Number Systems

Number Systems

Counting begins with \(1,2,3,\ldots\), but the equation \(x+4=4\) requires the number \(0\). The equation \(x+7=2\) requires a negative number, and \(3x=2\) requires a rational number. The equation \(x^2=2\) cannot be solved by any rational number, while \(x^2=-1\) cannot be solved by any real number.

Thus the number systems appear naturally in the hierarchy \[\mathbb N\subset\mathbb W\subset\mathbb Z\subset\mathbb Q\subset\mathbb R\subset\mathbb C.\]

The symbol \(A\subset B\) means that every number belonging to \(A\) also belongs to \(B\), and that \(A\ne B\).

Natural Numbers

The numbers \(1,2,3,4,\ldots\), used for counting, are called the natural numbers. We write \[\mathbb N:\qquad 1,2,3,\ldots\]

Conventions for \(\mathbb N\) differ. Throughout this book, \[\mathbb N=\{1,2,3,\ldots\},\] so \(0\) is not included among the natural numbers.

Natural numbers are sufficient for ordinary counting, addition, and multiplication; for example, \(5+8=13\) and \(5\cdot8=40\). They are not sufficient for every subtraction: \(3-7\) is not a natural number. Even the equation \(x+5=5\) has no natural-number solution.

Whole Numbers

To solve \(x+5=5\), we need \(x=0\). Hence we adjoin \(0\) to the natural numbers and obtain the whole numbers: \[\mathbb W:\qquad 0,1,2,3,\ldots\]

Now equations such as \(x+7=7\) are solvable. However, \(x+9=4\) requires \(x=-5\), which is still not a whole number.

Thus subtraction again forces an enlargement.

Integers

Positive integers, zero, and negative integers together form the integers: \[\mathbb Z:\qquad \ldots,-3,-2,-1,0,1,2,3,\ldots\]

Thus \[\mathbb N\subset\mathbb W\subset\mathbb Z.\]

The position of an integer relative to \(0\) determines its sign. The dots below represent integers; the coloured guide arrows show the relevant direction from \(0\).

Thus \(0\) is non-negative and non-positive, but it is neither positive nor negative.

Integers are sufficient for addition, subtraction, and multiplication. But division creates a new difficulty: \(2x=1\) has no integer solution, since solving it requires \(x=\frac12\).

Example 1.

Among \(-11,0,7,18\), identify the natural numbers, whole numbers, and integers.

Solution.

The positive integers \(7\) and \(18\) are natural numbers.

The numbers \(0,7,18\) are whole numbers, and all four numbers \(-11,0,7,18\) are integers.

Notice that a natural number is also a whole number and an integer. The number systems are nested rather than mutually exclusive.

Rational Numbers

The equation \(qx=p\), where \(q\ne0\), has solution \(x=\frac pq\).

A number is called rational if it can be written in the form \[\frac pq,\] where \(p\) and \(q\) are integers and \(q\ne0\).

The rational numbers are denoted by \(\mathbb Q\).

Examples are \[\frac37,\qquad -\frac{11}{4},\qquad 0=\frac05,\qquad 8=\frac81.\]

Every integer is rational because \(n=\frac n1\). Therefore \(\mathbb Z\subset\mathbb Q\).

A rational number may have many fractional representations: \[\frac12=\frac24=\frac{50}{100}.\]

Decimal Forms of Rational Numbers

A rational number can also be written as a decimal.

For example, \(\frac38=0.375\). The decimal ends after finitely many digits, so it is called a terminating decimal.

On the other hand, \(\frac13=0.3333\ldots=0.\overline3\) and \(\frac7{12}=0.58333\ldots=0.58\overline3\).

These decimals do not terminate, but a block of digits eventually repeats forever. They are called non-terminating recurring decimals.

Decimal form of a rational number

Every rational number has a decimal expansion that is either \[\text{terminating}\] or \[\text{non-terminating but eventually repeating}.\]

Conversely, every terminating decimal and every eventually repeating decimal represents a rational number.

Why \(0.999\ldots=1\)

The recurring decimal \(0.9999\ldots=0.\overline9\) does not represent a number slightly smaller than \(1\). It is another decimal representation of \(1\).

Let \(x=0.9999\ldots\). Multiplying by \(10\) gives \(10x=9.9999\ldots\). The recurring tails are identical, so subtraction gives \(10x-x=9\). Therefore \[9x=9, \qquad x=1.\] Hence \[0.9999\ldots=1.\]

The decimal \(0.999\ldots\) represents the same real number as \(1\)

The dots mean that the digit \(9\) continues forever. A finite decimal such as \(0.999\) is less than \(1\), but the infinite recurring decimal \(0.999\ldots\) is exactly equal to \(1\).

When Does a Rational Number Terminate?

Suppose \(\frac pq\) has been reduced to lowest terms with \(q>0\).

Its decimal expansion terminates exactly when \(q=2^m5^n\) for some non-negative integers \(m,n\).

In other words, the reduced denominator may contain only the prime factors \(2\) and \(5\).

For example, \(\frac7{40}=\frac7{2^3\cdot5}\) terminates, whereas \(\frac7{12}=\frac7{2^2\cdot3}\) does not terminate.

Reduce before testing

The denominator test must be applied only after the fraction is reduced to lowest terms.

For example, \(\frac6{15}=\frac25=0.4\). The original denominator \(15\) contains a factor \(3\), but that factor disappears after cancellation.

Example 2.

Without carrying out long division, decide which of \(\frac{21}{250}\), \(\frac{14}{75}\), and \(\frac{18}{45}\) have terminating decimal expansions.

Solution.

For \(\frac{21}{250}\), the denominator is \(250=2\cdot5^3\), so the decimal terminates. The fraction \(\frac{14}{75}\) is already reduced, and \(75=3\cdot5^2\); the factor \(3\) prevents termination. Finally, \(\frac{18}{45}=\frac25\), so its decimal terminates.

Therefore the first and third fractions terminate, while the second does not.

Converting a Repeating Decimal into a Fraction

The basic idea is to multiply by a suitable power of \(10\) so that the repeating portions line up and disappear upon subtraction.

Example 3.

Convert \(0.\overline{27}\) into a fraction.

Solution.

Let \(x=0.\overline{27}=0.272727\ldots\). Then \[\begin{aligned} 100x&=27.272727\ldots,\\ 100x-x&=27,\\ 99x&=27. \end{aligned}\] Therefore \[x=\frac3{11}.\]

Example 4.

Convert \(0.1\overline{6}\) into a fraction.

Solution.

Let \(x=0.16666\ldots\). Then \[\begin{aligned} 10x&=1.6666\ldots,\\ 100x&=16.6666\ldots,\\ 100x-10x&=15,\\ 90x&=15. \end{aligned}\] Hence \[x=\frac16.\]

Without doing long division, decide whether \(\frac{17}{120}\) has a terminating decimal expansion.

Answer. The fraction is already reduced and \(120=2^3\cdot3\cdot5\). Since the reduced denominator contains the prime factor \(3\), the decimal is non-terminating and recurring.

Integral and Fractional Parts

A real number can be separated into an integral part and a remaining fractional part.

Throughout this book, for a real number \(x\), the notation \([x]\) denotes the greatest integer not exceeding \(x\).

Thus \[[4.7]=4,\qquad [6]=6.\]

For negative numbers, care is needed: \([-2.3]=-3\), because \(-3\le-2.3\), while \(-2>-2.3\).

The fractional part of \(x\) is denoted by \(\{x\}\) and is defined here by \[\{x\}=x-[x].\]

Therefore \(0\le\{x\}<1\).

Example 5.

Find the integral and fractional parts of \(\frac{17}{4}\) and \(-\frac{17}{4}\).

Solution.

Since \(\frac{17}{4}=4.25\), we have \[\left[\frac{17}{4}\right]=4,\qquad \left\{\frac{17}{4}\right\}=\frac{17}{4}-4=\frac14.\]

Now \(-\frac{17}{4}=-4.25\). The greatest integer not exceeding \(-4.25\) is \(-5\). Hence \[\left[-\frac{17}{4}\right]=-5,\qquad \left\{-\frac{17}{4}\right\}=-\frac{17}{4}-(-5)=\frac34.\]

Integral part is not truncation toward zero

For a negative non-integer number, \([x]\) is not obtained merely by deleting the decimal digits. For example, \([-3.2]=-4\), not \(-3\).

Example 6.

A real number \(x\) satisfies \([x]=5\) and \(\{x\}=\frac37\). Find \(x\).

Solution.

Using \(x=[x]+\{x\}\), we obtain \[x=5+\frac37 =\frac{38}{7}.\]

Irrational Numbers

Rational numbers are still not sufficient for every equation.

Consider \[x^2=2.\] A positive solution exists geometrically: it is the length of the diagonal of a unit square. This number is written \[\sqrt2.\]

The important fact is that \(\sqrt2\) is not rational.

Why \(\sqrt2\) Is Irrational

Assume, for contradiction, that \[\sqrt2=\frac pq,\] where \(p,q\) are integers, \(q\ne0\), and the fraction has been reduced to lowest terms.

Squaring gives \[2=\frac{p^2}{q^2},\] so \[p^2=2q^2.\]

Therefore \(p^2\) is even. The square of an odd integer is odd; hence, if \(p^2\) is even, \(p\) must be even. Write \[p=2k.\] Then \[(2k)^2=2q^2,\] so \[4k^2=2q^2,\] and therefore \[q^2=2k^2.\] Hence \(q^2\) is even. By the same reasoning, \(q\) is also even.

Thus both \(p\) and \(q\) are divisible by \(2\), contradicting the assumption that \[\frac pq\] was in lowest terms.

Therefore \[\sqrt2\text{ is irrational}.\]

An irrational number is a real number that cannot be written as \[\frac pq\] with integers \(p,q\) and \(q\ne0\).

Its decimal expansion is non-terminating and does not eventually repeat.

Examples include \[\sqrt2,\qquad \sqrt3,\qquad \pi,\qquad e.\]

Rational and irrational numbers are interspersed on the same real number line. The coloured points below show several familiar examples; their positions are approximate.

Important Irrational Constants

The number \(\pi\) (pi) arises naturally from circles; it is the common ratio of circumference to diameter. The number e appears naturally in continuous growth, logarithms, calculus, and limiting processes.

Two elementary geometric examples illustrate how important irrational numbers arise naturally:

For a circle of diameter \(1\), the circumference is \(\pi\). For a right triangle with perpendicular sides \(1\) and \(1\), the Pythagorean theorem gives the hypotenuse \[\sqrt{1^2+1^2}=\sqrt2.\]

Useful decimal approximations are \[\sqrt2\approx1.414,\qquad \sqrt3\approx1.732,\] \[\pi\approx3.142,\qquad e\approx2.718.\]

These decimal values are approximations

The symbols \[\sqrt2,\quad \sqrt3,\quad \pi,\quad e\] represent exact numbers.

Statements such as \[\pi\approx3.142\] are approximations. It is incorrect to write \[\pi=3.142.\]

Example 7.

Decide whether each of \(\sqrt{49}\), \(\sqrt{18}\), and \(0.101001000100001\ldots\) is rational or irrational.

Solution.

First, \(\sqrt{49}=7\), so it is rational.

Next, \(\sqrt{18}=3\sqrt2\). If \(3\sqrt2\) were rational, division by the non-zero rational number \(3\) would make \(\sqrt2\) rational, which is impossible. Hence \(\sqrt{18}\) is irrational.

Finally, \(0.101001000100001\ldots\) does not terminate. The number of zeros between successive \(1\)’s is unbounded, so the decimal expansion cannot eventually become periodic. Hence the number is irrational.

Real Numbers

Rational and irrational numbers together form the real numbers, denoted by \(\mathbb R\). Thus \(\mathbb Q\subset\mathbb R\).

The real numbers are precisely the numbers represented by points on the real number line.

Every real number is either rational or irrational.

Square Roots of Non-Negative Real Numbers

Consider the equation \(x^2=a\).

If \(a>0\), this equation has two real solutions, \[x=\pm\sqrt a.\] If \(a=0\), its only solution is \(x=0\).

For \(a\ge0\), the symbol \(\sqrt a\) denotes the non-negative square root of \(a\). Thus \[(\sqrt a)^2=a,\qquad \sqrt a\ge0.\]

It is important to distinguish the value of a square-root expression from the solutions of an equation. For example, \(\sqrt{25}=5\), whereas \(x^2=25\implies x=\pm5\). The symbol \(\sqrt a\) denotes one non-negative number; it does not mean “plus or minus.” The sign \(\pm\) appears when both solutions of a quadratic equation are being stated.

For non-negative real numbers, square roots satisfy the product rule \[\sqrt{ab}=\sqrt a\,\sqrt b,\qquad a\ge0,\ b\ge0,\] and the quotient rule \[ \sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}, \qquad a\ge0,\ b>0. \] The stated conditions are essential: these real square-root rules are not being asserted for negative radicands.

Modulus of a Real Number

On the real number line, the distance of a number from \(0\) is always non-negative.

This distance is called its modulus or absolute value and is denoted by \(|x|\).

For example, \[|5|=5,\qquad |-5|=5,\qquad |0|=0.\]

Algebraically, \[ |x|= \begin{cases} x,&x\ge0,\\ -x,&x<0. \end{cases}\]

The modulus is connected directly with square roots by \[|x|=\sqrt{x^2}.\] Equivalently, \(\sqrt{x^2}=|x|\), which is not necessarily equal to \(x\). For example, \(\sqrt{(-5)^2}=5=|-5|\).

The modulus of a difference measures distance: \(|x-a|\) is the distance between the real numbers \(x\) and \(a\).

The first line shows that \(5\) and \(-5\) are equally far from \(0\). The second shows that \(|x-a|\) measures the distance between two arbitrary points \(a\) and \(x\).

Basic Modulus Equations

The solutions of \(|x|=a\) depend on the value of \(a\): \[ \begin{array}{ll} \text{no real solution},&a<0,\\ x=0,&a=0,\\ x=\pm a,&a>0. \end{array}\]

More generally, \(|x-c|=r\), where \(r>0\), means that \(x\) is at distance \(r\) from \(c\). Therefore \[x=c-r\text{ or }x=c+r.\]

Example 8.

Solve \(|2x-5|=7\).

Solution.

We have \(2x-5=7\) or \(2x-5=-7\). Thus \[2x=12 \qquad\text{or}\qquad 2x=-2,\] so \[x=6\text{ or }x=-1.\]

Intervals on the Real Line

Inequalities often describe a continuous portion of the real line. Interval notation gives a compact way to write such portions.

For example, \(2<x<7\) is written as \((2,7)\).

The round bracket means that the endpoint is not included.

The inequality \(2\le x\le7\) is written as \([2,7]\).

The square bracket means that the endpoint is included.

Similarly, \(2<x\le7\) is written as \((2,7]\), while \(2\le x<7\) is written as \([2,7)\).

Finite interval notation

For \(a<b\), \[\begin{aligned} (a,b) &\quad\Longleftrightarrow\quad a<x<b,\\ [a,b] &\quad\Longleftrightarrow\quad a\le x\le b,\\ (a,b] &\quad\Longleftrightarrow\quad a<x\le b,\\ [a,b) &\quad\Longleftrightarrow\quad a\le x<b. \end{aligned}\]

Number-Line Representation of Finite Intervals

A hollow dot means that an endpoint is excluded; it corresponds to a round bracket. A solid dot means that an endpoint is included; it corresponds to a square bracket.

Infinite Intervals

Some inequalities continue indefinitely.

For example, \(x>3\) is written as \((3,\infty)\), while \(x\ge3\) is written as \([3,\infty)\). Similarly, \(x<-2\) is written as \((-\infty,-2)\), while \(x\le-2\) is written as \((-\infty,-2]\).

The symbols \(\infty\) and \(-\infty\) indicate that the interval continues without bound.

The same endpoint convention applies on a number line. The arrow shows the direction in which the interval continues.

Infinity is never included as an endpoint

The symbol \(\infty\) does not represent an ordinary real number. Therefore an infinite end always uses a round bracket: \[(3,\infty),\qquad (-\infty,5].\]

Expressions such as \([3,\infty]\) are not used.

Example 9.

Write \(-2<x<3\) in interval notation and state the endpoint types.

Solution.

Both inequalities are strict, so neither endpoint is included. Both endpoints are represented by hollow dots, and the required interval is \[(-2,3).\]

Example 10.

Write \(x\le-4\) or \(x>1\) using interval notation.

Solution.

The part \(x\le-4\) is written as \((-\infty,-4]\), while \(x>1\) is written as \((1,\infty)\).

Thus the description consists of the two intervals \[(-\infty,-4]\text{ and }(1,\infty).\]

Complex Numbers

The real number system is large enough to solve equations such as \(x^2=2\), but it still does not solve every algebraic equation. Consider \(x^2=-1\). For every real number \(x\), \(x^2\ge0\), so no real number can satisfy this equation.

To make such equations solvable, we introduce a new number \(i\) defined by \[i^2=-1.\]

The symbol \(i\) is called the imaginary unit.

Numbers of the form \[a+ib,\] where \(a\) and \(b\) are real numbers, are called complex numbers.

The complex numbers are denoted by \(\mathbb C\). Thus \(\mathbb R\subset\mathbb C\).

A real number is already a complex number because \(a=a+i\cdot0\). Similarly, a purely imaginary number \(ib\) has real part \(0\).

Square Roots of Negative Real Numbers

Let \(a>0\). Since \(i^2=-1\), we have \((i\sqrt a)^2=i^2(\sqrt a)^2=-a\). Therefore the equation \(z^2=-a\) has the two solutions \[z=\pm i\sqrt a.\]

When we write \(\sqrt{-a}=i\sqrt a\) for \(a>0\), the symbol denotes the principal square root of \(-a\). Thus \(\sqrt{-16}=4i\), whereas the equation \(z^2=-16\) has the two solutions \(z=\pm4i\). Similarly, \(\sqrt{-9}=3i\) denotes the principal square root, while \(z^2=-9\) has the solutions \(z=\pm3i\).

Square-root rules require their stated conditions

The real-number identity \(\sqrt{ab}=\sqrt a\,\sqrt b\) was stated for \(a\ge0\) and \(b\ge0\). It must not be used blindly for negative or complex quantities. For example, \(\sqrt{(-1)(-1)}=\sqrt1=1\), but \(\sqrt{-1}\sqrt{-1}=i^2=-1\). Therefore \(\sqrt{ab}=\sqrt a\,\sqrt b\) is not valid in general for arbitrary complex numbers.

Example 11.

Write \(3+\sqrt{-16}\) and \(5-\sqrt{-9}\) in the form \(a+ib\).

Solution.

Since \(\sqrt{-16}=4i\), we get \(3+\sqrt{-16}=3+4i\). Similarly, \(\sqrt{-9}=3i\), so \(5-\sqrt{-9}=5-3i\).

Hence the required forms are \[3+4i,\qquad 5-3i.\]

The Number-System Hierarchy

We can now summarize the enlargements: \[ \mathbb N \subset \mathbb W \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C .\]

Each enlargement solves equations that could not always be solved in the previous system: \(x+5=5\) requires \(0\), \(x+7=2\) requires a negative integer, \(3x=2\) requires a rational number, \(x^2=2\) requires an irrational real number, and \(x^2=-1\) requires a complex number.

The hierarchy is therefore not an arbitrary list. It records successive extensions of the kinds of numbers available to us.