Number Systems \(\to\) Summary

Natural numbers (N) – \(1,2,3,4,5…\)
Whole Numbers (W) – \(0,1,2,3,4,5…\)
Integers (Z) – \(…-3,-2,-1,0,1,2,3,…\)
Rational Numbers (Q) – A number of the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q\neq 0.\)
Every integer is a rational number. For example, the integer \(5\) can be expressed as a rational number \(\frac51.\)
Every rational number need not be an integer. For example, \(\frac29\) is a rational number, but it is not an integer.
Every natural number is a whole number, but every whole number need not be a natural number, In fact, 0 is a whole number that is not a natural number.
\(N \to W\to Z\to Q\)

a gif on how the different number systems are related

Equivalent rational numbers – \(\frac24, \frac36,\frac48,…,…\) – All represent the same rational number \(\frac12\)
Convention: We usually express rational numbers in their simplified form. Thus, not only are \(p\) and \(q\) integers where \(q\neq 0,\) we also take \(p\) and \(q\) to be co-prime (no other common factors than \(1\)).
\(\frac{r+s}{2}\) is a rational number that lies between the integers \(r\) and \(s.\)
There are infinitely many rational numbers between any two integers, and hence between any two rational numbers.
Irrational number – A number that cannot be written in the form \(\frac{p}{q},\) where \(p\) and \(q\) are integers and \(q\neq 0.\)
- Examples: \(\sqrt2, \sqrt3, \sqrt7, \pi, 0.783610928…,9.6820523…\)
There are infinitely many irrational numbers.

A gif of how rational, irrational and real numbers are related

A real number is either rational or irrational.
Every real number can be represented by a unique point on the number line.
Also, every point on the number line corresponds to a unique real number.
To locate \(\sqrt{n}\) on the number line:
- We first locate \(\sqrt{n-1}.\)
- Then, constructing a right-triangle with \(\sqrt{n-1}\) as the base and \(1\) unit as the height, we get the hypotenuse of the triangle to be of \(\sqrt{(\sqrt{n-1})^2+(1)^2}=\sqrt{n-1+1}=\sqrt{n}\) units.
- Now, using the origin as the centre and \(\sqrt{n}\) as the radius, construct a circle that cuts the number line on the positive side. This point represents the real number \(\sqrt{n}.\)
  - Example: \(\sqrt5\)

Locating square root of 5 on the number line (with steps)

Terminating decimal
- The decimal expansion terminates (that is, the remainder becomes zero) after a finite number of steps.
  - Examples: \(\frac52=2.5,\frac48=0.5,\frac78=0.875,0.5639562,…\)
Non-terminating decimal
- The decimal expansion does not terminate (that is, the remainder never becomes zero).
  - Examples: \(\frac{10}{3}=3.333333…=3.\overline{3}, 1.920673598328…\)
Two types of non-terminating decimals
- Repeating – The decimal expansion does not terminate, but repeats itself.
  - Examples: \(\frac{10}{3}=3.333…=3.\overline{3}, 5.8633333…=5.86\overline{3}, 7.35353535…=7.\overline{35}, 4.123123123…=4.\overline{123}\)
- Non-repeating – The decimal expansion neither terminates nor repeats.
  - Examples: \(\pi=3.1415926…, \sqrt2=1.414213…, e=2.71828182846…\)

Classification of decimals with examples and their decimal expansions

A rational number is either terminating or non-terminating repeating. Also, a number whose decimal expansion is either terminating or non-terminating repeating is rational.
An irrational number is non-terminating non-repeating. Also, a number whose decimal expansion is non-terminating non-repeating is irrational.

Rational and irrational numbers and decimal expansions

Note:
- \(\pi=3.1415926…\) is a non-terminating non-repeating decimal, and hence is an irrational number.
- \(\frac{22}{7}=3.\overline{142857}\) is a non-terminating repeating decimal, and hence is a rational number.
- \(\frac{22}{7}\) is an approximation of \(\pi\) and \(\pi \neq \frac{22}{7}.\)
Expressing the non-terminating repeating decimal \(0.8888…=0.\overline{8}\) in the form \(\frac{p}{q},\) where \(p\) and \(q\) are integers and \(q\neq 0\):
- Let \(y=0.\overline{8}.\)
- \(\implies 10y=10\times 0.\overline{8}=8.\overline{8}\)
- \(8.\overline{8}=8+0.\overline{8}\)
- \(8.\overline{8}=8+y\)
- \(\implies 10y=8+y\implies 9y=8\implies y=\color{green}{\frac{8}{9}}\)
Successive magnification to represent real numbers on the number line
- Example: \(3.5\overline{6}\)….
  - To represent this rational number on the number line, we first note that it is located between the integers \(3\) and \(4.\)
  - We divide the portion between the integers into \(10\) equal parts.
  - Note that \(3.5\overline{6}\) lies between \(3.5\) and \(3.6.\)
  - We divide the portion between \(3.5\) and \(3.6\) into \(10\) equal parts and note that \(3.5\overline{6}\) lies between \(3.56\) and \(3.57.\)
  - Dividing the portion between \(3.56\) and \(3.57\) into \(10\) equal parts, we note that \(3.5\overline{6}\) lies between \(3.566\) and \(3.567.\)
  - As we keep repeating the process, the accuracy of representation increases.

3.566666.... on the number line using successive magnification — Visualisation of \(3.5\overline{6}\) upto 4 decimal places

The collection of irrationals (like that of rationals) obeys commutative, associative and distributive laws of addition and multiplication.
The sum, difference, product and quotient of irrational numbers need not always be irrational.
- Examples:
  - \(({\color{red}{-\sqrt2}})+({\color{red}{\sqrt2}})={\color{green}{0}}\)
  - \({\color{red}{\sqrt5}}-{\color{red}{\sqrt5}}={\color{green}{0}}\)
  - \({\color{red}{\sqrt2}}\times {\color{red}{\sqrt2}}={\color{green}{2}}\)
  - \(\frac{{\color{red}{\sqrt2}}}{{\color{red}{\sqrt2}}}={\color{green}{1}}\)
The sum or difference of a rational number and an irrational number is irrational.
- Examples:
  - \({\color{green}{2}}+{\color{red}{\sqrt3}}={\color{red}{2+\sqrt3}}={\color{red}{3.7320508075…}}\)
  - \({\color{green}{2}}-{\color{red}{\sqrt3}}={\color{red}{2-\sqrt3}}={\color{red}{0.267949192…}}\)
The product or quotient of a non-zero rational number and an irrational number is irrational.
- Examples:
  - \({\color{green}{2}}\times ({\color{red}{\sqrt3+\sqrt5}})={\color{red}{2\sqrt3+2\sqrt5}}={\color{red}{7.936237570…}}\)
  - \(\frac{{\color{green}{2}}}{{\color{red}{\sqrt3}}}={\color{red}{\frac{2}{\sqrt3}}}={\color{red}{1.154700538…}}\)
If \(a\gt 0\) is a real number such that \(\sqrt{a}=b,\) then \(a=b^2\) and \(b\gt 0.\)
Locating \(\sqrt{x}\) on the number line where \(x\) is a positive real number

Steps to locate square root of x on the number line (Part 1)

\(n^\text{th}\) root of \(a\) where \(a\) is a positive real:
- \(\sqrt4=2\to\) The square root of \(4\) is \(2.\)
  - \(2^2=4\)
- \(\sqrt[3]{64}=4\to\) The cube root of \(64\) is \(4.\)
  - \(4^3=64\)
- \(\sqrt[n]{a}=b\to\) The \(n^\text{th}\) root of \(a\) is \(b.\)
  - \(b^n=a,\) where \(b\gt 0\)
\(\sqrt{}\to\) Radical sign
Let \(a\) and \(b\) be positive reals. Then, we have:
- \(\sqrt{ab}=\sqrt{a}\sqrt{b}\)
- \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)
- \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=(\sqrt{a})^2-(\sqrt{b})^2=a-b\)
- \((\sqrt{a}+\sqrt{b})^2=(\sqrt{a})^2+2(\sqrt{a})(\sqrt{b})+(\sqrt{b})^2=a+2\sqrt{ab}+b\)
Rationalising the denominator of an irrational number to remove the radical sign in the denominator
- Examples:
  - To rationalise the denominator of \(\frac{1}{\sqrt3},\) we multiply both its numerator and denominator by \({\color{blue}{\sqrt3}}.\)
    - \(\frac{1}{\sqrt3}=\frac{1}{\sqrt3}\times \frac{{\color{blue}{\sqrt3}}}{{\color{blue}{\sqrt3}}}=\frac{\sqrt3}{(\sqrt3)^2}=\frac{\sqrt3}{3}\)
  - To rationalise the denominator of \(\frac{1}{\sqrt5+\sqrt3},\) multiply both numerator and denominator by \({\color{blue}{\sqrt5-\sqrt3}}.\)
    - \(\frac{1}{\sqrt5+\sqrt3}=\frac{1}{\sqrt5+\sqrt3} \times \frac{{\color{blue}{\sqrt5-\sqrt3}}}{{\color{blue}{\sqrt5-\sqrt3}}}=\frac{\sqrt5-\sqrt3}{(\sqrt5)^2-(\sqrt3)^2}=\frac{\sqrt5-\sqrt3}{5-3}=\frac{\sqrt5-\sqrt3}{2}\)
\(\sqrt[{\color{red}{n}}]{{\color{green}{a}}}={\color{green}{a}}^{\frac{1}{{\color{red}{n}}}},\) where \(a\) is a real number and \(n\) is a positive integer
Laws of exponents for real numbers:
- Let \(a\) and \(b\) be real numbers. Let \(p\) and \(q\) be rational numbers. Then, we have:
  - \((a^p)^q=a^{pq}=(a^q)^p\)
  - \(a^p\times a^q=a^{p+q}\)
  - \(\frac{a^p}{a^q}=a^{p-q}, a\neq 0\)
  - \((ab)^p=a^p\times b^p\)
  - \(a^0=1\)
Let \(a>0\) be a real number. Let \(m\) and \(n\) be integers such that \(n\neq 0.\) Then, we have:
- \(a^{\frac{m}{n}}={(a^m)}^{\frac{1}{n}}=\sqrt[n]{a^m}\)

Number Systems \(\to\) Summary

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