HSC MATHS FORMULA FOR BOARD EXAM.

MATHEMATICS

Class IX - Chapter 1: Number Systems

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Page 1: Introduction to Number Systems

Natural Numbers (\(\mathbb{N}\)): The collection of natural numbers is denoted by \(\mathbb{N}\).
\(\mathbb{N} = \{1, 2, 3, 4, \dots\}\)

Whole Numbers (\(\mathbb{W}\)): The collection of whole numbers is denoted by \(\mathbb{W}\).
\(\mathbb{W} = \{0, 1, 2, 3, \dots\}\)

Integers (\(\mathbb{Z}\)): The collection of integers is denoted by \(\mathbb{Z}\).
\(\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}\)

Rational Numbers (\(\mathbb{Q}\)): A number \(r\) is called a rational number if it can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). The collection of rational numbers is denoted by \(\mathbb{Q}\).
Examples: \(\frac{1}{2}, \frac{3}{4}, \frac{-5}{7}, 0, 5\) (since \(0 = \frac{0}{1}\) and \(5 = \frac{5}{1}\)).

Question: Is zero a rational number? Can you write it in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\)?
Answer: Yes, zero is a rational number. It can be written as \(\frac{0}{1}\), \(\frac{0}{2}\), \(\frac{0}{-3}\), etc., where \(p=0\) and \(q\) is any non-zero integer.

Example 1:

Are the following statements true or false? Give reasons for your answers.

1. Every whole number is a natural number.
   False, because 0 is a whole number but not a natural number.
2. Every integer is a rational number.
   True, because every integer \(m\) can be expressed as \(\frac{m}{1}\).
3. Every rational number is an integer.
   False, because \(\frac{3}{5}\) is a rational number but not an integer.

Rational numbers have decimal expansions that are either:

• Terminating (e.g., \(\frac{1}{2} = 0.5\), \(\frac{1}{4} = 0.25\))
• Non-terminating recurring (repeating) (e.g., \(\frac{1}{3} = 0.333\dots = 0.\bar{3}\), \(\frac{1}{7} = 0.\overline{142857}\))

Page 2: Finding Rational Numbers & Exercise 1.1

Example 2:

Find five rational numbers between 1 and 2.

Solution:

Method 1: To find a rational number between \(r\) and \(s\), we can use \(\frac{r+s}{2}\).
A rational number between 1 and 2 is \(\frac{1+2}{2} = \frac{3}{2}\).
A rational number between 1 and \(\frac{3}{2}\) is \(\frac{1 + \frac{3}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4}\).
A rational number between \(\frac{3}{2}\) and 2 is \(\frac{\frac{3}{2} + 2}{2} = \frac{\frac{7}{2}}{2} = \frac{7}{4}\).
Continuing this way, we can find: \(1 < \frac{5}{4} < \frac{3}{2} < \frac{7}{4} < 2\). We need two more. Let's find one between 1 and 5/4: \(\frac{1 + \frac{5}{4}}{2} = \frac{\frac{9}{4}}{2} = \frac{9}{8}\). And one between 7/4 and 2: \(\frac{\frac{7}{4} + 2}{2} = \frac{\frac{15}{4}}{2} = \frac{15}{8}\). So, five rational numbers are \(\frac{9}{8}, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, \frac{15}{8}\).

Method 2: Since we want five rational numbers, we write 1 and 2 as rational numbers with denominator \(5+1 = 6\).
\(1 = \frac{1 \times 6}{6} = \frac{6}{6}\)
\(2 = \frac{2 \times 6}{6} = \frac{12}{6}\)
Now, we can find five rational numbers between \(\frac{6}{6}\) and \(\frac{12}{6}\):
\(\frac{7}{6}, \frac{8}{6}, \frac{9}{6}, \frac{10}{6}, \frac{11}{6}\).

EXERCISE 1.1

1. Is zero a rational number? Can you write it in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\)?
   (Answered above: Yes, e.g., \(\frac{0}{1}\))
2. Find six rational numbers between 3 and 4.
   Using Method 2: \(n=6\), so denominator \(n+1=7\).
   \(3 = \frac{21}{7}\), \(4 = \frac{28}{7}\).
   Six rational numbers are \(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}\).
3. Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\).
   Using Method 2: \(n=5\), so multiply numerator and denominator by \(n+1=6\).
   \(\frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}\)
   \(\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}\)
   Five rational numbers are \(\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}\).
4. State whether the following statements are true or false. Give reasons for your answers.
   1. Every natural number is a whole number.
      True, because the collection of whole numbers contains all natural numbers.
   2. Every integer is a whole number.
      False, because negative integers (e.g., -2) are not whole numbers.
   3. Every rational number is a whole number.
      False, because \(\frac{1}{2}\) is a rational number but not a whole number.

Page 3: Irrational and Real Numbers

Irrational Numbers: A number \(s\) is called irrational if it cannot be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Irrational numbers have decimal expansions that are non-terminating and non-recurring.

Examples: \(\sqrt{2}, \sqrt{3}, \sqrt{15}, \pi, 0.10110111011110\dots\) Note: \(\pi \approx \frac{22}{7}\) or \(3.14\), but these are approximations. \(\pi\) is irrational.

Real Numbers (\(\mathbb{R}\)): The collection of all rational numbers and irrational numbers together make up what we call the collection of real numbers, denoted by \(\mathbb{R}\). Every real number is represented by a unique point on the number line. Also, every point on the number line represents a unique real number.

Example 3:

Locate \(\sqrt{2}\) on the number line.

Solution: Consider a unit square OABC with O at 0 on the number line, OA along the number line, and AB perpendicular to OA, each of length 1. By Pythagoras theorem, \(OB = \sqrt{OA^2 + AB^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\). Using a compass with center O and radius OB, draw an arc intersecting the number line at P. Then P corresponds to \(\sqrt{2}\).

Example 4:

Locate \(\sqrt{3}\) on the number line.

Solution: First, locate \(\sqrt{2}\) as above (point P). Construct a line BD of unit length perpendicular to OP at P. By Pythagoras theorem, \(OD = \sqrt{OP^2 + PD^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}\). Using a compass with center O and radius OD, draw an arc intersecting the number line at Q. Then Q corresponds to \(\sqrt{3}\).

Page 4: Exercise 1.2 & Representing Real Numbers

EXERCISE 1.2

1. State whether the following statements are true or false. Justify your answers.
   1. Every irrational number is a real number.
      True, as real numbers are composed of rational and irrational numbers.
   2. Every point on the number line is of the form \(\sqrt{m}\), where \(m\) is a natural number.
      False. Negative numbers are on the number line but cannot be expressed as \(\sqrt{m}\) for natural \(m\). Also, non-integer positive numbers like 0.5 are on the number line but \(\sqrt{m}\) for natural \(m\) like \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}=2\) does not cover all points. For example, 1.5 is not \(\sqrt{m}\) for natural \(m\).
   3. Every real number is an irrational number.
      False. Real numbers also include rational numbers (e.g., 2 is real but not irrational).
2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
   No. For example, \(\sqrt{4} = 2\), which is a rational number. \(\sqrt{9} = 3\), etc.
3. Show how \(\sqrt{5}\) can be represented on the number line.
   Construct a right triangle with base 2 units and height 1 unit. The hypotenuse will be \(\sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}\). Transfer this length to the number line using a compass with center at 0.

Representing Real Numbers on the Number Line (Successive Magnification)

To visualize the representation of real numbers with terminating or non-terminating recurring decimal expansions on the number line, we use the process of successive magnification.

Example 5:

Visualize the representation of \(5.3\overline{7}\) on the number line upto 5 decimal places, that is, up to \(5.37777\).

Solution:

1. \(5.37777\) lies between 5 and 6. Divide [5, 6] into 10 equal parts.
2. It lies between 5.3 and 5.4. Magnify the interval [5.3, 5.4] and divide it into 10 parts.
3. It lies between 5.37 and 5.38. Magnify [5.37, 5.38].
4. It lies between 5.377 and 5.378. Magnify [5.377, 5.378].
5. It lies between 5.3777 and 5.3778. Magnify [5.3777, 5.3778].
6. Locate 5.37777 within this final magnified interval.

(This process is usually shown with diagrams of number line segments being magnified).

Page 5: Exercise 1.4 & Operations on Real Numbers

EXERCISE 1.4

1. Visualize \(3.765\) on the number line, using successive magnification.
2. Visualize \(4.\overline{26}\) on the number line, up to 4 decimal places. (i.e., 4.2626)

(Solutions would involve steps similar to Example 5)

Operations on Real Numbers

• The sum or difference of a rational number and an irrational number is irrational.
• The product or quotient of a non-zero rational number with an irrational number is irrational.
• If we add, subtract, multiply, or divide two irrationals, the result may be rational or irrational.
  • Examples: \(\sqrt{6} + (-\sqrt{6}) = 0\) (rational)
  • \(\sqrt{2} - \sqrt{2} = 0\) (rational)
  • \(\sqrt{2} \cdot \sqrt{2} = 2\) (rational)
  • \(\frac{\sqrt{18}}{\sqrt{2}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3\) (rational)
  • \(\sqrt{2} + \sqrt{3}\) (irrational)
  • \(\sqrt{2} \cdot \sqrt{3} = \sqrt{6}\) (irrational)

Example (Checking rationality):

Check whether \(7\sqrt{5}\), \(\frac{7}{\sqrt{5}}\), \(\sqrt{2}+21\), \(\pi-2\) are irrational numbers or not.

Solution: \(\sqrt{5} = 2.236\dots\), \(\sqrt{2} = 1.414\dots\), \(\pi = 3.1415\dots\) are irrational.

• \(7\sqrt{5} = 7 \times 2.236\dots = 15.652\dots\) (Non-terminating, non-recurring) \(\implies\) Irrational.
• \(\frac{7}{\sqrt{5}} = \frac{7\sqrt{5}}{5} = \frac{7 \times 2.236\dots}{5} = \frac{15.652\dots}{5} = 3.1304\dots\) (Non-terminating, non-recurring) \(\implies\) Irrational.
• \(\sqrt{2}+21 = 1.414\dots + 21 = 22.414\dots\) (Non-terminating, non-recurring) \(\implies\) Irrational.
• \(\pi-2 = 3.1415\dots - 2 = 1.1415\dots\) (Non-terminating, non-recurring) \(\implies\) Irrational.

Page 6: Simplifying Expressions & Rationalising Denominators

Example (Simplification):

Simplify the following expressions:

1. \((5+\sqrt{7})(2+\sqrt{5})\)
   \( = 5(2+\sqrt{5}) + \sqrt{7}(2+\sqrt{5}) = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35} \)
2. \((5+\sqrt{5})(5-\sqrt{5})\)
   Using \((a+b)(a-b) = a^2 - b^2\): \( = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20 \)
3. \((\sqrt{3}+\sqrt{7})^2\)
   Using \((a+b)^2 = a^2 + 2ab + b^2\): \( = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{7}) + (\sqrt{7})^2 = 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21} \)
4. \((\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})\)
   Using \((a-b)(a+b) = a^2 - b^2\): \( = (\sqrt{11})^2 - (\sqrt{7})^2 = 11 - 7 = 4 \)

Identities for square roots:

Let \(a\) and \(b\) be positive real numbers. Then

• \(\sqrt{ab} = \sqrt{a}\sqrt{b}\)
• \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\)
• \((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a-b\)
• \((a+\sqrt{b})(a-\sqrt{b}) = a^2-b\)
• \((\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{d}) = \sqrt{ac} + \sqrt{ad} + \sqrt{bc} + \sqrt{bd}\)
• \((\sqrt{a}+\sqrt{b})^2 = a + 2\sqrt{ab} + b\)

Rationalising the Denominator

This means to convert a denominator containing a square root term into a rational number by multiplying numerator and denominator by a suitable factor.

Example:

Rationalise the denominator of \(\frac{1}{\sqrt{2}}\).

Solution: \(\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

Example:

Rationalise the denominator of \(\frac{1}{2+\sqrt{3}}\).

Solution: The conjugate of \(2+\sqrt{3}\) is \(2-\sqrt{3}\).
\(\frac{1}{2+\sqrt{3}} = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}\)

Example:

Rationalise the denominator of \(\frac{5}{\sqrt{3}-\sqrt{5}}\).

Solution: The conjugate of \(\sqrt{3}-\sqrt{5}\) is \(\sqrt{3}+\sqrt{5}\).
\(\frac{5}{\sqrt{3}-\sqrt{5}} = \frac{5}{\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} = \frac{5(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2} = \frac{5(\sqrt{3}+\sqrt{5})}{3-5} = \frac{5(\sqrt{3}+\sqrt{5})}{-2} = -\frac{5}{2}(\sqrt{3}+\sqrt{5})\)

Example:

Rationalise the denominator of \(\frac{1}{7+3\sqrt{2}}\).

Solution: The conjugate of \(7+3\sqrt{2}\) is \(7-3\sqrt{2}\).
\(\frac{1}{7+3\sqrt{2}} = \frac{1}{7+3\sqrt{2}} \times \frac{7-3\sqrt{2}}{7-3\sqrt{2}} = \frac{7-3\sqrt{2}}{7^2 - (3\sqrt{2})^2} = \frac{7-3\sqrt{2}}{49 - (9 \times 2)} = \frac{7-3\sqrt{2}}{49 - 18} = \frac{7-3\sqrt{2}}{31}\)

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