Chapter 1 Number Systems (Concepts)

Rational Numbers - Introduction and Classification

Welcome to Class 9 Mathematics! We begin our journey by delving deeper into the fascinating world of numbers. In your previous classes, you have encountered various types of numbers. This chapter will consolidate your understanding of these numbers and introduce you to new categories, starting with a more formal introduction to Rational Numbers and their place in the broader number system.

Review of Number Systems

Let's start by recalling the different collections of numbers you are already familiar with. Think of them as boxes, with smaller boxes fitting inside larger ones.

• Natural Numbers: These are the numbers we use for counting. They start from 1 and go on forever.

  Example: 1, 2, 3, 4, 10, 100, etc.

• Whole Numbers: This collection is simply the set of natural numbers with the number zero added to it.

  Example: 0, 1, 2, 3, 4, etc.

  So, every natural number is also a whole number. However, zero is a whole number but not a natural number.

• Integers: This collection includes all whole numbers and their negative counterparts. The word 'integer' comes from the Latin word meaning 'whole' or 'intact'.

  Example: ..., -3, -2, -1, 0, 1, 2, 3, ...

  This means that all natural numbers and all whole numbers are also integers.

Beyond integers, you have also worked with fractions, like $\frac{1}{2}$, $\frac{3}{4}$, and $-\frac{2}{5}$. These numbers, which can be expressed as a division of one integer by another, belong to a larger group called Rational Numbers.

Introduction to Rational Numbers

A number is defined as a Rational Number if it can be written in the form of a fraction $\frac{p}{q}$, where:

1. $p$ and $q$ are both integers.
2. The denominator $q$ is not equal to zero ($q \neq 0$).

The name "rational" comes from the word "ratio", as these numbers can be expressed as a ratio of two integers.

Let's see which numbers fit this definition:

• All Integers: Any integer can be written as a fraction by putting 1 in the denominator.
  • $5$ can be written as $\frac{5}{1}$. Here, $p=5$ and $q=1$. Both are integers and $q \neq 0$.
  • $-3$ can be written as $\frac{-3}{1}$. Here, $p=-3$ and $q=1$. Both are integers and $q \neq 0$.
  • $0$ can be written as $\frac{0}{1}$. Here, $p=0$ and $q=1$. Both are integers and $q \neq 0$.
  So, all integers (including natural and whole numbers) are rational numbers.
• All Fractions: Proper fractions like $\frac{1}{2}$, improper fractions like $\frac{7}{4}$, and mixed fractions like $2\frac{1}{3} (= \frac{7}{3})$ are all rational numbers by definition.
• Terminating Decimals: These are decimals that end. They can always be converted into the $\frac{p}{q}$ form.
  • $0.5 = \frac{5}{10} = \frac{1}{2}$
  • $2.75 = \frac{275}{100} = \frac{11}{4}$
• Non-Terminating Repeating (Recurring) Decimals: These are decimals that go on forever but have a repeating pattern. These can also be converted to the $\frac{p}{q}$ form.
  • $0.333\dots$ (written as $0.\overline{3}$) is equal to $\frac{1}{3}$.
  • $0.141414\dots$ (written as $0.\overline{14}$) is equal to $\frac{14}{99}$.

Classification of Numbers and the Role of Rational Numbers

The set of rational numbers is a large group that contains all the other number systems we've discussed. We can visualize this relationship as a series of nested containers:

Natural Numbers → Whole Numbers → Integers → Rational Numbers

Let's trace this path:

• The 'Natural Numbers' box is the smallest.
• If we add zero to this box, we get a slightly bigger box called 'Whole Numbers'.
• If we add all the negative whole numbers to this box, we get an even bigger box called 'Integers'.
• Finally, if we add all the fractions (like $\frac{1}{2}, -\frac{3}{4}$, etc.) and decimals that can be written as fractions, we get the largest box in our current discussion, the 'Rational Numbers'.

The following flowchart illustrates this classification:

It's important to remember that there are numbers that cannot be expressed in the form $\frac{p}{q}$. These numbers have decimal representations that are non-terminating and non-repeating (e.g., $\sqrt{2} = 1.41421356...$ or $\pi = 3.14159265...$). Such numbers are called Irrational Numbers, which we will explore in the next section.

Answer:

Representation of Rational Numbers on Real Number Line

The Real Number Line is a fundamental visual tool in mathematics. It is a straight line where every point corresponds to a unique real number, and conversely, every real number can be represented by a unique point on the line. Real numbers include both rational and irrational numbers. In this section, we will focus on how to represent rational numbers on this line.

Representing Integers on the Number Line

The construction of the real number line starts with representing integers. We select a point on the line and designate it as the origin, representing the number 0. Then, we choose a standard unit length. Points located at unit distances to the right of the origin represent positive integers (1, 2, 3, ...), and points located at unit distances to the left of the origin represent negative integers (-1, -2, -3, ...). The points are marked and labelled accordingly.

Representing Rational Numbers (Fractions $\frac{p}{q}$) on the Number Line

Rational numbers are numbers that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Since $p/q = p/(-q)$, we can assume that the denominator $q$ is always positive ($q > 0$) without losing generality when representing fractions.

To represent a rational number $\frac{p}{q}$ (where $q > 0$) on the number line, we determine its position relative to the integers.

• If the rational number is positive ($\frac{p}{q} > 0$), it lies to the right of 0.
• If the rational number is negative ($\frac{p}{q} < 0$), it lies to the left of 0.
• If the rational number is 0 ($\frac{0}{q} = 0$), it is located at the origin.

For a non-zero rational number $\frac{p}{q}$:

• If $|\frac{p}{q}| < 1$ (meaning the numerator $|p|$ is less than the denominator $q$), the number lies between -1 and 1 (specifically, between 0 and 1 if positive, and between -1 and 0 if negative).
• If $|\frac{p}{q}| = 1$, the number is either 1 or -1.
• If $|\frac{p}{q}| > 1$ (meaning the numerator $|p|$ is greater than the denominator $q$), the number is an improper fraction. Convert it to a mixed fraction: $\frac{p}{q} = a\frac{r}{q}$ (where $a$ is the integer part and $\frac{r}{q}$ is the proper fractional part, $0 \leq \frac{r}{q} < 1$). The number $\frac{p}{q}$ lies between the integers $a$ and $a+1$ (if positive) or between $a$ and $a-1$ (if negative).

General Steps to Represent $\frac{p}{q}$ (where $q > 0$):

1. Draw a real number line and mark the integers 0, 1, -1, 2, -2, etc., with equal spacing based on a chosen unit length.
2. Determine the two consecutive integers between which the rational number $\frac{p}{q}$ lies. If it's a positive number, it's between 0 and a positive integer. If it's negative, it's between a negative integer and 0. For improper fractions, convert to a mixed number $a\frac{r}{q}$; the number is between $a$ and $a+1$ (for positive) or $a$ and $a-1$ (for negative, where $a$ is the integer part including the sign).
3. Divide the segment of the number line between these two integers into $q$ (the absolute value of the denominator) equal parts.
4. Starting from the left endpoint of this segment (the smaller integer), count $p$ (the numerator) parts towards the right if the number is positive. The point you reach is $\frac{p}{q}$.
5. If the number is negative, say $-\frac{|p|}{q}$, locate $\frac{|p|}{q}$ to the right of 0 first. Then mark the point at the same distance from 0 on the left side. Alternatively, work with the negative number directly: start from the right endpoint of the segment (the larger integer, e.g., 0 for numbers between -1 and 0, or -1 for numbers between -2 and -1) and count $|p|$ parts towards the left. The point you reach is $-\frac{|p|}{q}$.

Answer:

Given number: $\frac{1}{2}$. Here $p=1, q=2$. It is positive, and $1 < 2$, so $0 < \frac{1}{2} < 1$. The number lies between 0 and 1.

Draw a number line and mark 0 and 1.

The denominator is 2, so divide the segment between 0 and 1 into 2 equal parts.

Starting from 0, count 1 part to the right (since the numerator is 1). The point marking the end of the first part from 0 represents $\frac{1}{2}$.

Answer:

Given number: $\frac{3}{4}$. Here $p=3, q=4$. It is positive, and $3 < 4$, so $0 < \frac{3}{4} < 1$. The number lies between 0 and 1.

Draw a number line and mark 0 and 1.

The denominator is 4, so divide the segment between 0 and 1 into 4 equal parts.

Starting from 0, count 3 parts to the right (since the numerator is 3). The point marking the end of the third part from 0 represents $\frac{3}{4}$.

Answer:

Given number: $\frac{7}{3}$. Here $p=7, q=3$. It is positive, and $7 > 3$, so it is an improper fraction. Convert to a mixed fraction:

$\frac{7}{3} = 2 \frac{1}{3}$

The integer part is 2. So, $\frac{7}{3}$ lies between 2 and $2+1=3$.

Draw a number line and mark 2 and 3.

The denominator of the fractional part is 3, so divide the segment between 2 and 3 into 3 equal parts.

Starting from 2, count 1 part to the right (since the numerator of the fractional part is 1). The point marking the end of the first part from 2 represents $2\frac{1}{3}$ or $\frac{7}{3}$.

Answer:

Given number: $\frac{-5}{3}$. This is a negative rational number. The absolute value is $\frac{5}{3}$. We know $\frac{5}{3} = 1\frac{2}{3}$, which is between 1 and 2 on the positive side.

So, $\frac{-5}{3}$ will be between -2 and -1 on the negative side.

Draw a number line and mark -1 and -2.

The denominator is 3, so divide the segment between -2 and -1 into 3 equal parts.

Starting from -1 (the integer to the right), count 2 parts to the left (corresponding to the numerator 2 in $1\frac{2}{3}$). The point marking the end of the second part from -1 towards -2 represents $-1\frac{2}{3}$ or $\frac{-5}{3}$.

Finding Rational Numbers Between Two Rational Numbers

One of the interesting properties of the set of rational numbers ($\mathbb{Q}$) is its density. Unlike integers, where there might be a finite number of integers (or no integers) between two given integers, between any two distinct rational numbers, there exist infinitely many other rational numbers.

Density Property of Rational Numbers

Between any two distinct rational numbers, no matter how close they are, there always exists another rational number. Because of this, you can find infinitely many rational numbers between any two given distinct rational numbers.

Let $r_1$ and $r_2$ be two distinct rational numbers, say with $r_1 < r_2$. We will explore methods to find rational numbers that lie between $r_1$ and $r_2$.

Method 1: Using the Mean (Average)

The average or arithmetic mean of two numbers always lies between the two numbers. If $a$ and $b$ are any two distinct rational numbers, then their mean $\frac{a+b}{2}$ is also a rational number and lies strictly between $a$ and $b$.

Since $\frac{a+b}{2}$ is a rational number (because the sum of two rational numbers is rational, and dividing by 2 is multiplying by $\frac{1}{2}$, which is rational, and the product of two rational numbers is rational), this method always yields a rational number between the given two. You can repeat this process indefinitely. For example, once you find $r_3 = \frac{r_1+r_2}{2}$ between $r_1$ and $r_2$, you can find another rational number between $r_1$ and $r_3$ by calculating $\frac{r_1+r_3}{2}$, and so on. This shows that there are infinitely many rational numbers between any two distinct rational numbers.

Answer:

Method 2: Creating Equivalent Fractions (with Sufficiently Large Denominators)

This method is more systematic when you need to find a specific number of rational numbers between two given rational numbers. The idea is to rewrite the given fractions as equivalent fractions with a common denominator that is large enough to create "space" (integers) between the numerators.

Steps:

1. Find a common denominator for the two given rational numbers. The Least Common Multiple (LCM) of the denominators is a good choice. Rewrite both fractions as equivalent fractions with this common denominator. Let the fractions be $\frac{p_1}{k}$ and $\frac{p_2}{k}$, where $k$ is the common denominator.
2. Check if there are enough integers between the numerators $p_1$ and $p_2$ to form the desired number of rational numbers. The number of integers strictly between $p_1$ and $p_2$ is $|p_1 - p_2| - 1$.
3. If there are not enough integers between $p_1$ and $p_2$, multiply the numerator and denominator of both equivalent fractions ($\frac{p_1}{k}$ and $\frac{p_2}{k}$) by a suitable integer $m$. A value of $m$ slightly larger than the number of rational numbers you need to find is usually sufficient (e.g., if you need $N$ numbers, try $m = N+1$ or 10). The new equivalent fractions are $\frac{m \times p_1}{m \times k}$ and $\frac{m \times p_2}{m \times k}$. Now, there will be more integers between the new numerators $m \times p_1$ and $m \times p_2$.
4. The rational numbers between the original two numbers are fractions with the new common denominator ($m \times k$) and numerators that are the integers strictly between $m \times p_1$ and $m \times p_2$.

Answer:

Answer:

Irrational Numbers - Introduction and Classification

In the previous sections, we have extensively discussed rational numbers ($\mathbb{Q}$), which include all integers, fractions, terminating decimals, and non-terminating repeating decimals. However, the set of rational numbers does not cover all the numbers that exist on the real number line. There are numbers on the number line that cannot be expressed in the form $\frac{p}{q}$. These are called irrational numbers.

Introduction to Irrational Numbers ($\mathbb{I}$ or $\mathbb{Q}^c$)

A number is called an irrational number if it cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. In other words, an irrational number is a real number that is not rational.

The set of irrational numbers is often denoted by $\mathbb{I}$ or $\mathbb{Q}^c$ (the complement of the set of rational numbers within the real numbers).

Examples of irrational numbers:

• Square roots of positive non-perfect squares: For any positive integer $n$ that is not a perfect square (i.e., $n$ is not equal to the square of any integer), its square root $\sqrt{n}$ is an irrational number. Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}$, etc. Note that $\sqrt{4} = 2$, which is rational. $\sqrt{9} = 3$, which is rational.
• Cube roots (and other roots) of integers that are not perfect cubes (or appropriate powers): For example, $\sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{9}, \sqrt[3]{16}$, etc., are irrational numbers because 2, 4, 9, 16 are not perfect cubes. Note that $\sqrt[3]{8} = 2$, which is rational.
• Certain special mathematical constants:
  • $\pi$ (pi): The ratio of the circumference of any circle to its diameter. Its value is approximately 3.14159265... Its decimal expansion is non-terminating and non-repeating.
  • $e$ (Euler's number): The base of the natural logarithm, approximately 2.71828... Its decimal expansion is non-terminating and non-repeating.
• Numbers with decimal expansions that are non-terminating and non-recurring (non-repeating): As we will see in a later section, rational numbers have decimal expansions that either terminate (like 0.5) or are non-terminating but repeat in a pattern (like $0.333...$). Irrational numbers have decimal expansions that go on forever without repeating any sequence of digits. Examples: $0.10110111011110...$ (where the number of 1s between the 0s increases), $1.23456789101112...$ (concatenating all natural numbers).

The existence of irrational numbers was a major discovery in ancient Greek mathematics, proving that not all lengths could be measured using simple ratios of integers. This is famously illustrated by the diagonal of a unit square.

Proof that $\sqrt{2}$ is Irrational (Proof by Contradiction)

This is a classic proof showing that a seemingly simple number like $\sqrt{2}$ is not rational. The method used is called proof by contradiction. We start by assuming the opposite of what we want to prove (that $\sqrt{2}$ is rational) and show that this assumption leads to a logical inconsistency.

Given: The number $\sqrt{2}$.

To Prove: $\sqrt{2}$ is an irrational number.

Proof:

Assume, for the sake of contradiction, that $\sqrt{2}$ is a rational number.

If $\sqrt{2}$ is rational, it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and the fraction $\frac{p}{q}$ is in its simplest form (i.e., $p$ and $q$ have no common factors other than 1, they are coprime).

Squaring both sides of the equation (i):

Multiply both sides by $q^2$:

Equation (ii) tells us that $p^2$ is equal to 2 times some integer ($q^2$). This means $p^2$ is an even number.

If the square of a number ($p^2$) is even, then the number itself ($p$) must also be even. (Because: if $p$ were odd, $p^2$ would be odd; if $p$ were even, $p^2$ would be even. Only the even case results in an even square).

So, $p$ is an even integer. This means $p$ can be written in the form $2k$ for some integer $k$.

Substitute $p = 2k$ into equation (ii):

Divide both sides by 2:

Equation (iii) tells us that $q^2$ is equal to 2 times some integer ($k^2$). This means $q^2$ is an even number.

If the square of a number ($q^2$) is even, then the number itself ($q$) must also be even. (Using the same logic as for $p$).

So, $q$ is an even integer. This means $q$ can be written in the form $2j$ for some integer $j$.

From our derivation, we have concluded that both $p$ and $q$ are even numbers. An even number is a number divisible by 2. This implies that $p$ and $q$ have a common factor of 2.

However, earlier, we assumed that $\frac{p}{q}$ was in its simplest form, meaning $p$ and $q$ have no common factors other than 1.

The conclusion that $p$ and $q$ have a common factor of 2 contradicts our initial assumption that $p$ and $q$ are coprime.

This contradiction means that our original assumption (that $\sqrt{2}$ is rational) must be false.

Therefore, $\sqrt{2}$ is an irrational number.

This method of proof by contradiction is very common in mathematics. Similar proofs can be constructed to show that $\sqrt{3}$, $\sqrt{5}$, and the square roots of other non-perfect squares are irrational.

Real Numbers ($\mathbb{R}$)

The set of all rational numbers ($\mathbb{Q}$) and the set of all irrational numbers ($\mathbb{I}$) together form the set of Real Numbers. The set of real numbers is denoted by $\mathbb{R}$.

The set of rational numbers and the set of irrational numbers are disjoint sets, meaning a number cannot be both rational and irrational simultaneously ($\mathbb{Q} \cap \mathbb{I} = \emptyset$).

Every real number can be represented by a unique point on the real number line, and conversely, every point on the real number line represents a unique real number. The real number line is often called the continuous line because the combined set of rational and irrational numbers fills up the entire line without any gaps.

In summary, the number system expands from natural numbers to whole numbers, integers, rational numbers, and finally, real numbers (which include both rational and irrational numbers).

Representation of Irrational Numbers on Real Number Line

We know that rational numbers can be represented by unique points on the real number line. What about irrational numbers? Do they also correspond to points on the number line? Yes, they do. Every irrational number can be represented by a unique point on the real number line, and these points fill the 'gaps' between the rational numbers.

We can use geometric constructions based on the Pythagorean Theorem to represent certain irrational numbers, particularly square roots of positive integers, on the number line.

Using the Pythagorean Theorem for Construction

The Pythagorean Theorem applies to right-angled triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (the legs or perpendicular sides). If the lengths of the legs are $a$ and $b$, and the length of the hypotenuse is $c$, then:

Taking the square root of both sides (and considering positive lengths):

If we can construct a right-angled triangle on the number line such that the lengths of its legs are rational numbers whose squares sum up to a non-perfect square, the length of the hypotenuse will be an irrational number. We can then transfer this length to the number line using a compass.

Representing $\sqrt{2}$ on the Number Line

We know that $1^2 + 1^2 = 1 + 1 = 2$. So, $\sqrt{2}$ is the hypotenuse of a right-angled triangle with legs of length 1 unit each. We can use this fact to represent $\sqrt{2}$ on the number line.

Construction Steps:

1. Draw a number line and mark the origin O representing 0. Choose a convenient unit length and mark the point A to the right of O, representing the number 1. The distance OA is 1 unit.

2. At point A (representing 1), construct a perpendicular line segment AB of length 1 unit, upwards from the number line. So, OA = 1 unit and AB = 1 unit. $\angle$OAB is a right angle ($90^\circ$).

3. Join the points O and B with a straight line segment. Triangle OAB is a right-angled triangle with legs OA and AB.

4. By the Pythagorean Theorem in $\triangle$OAB, the length of the hypotenuse OB is given by $OB^2 = OA^2 + AB^2$.

   $OB^2 = 1^2 + 1^2 = 1 + 1 = 2$

   $OB = \sqrt{2}$ units

   [Taking positive square root as length must be positive]

   The length of the segment OB is $\sqrt{2}$ units.
5. With O as the centre and OB as the radius, draw an arc that intersects the number line to the right of O.

6. The point where the arc intersects the number line represents the number whose distance from the origin is $\sqrt{2}$ units. This point represents the irrational number $\sqrt{2}$. Let's label this point P.

Representing $\sqrt{3}$ on the Number Line

We can extend the previous construction to represent $\sqrt{3}$. We know that $(\sqrt{2})^2 + 1^2 = 2 + 1 = 3$. So, $\sqrt{3}$ is the hypotenuse of a right-angled triangle with legs of length $\sqrt{2}$ units and 1 unit. Since we have already located $\sqrt{2}$ on the number line, we can use it as one leg of the triangle.

Construction Steps:

1. First, locate $\sqrt{2}$ on the number line as described above. Let the point representing $\sqrt{2}$ be P. The distance OP from the origin is $\sqrt{2}$ units.

2. At point P (representing $\sqrt{2}$), construct a perpendicular line segment PQ of length 1 unit, upwards from the number line. So, OP = $\sqrt{2}$ units and PQ = 1 unit. $\angle$OPQ is a right angle ($90^\circ$).

3. Join the points O and Q with a straight line segment. Triangle OPQ is a right-angled triangle with legs OP and PQ.

4. By the Pythagorean Theorem in $\triangle$OPQ, the length of the hypotenuse OQ is given by $OQ^2 = OP^2 + PQ^2$.

   $OQ^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$

   $OQ = \sqrt{3}$ units

   The length of the segment OQ is $\sqrt{3}$ units.
5. With O as the centre and OQ as the radius, draw an arc that intersects the number line to the right of O.

6. The point where the arc intersects the number line represents the number whose distance from the origin is $\sqrt{3}$ units. This point represents the irrational number $\sqrt{3}$. Let's label this point R.

The Square Root Spiral

This method can be generalised. To represent $\sqrt{n}$ (for any positive integer $n > 1$) on the number line, we can construct a right-angled triangle with one leg of length $\sqrt{n-1}$ units and the other leg of length 1 unit. The hypotenuse will have a length of $\sqrt{(\sqrt{n-1})^2 + 1^2} = \sqrt{n-1+1} = \sqrt{n}$ units.

Starting with $\sqrt{1}=1$, we can construct $\sqrt{2}$ (using legs 1 and 1). Then using $\sqrt{2}$ as a leg and 1 as the other leg, we construct $\sqrt{3}$. Using $\sqrt{3}$ and 1, we construct $\sqrt{4}=2$. Using $\sqrt{4}$ and 1, we construct $\sqrt{5}$, and so on. This process creates a spiral shape made of hypotenuses, known as the "Square Root Spiral". The points where the arcs from the origin intersect the number line represent the square roots of consecutive integers.

This shows that all irrational numbers of the form $\sqrt{n}$ (where $n$ is a non-perfect square positive integer) can be located on the real number line, confirming that points on the number line exist for irrational numbers.

Representing $\sqrt{x}$ on Real Number Line (where x can be any positive real number)

In the previous section, we learned how to represent square roots of positive integers (specifically non-perfect squares like $\sqrt{2}, \sqrt{3}$) on the number line using geometric constructions based on the Pythagorean theorem. This method involves constructing right-angled triangles with integer side lengths or side lengths that are square roots of integers. We can extend this idea to represent the square root of *any* positive real number $x$, not just integers, on the number line.

Construction Steps to Represent $\sqrt{x}$ (where $x > 0$)

This construction uses a geometric property involving a semicircle and an altitude. The altitude from the right angle vertex to the hypotenuse in a right triangle inscribed in a semicircle is the geometric mean of the two segments it divides the hypotenuse into. If the segments of the diameter are $x$ and 1, the altitude will have length $\sqrt{x \times 1} = \sqrt{x}$.

Let's use this property to construct a segment of length $\sqrt{x}$ and then transfer this length to the number line starting from the origin (0).

Construction Steps:

1. Draw a line and mark a point A on it. Let this point A represent the number 0 on the number line.

2. From A, mark a point B on the line such that the distance AB is equal to $x$ units. Since $x > 0$, B will be to the right of A. So, the point B represents the number $x$ on the number line relative to A as 0.

3. Extend the line segment AB further to the right to a point C such that the distance BC is exactly 1 unit. So, the length of segment AC is AB + BC = $x + 1$ units.

4. Find the midpoint of the segment AC. Let the midpoint be M. The length $AM = MC = \frac{x+1}{2}$ units. You can find the midpoint by constructing the perpendicular bisector of AC or by calculation if A is at 0, C is at $x+1$, so M is at $\frac{x+1}{2}$.

5. With M as the centre and radius equal to AM (or MC or even calculate the radius as $\frac{x+1}{2}$), draw a semicircle above the line AC.

6. Draw a line perpendicular to the line AC passing through point B. Let this perpendicular line intersect the semicircle at point D.

7. The length of the segment BD is the square root of $x$, i.e., BD = $\sqrt{x}$ units. (The proof is given below).
8. Finally, to represent $\sqrt{x}$ starting from the origin (A), use a compass. With A (representing 0) as the centre and BD as the radius, draw an arc that intersects the line (number line) to the right of A. Let this intersection point be P.

9. The point P on the number line represents the number $\sqrt{x}$.

Proof that BD = $\sqrt{x}$

In the construction, AC is the diameter of the semicircle, and $\angle$ADC is the angle in a semicircle, so $\angle$ADC = $90^\circ$. Thus, $\triangle$ADC is a right-angled triangle.

BD is the altitude from vertex D to the hypotenuse AC in the right triangle ADC. (Although D is on the semicircle, A, B, C are collinear on the diameter). More directly, consider $\triangle$ABD and $\triangle$DBC. Both are right-angled triangles at B (by construction). $\angle$BDA and $\angle$CDB sum to $\angle$ADC. In right triangle ADC, BD is the altitude to the hypotenuse if D was the vertex with the right angle, which is not the case here. Let's use similar triangles formed by the altitude to the hypotenuse in a right-angled triangle.

Draw lines AD and CD. $\triangle$ABD and $\triangle$DBC are right triangles. In $\triangle$ADC, the angle subtended by the diameter AC at any point D on the semicircle is $90^\circ$. Thus, $\triangle$ADC is a right-angled triangle at D.

In the right-angled triangle ADC, BD is the altitude from the vertex D to the hypotenuse AC. When an altitude is drawn to the hypotenuse of a right triangle, it creates two smaller triangles that are similar to the original triangle and to each other. So, $\triangle$ADB is similar to $\triangle$BDC ($\triangle ADB \sim \triangle BDC$).

From the property of similar triangles, the ratio of corresponding sides is equal. Specifically, for $\triangle$ADB and $\triangle$BDC:

Cross-multiplying gives:

From our construction steps, we have AB = $x$ and BC = 1 unit.

Taking the square root of both sides (and considering only the positive length for BD):

Thus, the length of the perpendicular segment BD is indeed $\sqrt{x}$ units. Transferring this length from the origin A (representing 0) gives the point P representing $\sqrt{x}$ on the number line.

Decimal Expansion of Rational Numbers and Irrational Numbers

Every real number can be represented by a decimal expansion. The nature of this decimal expansion provides a definitive way to distinguish between rational and irrational numbers. Performing division allows us to find the decimal form of a rational number, and observing the pattern of the decimal helps us identify whether it's rational or irrational.

Decimal Expansion of Rational Numbers

To find the decimal expansion of a rational number given in the form $\frac{p}{q}$ (where $p, q$ are integers and $q \neq 0$), we perform long division of the numerator $p$ by the denominator $q$. When we perform this division, there are only two possible outcomes for the decimal representation:

1. Terminating Decimal Expansion: This occurs when the long division process eventually results in a remainder of 0. The decimal representation stops or terminates after a finite number of digits.

   Examples:

   • $\frac{1}{2}$: Performing $1 \div 2$, we get 0.5. The remainder is 0. This is a terminating decimal.
   • $\frac{3}{4}$: Performing $3 \div 4$, we get 0.75. The remainder is 0. This is a terminating decimal.
   • $\frac{5}{8}$: Performing $5 \div 8$, we get 0.625. The remainder is 0. This is a terminating decimal.
   • $\frac{7}{10}$: Performing $7 \div 10$, we get 0.7. The remainder is 0. This is a terminating decimal.

   In a fraction $\frac{p}{q}$ in simplest form, the decimal expansion is terminating if and only if the prime factors of the denominator $q$ are only 2s and/or 5s. For example, in $\frac{3}{4}$, $q=4=2^2$ (only 2s). In $\frac{5}{8}$, $q=8=2^3$ (only 2s). In $\frac{7}{10}$, $q=10=2 \times 5$ (only 2s and 5s).

2. Non-terminating Recurring (Repeating) Decimal Expansion: This occurs when the long division process never results in a remainder of 0. However, the sequence of remainders starts repeating after a certain stage. Consequently, the digits in the quotient also start repeating in a block or pattern periodically. Such a decimal is called a non-terminating recurring or repeating decimal. We indicate the repeating block of digits by placing a bar over it.

   Examples:

   • $\frac{1}{3}$: Performing $1 \div 3$, we get $0.3333\dots$. The digit 3 repeats infinitely. This is written as $0.\overline{3}$. The remainder never becomes 0.
   • $\frac{1}{7}$: Performing $1 \div 7$, we get $0.142857142857\dots$. The block of digits 142857 repeats infinitely. This is written as $0.\overline{142857}$. The remainder sequence is 3, 2, 6, 4, 5, 1, then it repeats.
   • $\frac{10}{3}$: Performing $10 \div 3$, we get $3.3333\dots$. The digit 3 repeats after the decimal point. This is written as $3.\overline{3}$.
   • $\frac{1}{11}$: Performing $1 \div 11$, we get $0.090909\dots$. The block of digits 09 repeats infinitely. This is written as $0.\overline{09}$.

   In a fraction $\frac{p}{q}$ in simplest form, the decimal expansion is non-terminating recurring if the denominator $q$ has prime factors other than 2s and/or 5s. For example, in $\frac{1}{3}$, $q=3$ (prime factor is 3). In $\frac{1}{7}$, $q=7$ (prime factor is 7). In $\frac{1}{11}$, $q=11$ (prime factor is 11).

Conclusion for Rational Numbers: The decimal expansion of every rational number is either terminating or non-terminating recurring. Conversely, every number whose decimal expansion is either terminating or non-terminating recurring is a rational number.

Decimal Expansion of Irrational Numbers

Irrational numbers are real numbers that cannot be expressed in the form $\frac{p}{q}$. How do their decimal expansions behave?

The decimal expansion of an irrational number is always non-terminating and non-recurring (non-repeating).

This means the digits after the decimal point go on infinitely (non-terminating), and there is no block or sequence of digits that repeats periodically (non-recurring).

Examples:

• $\sqrt{2} = 1.414213562373095\dots$. The digits continue infinitely without a repeating pattern.
• $\sqrt{3} = 1.732050810756887\dots$. The digits continue infinitely without a repeating pattern.
• $\pi = 3.141592653589793\dots$. The digits continue infinitely without a repeating pattern.
• A number like $0.10110111011110\dots$, where there is one 0 followed by one 1, then one 0 followed by two 1s, then one 0 followed by three 1s, and so on. The pattern of increasing numbers of 1s prevents the decimal from repeating periodically, so it is non-recurring. Since it continues indefinitely (indicated by $\dots$), it is non-terminating. Thus, this number is irrational.

Conclusion for Irrational Numbers: The decimal expansion of every irrational number is non-terminating and non-recurring. Conversely, every number whose decimal expansion is non-terminating and non-recurring is an irrational number.

Summary Table: Decimal Expansion Types

The nature of the decimal expansion is the key characteristic that separates rational and irrational numbers within the real number system.

This table provides a clear way to use decimal expansions to identify the type of real number.

Expressing Terminating and Non-Terminating Recurring Decimal Numbers in the form of $\frac{p}{q}$

We've established that a number is rational if and only if its decimal expansion is either terminating or non-terminating recurring. This section provides a detailed guide on how to convert any decimal number with such an expansion back into its fundamental fractional form $\frac{p}{q}$, where p and q are integers and q $\neq$ 0.

Converting Terminating Decimals to $\frac{p}{q}$ Form

Converting a terminating decimal into a fraction is the more straightforward of the two cases. The process is based on the concept of decimal place values (tenths, hundredths, thousandths, etc.). Essentially, a terminating decimal is a shorthand for a fraction whose denominator is a power of 10.

Steps to Convert:

1. Write the digits of the number as they appear, but without the decimal point. This becomes the numerator of your fraction.
2. Count the number of digits ($n$) that are to the right of the decimal point.
3. The denominator of the fraction will be 1 followed by $n$ zeros. Mathematically, this is $10^n$.
4. Simplify the resulting fraction $\frac{\text{numerator}}{\text{denominator}}$ by dividing both the numerator and the denominator by their Greatest Common Factor (GCF) to express it in its lowest terms.

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Converting Non-Terminating Recurring Decimals to $\frac{p}{q}$ Form

Converting recurring decimals involves an algebraic method. The core strategy is to manipulate the decimal algebraically to isolate and then eliminate the infinitely repeating part. We achieve this by creating two equations where the digits after the decimal point are perfectly aligned and identical, allowing them to cancel out upon subtraction.

Case 1: Purely Recurring Decimal (All digits after the decimal point repeat)

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Case 2: Mixed Recurring Decimal (Some digits do not repeat, followed by digits that do)

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Shortcut Method for Converting Recurring Decimals

For quick calculations or verification, a direct formula can be used. This method mechanically follows the logic of the algebraic subtraction we performed earlier.

The Rule:

To find the Numerator:

(Take all the digits of the number without the decimal point) - (Take the digits of the non-repeating part)

To find the Denominator:

Write a '9' for each digit in the repeating block, followed by a '0' for each non-repeating digit after the decimal point.

Alternate Solution for Example 5 ($0.\overline{123}$):

Alternate Solution for Example 7 ($2.5\overline{16}$):

Representing Terminating and Non-Terminating Recurring Decimal Numbers on Real Number Line upto a certain number of decimal places (Successive Magnification Process)

We know that rational numbers correspond to unique points on the real number line. We've also seen how to locate simple fractions and square roots of integers. When dealing with decimal numbers, especially those with many decimal places (terminating) or non-terminating recurring decimals, we can visualize their exact position on the number line using a process called Successive Magnification.

This process is like using a magnifying glass to look at smaller and smaller segments of the number line, zooming in on the location of the number digit by digit.

Representing a Terminating Decimal using Successive Magnification

Let's take an example of a terminating decimal, say 2.665. We want to pinpoint its location on the number line.

Steps for Successive Magnification (Example: Locating 2.665):

1. Step 1 (Units Place): Look at the integer part of the number. The number 2.665 lies between the integers 2 and 3. Draw a section of the number line showing the integers 2 and 3.

2. Step 2 (First Decimal Place): Consider the segment between 2 and 3. Divide this segment into 10 equal parts. Each part represents $\frac{1}{10} = 0.1$. Mark these divisions as 2.1, 2.2, ..., 2.9. The number 2.665 starts with 2.6, so it lies between 2.6 and 2.7. Magnify the segment between 2.6 and 2.7.

3. Step 3 (Second Decimal Place): Consider the segment between 2.6 and 2.7. Divide this segment into 10 equal parts. Each part represents $\frac{0.1}{10} = 0.01$. Mark these divisions as 2.61, 2.62, ..., 2.69. The number 2.665 starts with 2.66, so it lies between 2.66 and 2.67. Magnify the segment between 2.66 and 2.67.

4. Step 4 (Third Decimal Place): Consider the segment between 2.66 and 2.67. Divide this segment into 10 equal parts. Each part represents $\frac{0.01}{10} = 0.001$. Mark these divisions as 2.661, 2.662, ..., 2.669. The number 2.665 is exactly at the 5th marking from 2.66. Mark the point 2.665.

This process allows us to pinpoint the exact location of a terminating decimal on the number line with increasing accuracy, digit by digit.

Representing a Non-Terminating Recurring Decimal up to a Certain Number of Decimal Places

A non-terminating recurring decimal like $5.37\overline{7}$ continues infinitely. To represent it on the number line up to a certain number of decimal places (say, 4 decimal places), we first approximate the number by rounding it to that many decimal places.

Given $5.37\overline{7} = 5.377777\dots$. To round to 4 decimal places, we look at the 5th decimal place, which is 7. Since 7 is 5 or greater, we round up the 4th decimal place.

Now we locate the terminating decimal approximation 5.3778 on the number line using successive magnification, just like the previous example.

Steps for Successive Magnification (Example: Locating 5.3778):

1. Step 1 (Units Place): The number 5.3778 lies between the integers 5 and 6. Draw the segment [5, 6].

2. Step 2 (First Decimal Place): Divide [5, 6] into 10 equal parts. Mark 5.1, 5.2, ..., 5.9. The number 5.3778 starts with 5.3, so it's between 5.3 and 5.4. Magnify [5.3, 5.4].

3. Step 3 (Second Decimal Place): Divide [5.3, 5.4] into 10 equal parts. Mark 5.31, 5.32, ..., 5.39. The number 5.3778 starts with 5.37, so it's between 5.37 and 5.38. Magnify [5.37, 5.38].

4. Step 4 (Third Decimal Place): Divide [5.37, 5.38] into 10 equal parts. Mark 5.371, ..., 5.379. The number 5.3778 starts with 5.377, so it's between 5.377 and 5.378. Magnify [5.377, 5.378].

5. Step 5 (Fourth Decimal Place): Divide [5.377, 5.378] into 10 equal parts. Mark 5.3771, ..., 5.3779. The number 5.3778 is exactly at the 8th marking from 5.377. Mark this point.

Even though the original number is non-terminating, we can represent its location on the number line to any desired level of accuracy by continuing the successive magnification process. Each step zooms in 10 times on a smaller segment of the number line, corresponding to the next decimal place.

Algebraic Operations on Real Numbers

The set of real numbers ($\mathbb{R}$) is formed by combining all rational numbers ($\mathbb{Q}$) and all irrational numbers ($\mathbb{I}$). Just like with simpler numbers, we can perform the basic algebraic operations: addition, subtraction, multiplication, and division. Real numbers obey the standard rules of algebra (commutative, associative, and distributive laws). However, a key point of interest is understanding the nature of the result—whether it turns out to be rational or irrational—when we combine different types of real numbers.

Operations with Two Rational Numbers

The set of rational numbers is closed under addition, subtraction, multiplication, and division (by a non-zero rational number). In simple terms, this means that if you combine any two rational numbers using these operations, the answer will always be another rational number.

Let $a$ and $b$ be two rational numbers.

• Sum: $a + b$ is always rational. (e.g., $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$)
• Difference: $a - b$ is always rational. (e.g., $5 - \frac{3}{2} = \frac{7}{2}$)
• Product: $a \times b$ is always rational. (e.g., $0.5 \times 4 = 2$)
• Quotient: $a \div b$ is always rational, provided $b \neq 0$. (e.g., $\frac{2}{3} \div \frac{1}{3} = 2$)

Operations Involving Irrational Numbers

When an operation involves at least one irrational number, the outcome becomes more interesting. The results are not always predictable and depend on the specific numbers involved.

1. Operations between a Rational and an Irrational Number

This case is very consistent and follows a strict rule.

• Sum or Difference: The sum or difference of a rational number and an irrational number is always irrational.

  Reasoning: An irrational number has a non-terminating, non-repeating decimal expansion. Adding or subtracting a rational number (which has a terminating or repeating decimal) cannot cancel out the infinite non-repeating part. The irrationality persists.

• Product or Quotient: The product or quotient of a non-zero rational number and an irrational number is always irrational.

  Reasoning: Multiplying or dividing an infinitely non-repeating decimal by a fixed, non-zero number scales its value but does not change its non-repeating nature.

  Important Note: The rational number must be non-zero. If we multiply by zero, the result is zero, which is rational: $0 \times \sqrt{2} = 0$.

2. Operations between Two Irrational Numbers

This is where the outcome is unpredictable. The sum, difference, product, or quotient of two irrational numbers can be either rational or irrational. You must perform the calculation to determine the nature of the result.

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Simplifying Expressions with Radicals

When working with expressions involving square roots (radicals), we can simplify them by combining like terms and using algebraic identities. Treat radicals with the same number inside (e.g., $\sqrt{3}$) as "like terms", similar to how you would treat a variable like 'x'.

Key Properties and Identities for Radicals: For $a, b \ge 0$

• $\sqrt{ab} = \sqrt{a}\sqrt{b}$
• $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ (where $b \neq 0$)
• $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$
• $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$
• $(\sqrt{a} \pm \sqrt{b})^2 = a \pm 2\sqrt{ab} + b$

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Summary of Results

The following table summarizes the nature of the result when performing algebraic operations on real numbers.

Some Identities Related to Real Numbers

In addition to the standard algebraic identities you have learned, such as $(a+b)^2$ and $(a-b)(a+b)$, there are specific versions of these identities that are extremely useful when working with expressions involving square roots. These are not new rules, but rather applications of the old rules to numbers with radicals. For these identities to be valid with real numbers, any number under a square root sign must be non-negative.

Fundamental Identities for Radicals

Let $a$ and $b$ be positive real numbers. The following two properties form the basis for simplifying most radical expressions.

1. Product of Square Roots:

   The square root of a product is the product of the square roots.

   $\sqrt{ab} = \sqrt{a} \sqrt{b}$

   ... (i)

   Proof: Consider the square of the expression $\sqrt{a} \sqrt{b}$:

   $(\sqrt{a} \sqrt{b})^2 = (\sqrt{a} \sqrt{b}) \times (\sqrt{a} \sqrt{b})$

   $= (\sqrt{a} \times \sqrt{a}) \times (\sqrt{b} \times \sqrt{b})$

   $= a \times b = ab$

   Since $(\sqrt{a} \sqrt{b})^2 = ab$ and square roots are non-negative, by definition, $\sqrt{ab} = \sqrt{a} \sqrt{b}$.

   Example 1. Simplify (a) $\sqrt{72}$ and (b) $\sqrt{2} \times \sqrt{8}$.

   Answer:

   (a) To simplify $\sqrt{72}$, we look for the largest perfect square that divides 72. That is 36.

   $\sqrt{72} = \sqrt{36 \times 2}$

   $= \sqrt{36} \times \sqrt{2}$

   $= 6\sqrt{2}$

   (b) Using the identity in reverse:

   $\sqrt{2} \times \sqrt{8} = \sqrt{2 \times 8}$

   $= \sqrt{16}$

   $= 4$

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2. Quotient of Square Roots:

   The square root of a quotient is the quotient of the square roots.

   $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

   [For $a \ge 0, b > 0$]           ... (ii)

   Proof: Consider $(\frac{\sqrt{a}}{\sqrt{b}})^2 = \frac{(\sqrt{a})^2}{(\sqrt{b})^2} = \frac{a}{b}$. By definition of the square root, this means $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.

   Example 2. Simplify (a) $\sqrt{\frac{25}{16}}$ and (b) $\frac{\sqrt{48}}{\sqrt{3}}$.

   Answer:

   (a) $\sqrt{\frac{25}{16}} = \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4}$.

   (b) Using the identity in reverse:

   $\frac{\sqrt{48}}{\sqrt{3}} = \sqrt{\frac{48}{3}}$

   $= \sqrt{16} = 4$

Algebraic Identities with Radicals

The following are direct applications of standard algebraic identities to expressions involving radicals.

3. Difference of Squares (Type 1):

   This comes from the standard identity $(x+y)(x-y)=x^2-y^2$.

   $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$

   ... (iii)

   This identity is the foundation for the technique of rationalising the denominator.

   Example 3. Simplify $(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2})$.

   Answer:

   Here, $a=7$ and $b=2$. Using the identity:

   $(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2}) = (\sqrt{7})^2 - (\sqrt{2})^2$

   $= 7 - 2 = 5$

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4. Difference of Squares (Type 2):

   $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$

   ... (iv)

   Example 4. Simplify $(5 + \sqrt{3})(5 - \sqrt{3})$.

   Answer:

   Here, $a=5$ and $b=3$. Using the identity:

   $(5 + \sqrt{3})(5 - \sqrt{3}) = (5)^2 - (\sqrt{3})^2$

   $= 25 - 3 = 22$

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5. Product of Two Binomials (FOIL Method):

   $(\sqrt{a} + \sqrt{b})(\sqrt{c} + \sqrt{d}) = \sqrt{ac} + \sqrt{ad} + \sqrt{bc} + \sqrt{bd}$

   ... (v)

   Example 5. Multiply $(\sqrt{3} + \sqrt{2})$ by $(\sqrt{5} + \sqrt{7})$.

   Answer:

   Using the distributive property:

   $(\sqrt{3} + \sqrt{2})(\sqrt{5} + \sqrt{7}) = \sqrt{3}(\sqrt{5} + \sqrt{7}) + \sqrt{2}(\sqrt{5} + \sqrt{7})$

   $= \sqrt{3}\sqrt{5} + \sqrt{3}\sqrt{7} + \sqrt{2}\sqrt{5} + \sqrt{2}\sqrt{7}$

   $= \sqrt{15} + \sqrt{21} + \sqrt{10} + \sqrt{14}$

   Since none of the radicals can be simplified further and all are different, this is the final answer.

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6. Square of a Binomial Sum/Difference (Type 1):

   These come from $(x \pm y)^2 = x^2 \pm 2xy + y^2$.

   $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$

   ... (vi)

   $(\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$

   ... (vii)

   Example 6. Simplify (a) $(\sqrt{5} + \sqrt{3})^2$ and (b) $(\sqrt{6} - \sqrt{2})^2$.

   Answer:

   (a) Here, $a=5$ and $b=3$. Using identity (vi):

   $(\sqrt{5} + \sqrt{3})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2$

   $= 5 + 2\sqrt{15} + 3$

   $= 8 + 2\sqrt{15}$

   (b) Here, $a=6$ and $b=2$. Using identity (vii):

   $(\sqrt{6} - \sqrt{2})^2 = (\sqrt{6})^2 - 2(\sqrt{6})(\sqrt{2}) + (\sqrt{2})^2$

   $= 6 - 2\sqrt{12} + 2$

   $= 8 - 2\sqrt{12}$

   $= 8 - 2\sqrt{4 \times 3} = 8 - 2(2\sqrt{3})$

   $= 8 - 4\sqrt{3}$

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7. Square of a Binomial Sum/Difference (Type 2):

   $(a + \sqrt{b})^2 = a^2 + b + 2a\sqrt{b}$

   ... (viii)

   $(a - \sqrt{b})^2 = a^2 + b - 2a\sqrt{b}$

   ... (ix)

   Example 7. Simplify $(4 - \sqrt{5})^2$.

   Answer:

   Here, $a=4$ and $b=5$. Using identity (ix):

   $(4 - \sqrt{5})^2 = (4)^2 - 2(4)(\sqrt{5}) + (\sqrt{5})^2$

   $= 16 - 8\sqrt{5} + 5$

   $= 21 - 8\sqrt{5}$

Mastering these identities is fundamental for simplifying complex algebraic expressions containing square roots and for the important technique of rationalisation.

Rationalisation

In mathematics, particularly when dealing with expressions involving square roots, it is often convenient to remove the square root from the denominator of a fraction. The process of converting an irrational denominator into a rational denominator is called rationalisation.

To rationalise the denominator of a fraction, we multiply both the numerator and the denominator by a suitable irrational number or expression. This multiplier is called the rationalising factor. The choice of the rationalising factor depends on the form of the irrational denominator.

Rationalising a Denominator with a Single Term ($\frac{1}{\sqrt{a}}$ or $\frac{1}{b\sqrt{a}}$)

If the denominator is a single term involving a square root, like $\sqrt{a}$ (where $a$ is a positive real number), the rationalising factor is simply $\sqrt{a}$. Multiplying $\sqrt{a}$ by itself results in a rational number $a$.

If the denominator is of the form $b\sqrt{a}$ (where $b$ is a non-zero rational number), the rationalising factor is still $\sqrt{a}$ (or $b\sqrt{a}$, but $\sqrt{a}$ is sufficient to make the denominator rational).

Since $a$ and $b$ are (usually rational) numbers, the new denominator $a$ or $ba$ will be rational.

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Rationalising a Denominator with Two Terms (Binomial Denominators)

If the denominator is a binomial involving square roots, like $(a + \sqrt{b})$, $(a - \sqrt{b})$, $(\sqrt{a} + \sqrt{b})$, or $(\sqrt{a} - \sqrt{b})$, where $a$ and $b$ are positive real numbers, we use the identity $(x+y)(x-y) = x^2 - y^2$. This identity is helpful because the product of a sum and a difference of the same two terms results in the difference of their squares, which eliminates the square root if the terms are square roots.

The rationalising factor for a binomial denominator is its conjugate. The conjugate of a binomial of the form $(X+Y)$ is $(X-Y)$, and the conjugate of $(X-Y)$ is $(X+Y)$.

• The conjugate of $(a + \sqrt{b})$ is $(a - \sqrt{b})$.
• The conjugate of $(a - \sqrt{b})$ is $(a + \sqrt{b})$.
• The conjugate of $(\sqrt{a} + \sqrt{b})$ is $(\sqrt{a} - \sqrt{b})$.
• The conjugate of $(\sqrt{a} - \sqrt{b})$ is $(\sqrt{a} + \sqrt{b})$.

To rationalize a binomial denominator, multiply both the numerator and the denominator by the conjugate of the denominator.

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Laws of Exponents for Real Numbers

In previous classes, we learned about exponents and powers with integer bases and integer exponents, and the laws they follow. These laws are not limited to integer exponents; they can be extended to include rational exponents as well, applied to positive real number bases.

Meaning of Rational Exponents

For a positive real number $a$, and a rational exponent $\frac{m}{n}$ (where $m$ is an integer and $n$ is a positive integer, $n > 0$), the expression $a^{\frac{m}{n}}$ is defined as the $n$-th root of $a$ raised to the power $m$.

Equivalently, it can be defined as the $n$-th root of $a$, which is then raised to the power $m$:

For example, $8^{\frac{2}{3}}$ means either the cube root of $8^2$, or the cube root of 8, squared. The second approach is usually simpler.

$8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = (2)^2 = 4$.

Alternatively, $8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$. Both definitions give the same result.

When the exponent is of the form $\frac{1}{n}$, it represents the $n$-th root: $a^{\frac{1}{n}} = \sqrt[n]{a}$. For instance, $a^{\frac{1}{2}} = \sqrt{a}$ (the square root) and $a^{\frac{1}{3}} = \sqrt[3]{a}$ (the cube root).

Note: We restrict the base $a$ to be positive when dealing with rational exponents to avoid complexities. For example, $\sqrt{-4}$ is not a real number. Also, $(-8)^{\frac{1}{3}} = -2$, but $(-8)^{\frac{2}{6}} = ((-8)^2)^{\frac{1}{6}} = 64^{\frac{1}{6}} = 2$. Using a positive base ensures that the laws of exponents are always consistent.

Laws of Exponents for Real Numbers

The laws of exponents hold for any positive real base and any rational exponents. Let $a > 0$ and $b > 0$ be real numbers, and let $p$ and $q$ be rational numbers.

1. Product of Powers:

   When multiplying powers with the same base, we add the exponents.

   $a^p \cdot a^q = a^{p+q}$

   ... (iii)

   Example: $3^{\frac{1}{2}} \cdot 3^{\frac{1}{4}} = 3^{\frac{1}{2} + \frac{1}{4}} = 3^{\frac{2}{4} + \frac{1}{4}} = 3^{\frac{3}{4}}$.

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2. Power of a Power:

   When raising a power to another exponent, we multiply the exponents.

   $(a^p)^q = a^{pq}$

   ... (iv)

   Example: $(7^{\frac{1}{5}})^4 = 7^{\frac{1}{5} \times 4} = 7^{\frac{4}{5}}$.

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3. Quotient of Powers:

   When dividing powers with the same base, we subtract the exponents.

   $\frac{a^p}{a^q} = a^{p-q}$

   ... (v)

   Example: $\frac{4^{\frac{2}{3}}}{4^{\frac{1}{6}}} = 4^{\frac{2}{3} - \frac{1}{6}} = 4^{\frac{4}{6} - \frac{1}{6}} = 4^{\frac{3}{6}} = 4^{\frac{1}{2}} = \sqrt{4} = 2$.

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4. Power of a Product:

   When the bases are different but the exponents are the same, we can multiply the bases first.

   $a^p b^p = (ab)^p$

   ... (vi)

   Example: $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \times 8)^{\frac{1}{2}} = 56^{\frac{1}{2}} = \sqrt{56}$.

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5. Power of a Quotient:

   A power of a fraction can be distributed to the numerator and the denominator.

   $(\frac{a}{b})^p = \frac{a^p}{b^p}$

   ... (vii)

   Example: $(\frac{4}{9})^{\frac{1}{2}} = \frac{4^{\frac{1}{2}}}{9^{\frac{1}{2}}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$.

The definitions of negative exponents ($a^{-p} = \frac{1}{a^p}$) and the zero exponent ($a^0 = 1$) also hold true for positive real bases and rational exponents.

Examples using the Laws of Exponents

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