\({{p^2+1}\over{p}}=\frac{p^2}{p}+\frac{1}{p} = p+{{1}\over{p}}\)

Answer:

\(p+{{1}\over{p}}\)

Multiplication and division of fractions

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


The product of two fractions is the rational number which can be written as a fraction having an numerator that is equal to the product of the numerators and with a denominator that is equal to the product of the denominator.

In a formula, for integers \(a, b, c \) and \(d\) with \(b\) and \(d\) distinct from \(0\):

\(\begin{array}{rcl}\displaystyle \frac{a}{b}\cdot \frac{c}{d} &=&\displaystyle \frac{a\cdot c}{b\cdot d}\end {array}\)

The special case where \(c=b\) and \(d=a\) gives

\(\begin{array}{rcl}\displaystyle \frac{a}{b}\cdot \frac{b}{a} &=&1\tiny.\end{array}\)

The fraction \(\frac{b}{a}\) is called the inverse of \(\frac{a}{b}\).


For dividing fractions on each other:

Dividing by a fraction is the same as multiplying by the inverse fraction.

In a formula, for integers \(a, b, c \) and \(d\) with \(b\) and \(d\) distinct from \(0\):

\(\begin{array}{rcl}\frac{\ \displaystyle\frac{a}{b}\ }{\ \displaystyle\frac{c}{d}\ } &=&\displaystyle \frac{a}{b}\cdot \frac{d}{c}=\frac{a\cdot d}{b\cdot c}\end {array}\)


Let’s practice!

Calculate \(\frac{11}{7}\times\frac{5}{8}\) and simplify the answer as much as possible.

Answer:

\(\frac{11}{7}\times\frac{5}{8}=\frac{11\times 5}{7\times 8}=\frac{55}{56}\)


Let’s do one more!

Calculate \(\frac{17}{2}/\frac{19}{5}\) and simplify the answer as far as possible.

\(\frac{17}{2}/\frac{19}{5}=\frac{17}{2}\times\frac{5}{19}=\frac{17\times 5}{2\times 19}=\frac{85}{38}\tiny.\)

Answer:

\({{85}\over{38}}\)

Addition and subtraction of fractions

For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!


Both the sum and the difference of two rational numbers is a rational number.

Let \(a, b, c, d \) and \( n\) be integers with \(b\ne0, d\ne0 \) and \(n\ne0\).

A general formula for addition of the fractions \(\frac{a}{b}\) and \(\frac{c}{d}\) is

\(\frac{a}{b}+\frac{c}{d} = \frac{a\cdot d+b\cdot c}{b\cdot d} \tiny.\).In the case of common denominators, we have

\(\frac{a}{n}+\frac{b}{n}=\frac{a+b}{n}\tiny.\)


A formula for the subtraction of fractions is obtained from the given formulas by replacing \(c\) by \(-c\) in the formula for addition of fractions:

\(\frac{a}{b}-\frac{c}{d} = \frac{a\cdot d-b\cdot c}{b\cdot d} \tiny.\)


Let’s practice!

Calculate \(\frac{3}{5}-\frac{3}{4}+\frac{1}{3}\) and simplify as much as possible.

The least common multiple of the denominators \(5, 4 \) and \(3\) equals \(60\).
We put the fractions over this common denominator

\(\frac{3}{5}=\frac{3\times 12}{5\times 12}=\frac{36}{60}{\tiny,}\qquad\frac{3}{4}=\frac{3\times 15}{4\times 15}=\frac{45}{60}\quad\mathrm{en}\quad\frac{1}{3}=\frac{1\times 20}{3\times 20}=\frac{20}{60}\tiny.\)

The rest is simple arithmetic:

\(\frac{3}{5}-\frac{3}{4}+\frac{1}{3}=\frac{36}{60}-\frac{45}{60}+\frac{20}{60}=\frac{36-45+20}{60}=\frac{11}{60}\tiny.\)

Answer:

\(\frac{3}{5}-\frac{3}{4}+\frac{1}{3}=\frac{11}{60}\)