Number Systems and Real Numbers

Classification of Real Numbers

In mathematics, numbers can be grouped into different sets based on their properties. Learning how to classify numbers helps you understand where a number belongs and how it behaves in calculations.

In this lesson, you will learn the five main groups of real numbers: natural numbers, whole numbers, integers, rational numbers, and irrational numbers. You will also see how these groups fit inside one another.

All of these groups together make up the set of real numbers. Real numbers are all the numbers that can be placed on a number line.

1. Natural Numbers

Natural numbers are the counting numbers we use to count objects.

They are:

$$1, 2, 3, 4, 5, \dots$$

Some books include 0 as a natural number, but in many school courses, natural numbers start at 1. In this lesson, we will use:

$$\text{Natural numbers} = \{1,2,3,4,\dots\}$$

2. Whole Numbers

Whole numbers are the natural numbers together with 0.

$$0, 1, 2, 3, 4, 5, \dots$$

So every natural number is a whole number, but 0 is a whole number that is not a natural number in this lesson.

3. Integers

Integers include all positive whole numbers, 0, and negative whole numbers.

$$\dots, -4, -3, -2, -1, 0, 1, 2, 3, 4, \dots$$

Integers do not include fractions or decimals unless the decimal is equal to a whole number. For example, \(3.0\) is an integer because it is equal to 3, but \(3.5\) is not.

4. Rational Numbers

A rational number is any number that can be written as a fraction of the form

$$\frac{a}{b}$$

where \(a\) and \(b\) are integers and \(b \ne 0\).

This means rational numbers include:

• fractions such as \(\frac{3}{4}\)
• integers such as \(-5\), since \(-5 = \frac{-5}{1}\)
• terminating decimals such as \(0.8\), since \(0.8 = \frac{4}{5}\)
• repeating decimals such as \(0.333\dots\), since \(0.333\dots = \frac{1}{3}\)

So, a decimal is rational if it ends or repeats in a pattern.

5. Irrational Numbers

Irrational numbers are real numbers that cannot be written as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \ne 0\).

The decimal form of an irrational number goes on forever without ending and without repeating a pattern.

Common examples are:

• \(\sqrt{2}\)
• \(\sqrt{3}\)
• \(\pi\)

For example, \(\sqrt{2} = 1.4142135\dots\) continues forever and does not repeat in a fixed pattern.

6. The Real Number System as a Hierarchy

The sets of numbers are nested inside one another. This means smaller sets are contained inside larger sets.

$$\text{Natural} \subset \text{Whole} \subset \text{Integers} \subset \text{Rational} \subset \text{Real}$$

Irrational numbers are also part of the real numbers, but they are not rational.

So the real numbers can be divided into two big groups:

$$\text{Real numbers} = \text{Rational numbers} \cup \text{Irrational numbers}$$

And rational numbers and irrational numbers do not overlap.

Important idea: A number can belong to more than one set. For example, \(4\) is a natural number, a whole number, an integer, a rational number, and a real number.

7. How to Classify a Number

When classifying a number, ask these questions:

1. Is it a counting number? Then it is natural.
2. Is it 0 or a counting number? Then it is whole.
3. Is it a negative or positive whole number, or 0? Then it is an integer.
4. Can it be written as a fraction of integers? Then it is rational.
5. If it cannot be written as such a fraction and its decimal neither ends nor repeats, then it is irrational.

It is often helpful to name the smallest set the number belongs to, but sometimes you may be asked to list all the sets it belongs to.

8. Worked Examples

Example 1: Classify \(7\)

The number \(7\) is a counting number, so it is a natural number.

Because all natural numbers are also whole numbers, integers, rational numbers, and real numbers, \(7\) belongs to all of these sets.

Answer: \(7\) is natural, whole, integer, rational, and real.

Example 2: Classify \(0\)

The number \(0\) is not a natural number in this lesson, but it is a whole number.

It is also an integer. Since it can be written as

$$0 = \frac{0}{1}$$

it is rational, and therefore real.

Answer: \(0\) is whole, integer, rational, and real.

Example 3: Classify \(-\frac{5}{2}\)

This number is a fraction, so it is not natural, not whole, and not an integer.

However, it is already written as a ratio of two integers:

$$-\frac{5}{2}$$

So it is a rational number. Every rational number is also a real number.

Answer: \(-\frac{5}{2}\) is rational and real.

Example 4: Classify \(\sqrt{16}\) and \(\sqrt{5}\)

First, simplify each number.

$$\sqrt{16} = 4$$

Since \(4\) is a counting number, it is natural, whole, integer, rational, and real.

Now look at \(\sqrt{5}\). Since 5 is not a perfect square, \(\sqrt{5}\) cannot be written as a fraction of integers. Its decimal goes on forever without repeating.

So \(\sqrt{5}\) is irrational, and therefore real.

Answer:

• \(\sqrt{16}\) is natural, whole, integer, rational, and real.
• \(\sqrt{5}\) is irrational and real.

9. Common Mistakes to Avoid

• Thinking all decimals are irrational: Decimals that end or repeat are rational.
• Forgetting that integers are rational: Any integer \(n\) can be written as \(\frac{n}{1}\).
• Confusing whole numbers and integers: Whole numbers are \(0,1,2,3,\dots\), while integers also include negatives.
• Assuming every square root is irrational: The square root of a perfect square, such as \(\sqrt{25}=5\), is rational.

10. Quick Classification Practice

Try thinking about these on your own:

• \(-8\) is an integer, rational, and real.
• \(2.75\) is rational and real because \(2.75 = \frac{11}{4}\).
• \(0.121212\dots\) is rational because the block 12 repeats.
• \(\pi\) is irrational and real.

Summary

Real numbers include all the numbers on the number line. They are divided into rational and irrational numbers.

Inside the rational numbers are the integers, inside the integers are the whole numbers, and inside the whole numbers are the natural numbers.

If a number can be written as a fraction of integers, it is rational. If it cannot, and its decimal does not end or repeat, it is irrational.

Understanding this structure makes it easier to identify and compare numbers in algebra and other areas of mathematics.

Euclid's Division Lemma is a very important idea in number systems. It helps us describe what happens when one integer is divided by another.

This lemma is not just about division. It is also used to prove many useful results in mathematics, such as the forms of even numbers, odd numbers, and possible forms of square numbers.

In this lesson, we will understand the statement of Euclid's Division Lemma, learn what each part means, and see how to apply it in examples and simple proofs.

1. What is Euclid's Division Lemma?

If we divide an integer \(a\) by another positive integer \(b\), then we can always write:

$$a = bq + r$$

where:

• \(a\) is the dividend
• \(b\) is the divisor, with \(b > 0\)
• \(q\) is the quotient
• \(r\) is the remainder

The remainder must satisfy:

$$0 \le r < b$$

This means the remainder is always a whole number starting from 0, but it must be smaller than the divisor.

2. Understanding the meaning

Suppose you divide 17 by 5. Then 5 goes into 17 three times, and 2 is left over.

So we write:

$$17 = 5 \times 3 + 2$$

Here:

• \(a = 17\)
• \(b = 5\)
• \(q = 3\)
• \(r = 2\)

This follows Euclid's Division Lemma because \(0 \le 2 < 5\).

3. Why is this lemma useful?

Euclid's Division Lemma helps us express integers in different forms depending on the divisor.

For example, if an integer is divided by 2, the remainder can only be 0 or 1, because the remainder must be less than 2.

So every integer \(n\) can be written as either:

$$n = 2q$$

or

$$n = 2q + 1$$

This immediately gives us the forms of:

• even numbers: \(2q\)
• odd numbers: \(2q+1\)

This is one of the most common uses of the lemma.

4. General idea using a divisor

If a number \(a\) is divided by a positive integer \(b\), then the possible remainders are:

$$0, 1, 2, 3, \dots, b-1$$

So the possible forms of \(a\) are:

$$a = bq,\; bq+1,\; bq+2,\; \dots,\; bq+(b-1)$$

For example, when dividing by 3, the possible remainders are 0, 1, and 2. So every integer can be written as one of these forms:

$$a = 3q, \quad 3q+1, \quad 3q+2$$

5. Worked Example 1: Writing a number using the lemma

Question: Write 29 in the form \(a = bq + r\) when divided by 6.

Solution:

We divide 29 by 6.

\(6 \times 4 = 24\), and the remainder is \(29 - 24 = 5\).

So:

$$29 = 6 \times 4 + 5$$

Thus:

• quotient \(q = 4\)
• remainder \(r = 5\)

This is correct because \(0 \le 5 < 6\).

6. Worked Example 2: Forms of integers when divided by 2

Question: Show that every integer is either even or odd.

Solution:

Apply Euclid's Division Lemma with divisor \(2\).

For any integer \(n\), we can write:

$$n = 2q + r$$

where

$$0 \le r < 2$$

The only possible values of \(r\) are 0 and 1.

So there are two cases:

1. If \(r = 0\), then \(n = 2q\). This is an even number.
2. If \(r = 1\), then \(n = 2q + 1\). This is an odd number.

Therefore, every integer is either even or odd.

7. Worked Example 3: Possible forms of squares

Question: Show that the square of any positive integer is either of the form \(3m\) or \(3m+1\).

Solution:

Let any positive integer be \(n\). When divided by 3, by Euclid's Division Lemma, \(n\) can be written in one of these forms:

$$n = 3q, \quad 3q+1, \quad 3q+2$$

Now square each form.

Case 1: \(n = 3q\)

$$n^2 = (3q)^2 = 9q^2 = 3(3q^2)$$

So \(n^2\) is of the form \(3m\).

Case 2: \(n = 3q+1\)

$$n^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2+2q) + 1$$

So \(n^2\) is of the form \(3m+1\).

Case 3: \(n = 3q+2\)

$$n^2 = (3q+2)^2 = 9q^2 + 12q + 4$$

$$= 9q^2 + 12q + 3 + 1$$

$$= 3(3q^2 + 4q + 1) + 1$$

So \(n^2\) is again of the form \(3m+1\).

Therefore, the square of any positive integer is either:

$$3m \quad \text{or} \quad 3m+1$$

It can never be of the form \(3m+2\).

8. Worked Example 4: Possible forms of numbers divided by 5

Question: What are the possible forms of an integer when divided by 5?

Solution:

Let the integer be \(n\). By Euclid's Division Lemma:

$$n = 5q + r$$

where

$$0 \le r < 5$$

So the possible values of \(r\) are:

$$0, 1, 2, 3, 4$$

Therefore, the possible forms of \(n\) are:

$$n = 5q, \quad 5q+1, \quad 5q+2, \quad 5q+3, \quad 5q+4$$

9. How to apply Euclid's Division Lemma in proofs

When a question asks you to prove a property using this lemma, follow these steps:

1. Choose the divisor given in the question.
2. Write the number in the form \(a = bq + r\).
3. List all possible values of the remainder \(r\).
4. Consider each case one by one.
5. Simplify and show the required result.

This method is very useful in proofs involving even/odd numbers, divisibility, and square numbers.

10. Common mistakes to avoid

• Forgetting the condition on remainder: Always remember that \(0 \le r < b\).
• Using a remainder equal to the divisor: This is not allowed. For example, when dividing by 4, remainder cannot be 4.
• Missing one of the cases: If dividing by 3, you must check all three forms: \(3q\), \(3q+1\), and \(3q+2\).
• Confusing lemma with algorithm: The lemma gives the form \(a = bq+r\). It is often used as the starting point for proofs.

11. Quick recap

• Euclid's Division Lemma states that for integers \(a\) and positive integer \(b\),

$$a = bq + r, \quad 0 \le r < b$$

• The remainder is always less than the divisor.
• When dividing by 2, every integer is of the form \(2q\) or \(2q+1\).
• When dividing by 3, every integer is of the form \(3q\), \(3q+1\), or \(3q+2\).
• This lemma helps prove properties of integers, especially about even numbers, odd numbers, and squares.

12. Final summary

Euclid's Division Lemma tells us that any integer can be written in the form \(a = bq + r\), where the remainder \(r\) is always between 0 and \(b-1\).

Its real power is in proofs. By listing all possible remainders for a chosen divisor, we can describe all possible forms of an integer and then study its properties.

Once you are comfortable with this idea, many number system proofs become easier and more organized.

Applications of Euclid's Algorithm

When we need to find the greatest common divisor (GCD) or highest common factor (HCF) of two numbers, listing all factors can be slow, especially for large numbers.

Euclid's Algorithm is a fast and reliable method for finding the HCF of two positive integers. It is based on the idea of repeated division.

This method is an important application of the division lemma, which says that for any two integers \(a\) and \(b\), with \(a > b > 0\), we can write

Here, \(q\) is the quotient and \(r\) is the remainder.

The key idea behind Euclid's Algorithm is:

where \(r\) is the remainder when \(a\) is divided by \(b\).

So instead of finding the HCF of the original pair, we replace the larger problem with a smaller one. We keep repeating this process until the remainder becomes 0. The last non-zero remainder is the HCF.

Why does this work?

If a number divides both \(a\) and \(b\), then it must also divide \(a-bq\). But \(a-bq = r\), so any common divisor of \(a\) and \(b\) is also a common divisor of \(b\) and \(r\).

Similarly, any common divisor of \(b\) and \(r\) also divides \(a = bq+r\). So the common divisors of both pairs are the same, and their greatest common divisor is also the same.

Steps of Euclid's Algorithm

1. Take two positive integers, with the larger one first.
2. Divide the larger number by the smaller number.
3. Write down the remainder.
4. Replace the pair by: smaller number, remainder.
5. Repeat until the remainder becomes 0.
6. The last non-zero remainder is the HCF.

Worked Example 1: Find the HCF of 48 and 18

We divide 48 by 18:

Now divide 18 by 12:

Now divide 12 by 6:

The remainder is now 0, so the last non-zero remainder is \(6\).

Therefore,

Worked Example 2: Find the HCF of 119 and 544

First place the larger number first:

Now divide 119 by 68:

Now divide 68 by 51:

Now divide 51 by 17:

The last non-zero remainder is \(17\).

So,

Worked Example 3: Use Euclid's Algorithm to find the HCF of 867 and 255

Divide 867 by 255:

Now divide 255 by 102:

Now divide 102 by 51:

The last non-zero remainder is \(51\).

Therefore,

Application: Finding whether numbers are co-prime

Two numbers are called co-prime if their HCF is 1.

Euclid's Algorithm helps us check this quickly.

Worked Example 4: Are 35 and 64 co-prime?

Divide 64 by 35:

Now divide 35 by 29:

Now divide 29 by 6:

Now divide 6 by 5:

Now divide 5 by 1:

The last non-zero remainder is \(1\).

So,

Therefore, 35 and 64 are co-prime.

Another important application: Writing the HCF in the form \(ax+by\)

Sometimes, after finding the HCF by Euclid's Algorithm, we write it as a combination of the two given numbers. This means expressing the HCF in the form

where \(x\) and \(y\) are integers.

This is done by working backward from the division steps.

Let us use Example 2, where the HCF of 544 and 119 is 17.

We had:

Now work backward.

From the last equation:

From the second equation, \(51 = 119 - 68\). Substitute this:

From the first equation, \(68 = 544 - 119 \times 4\). Substitute this:

So we have written the HCF as

This shows that the HCF can be expressed using the given numbers.

Why is Euclid's Algorithm useful?

• It is much faster than listing factors.
• It works well even for large numbers.
• It helps us check if numbers are co-prime.
• It helps us express the HCF in terms of the given numbers.
• It is a basic tool in number systems and higher mathematics.

Common mistakes to avoid

• Do not stop too early. Continue until the remainder is 0.
• The HCF is not the quotient. It is the last non-zero remainder.
• Always divide the larger number by the smaller number first.
• Be careful while writing each division step.

Quick Recap

• Use the division lemma: \(a = bq + r\).
• Replace \((a,b)\) by \((b,r)\).
• Repeat until remainder becomes 0.
• The last non-zero remainder is the HCF.
• If the HCF is 1, the numbers are co-prime.

Summary

Euclid's Algorithm is a method to find the HCF of two integers by repeated division. At each step, we divide and use the remainder to form a smaller pair of numbers. This process continues until the remainder is 0, and the last non-zero remainder gives the HCF.

This algorithm is useful for large numbers, for checking co-prime numbers, and for expressing the HCF in the form \(ax+by\). With practice, it becomes a quick and powerful tool in number systems.

Introduction

In number systems, one of the most important ideas is that every whole number greater than 1 is built from prime numbers. This idea is called the Fundamental Theorem of Arithmetic.

This theorem helps us understand how numbers are formed, how to test divisibility, how to find HCF and LCM, and why prime factorization is such a powerful tool in mathematics.

In this lesson, you will learn what the theorem says, why it matters, and how to use it correctly.

1. Review: Prime and Composite Numbers

Before learning the theorem, let us recall two important types of numbers.

• A prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.
• A composite number is a natural number greater than 1 that has more than two factors.

Examples:

• 2, 3, 5, 7, 11 are prime numbers.

• 4, 6, 8, 9, 10, 12 are composite numbers.

The number 1 is neither prime nor composite.

2. Statement of the Fundamental Theorem of Arithmetic

The theorem says:

Every composite number can be expressed as a product of prime numbers, and this factorization is unique except for the order of the primes.

This means two things:

1. Existence: Every composite number can be broken into prime factors.

2. Uniqueness: The prime factors you get are always the same, though you may write them in a different order.

For example, consider 60.

We can write

$$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$$

So the prime factorization of 60 is

$$60 = 2^2 \times 3 \times 5$$

No matter how you start factoring 60, you will always end up with the same prime factors: two 2s, one 3, and one 5.

3. What Does “Unique Except for Order” Mean?

The order of multiplication does not matter. So these are all the same prime factorization:

$$2 \times 2 \times 3 \times 5$$

$$3 \times 2 \times 5 \times 2$$

$$5 \times 3 \times 2 \times 2$$

Each expression contains the same prime factors. Only the order has changed.

4. Why Is This Theorem Important?

The Fundamental Theorem of Arithmetic is important because it gives a basic structure for all whole numbers greater than 1.

It helps us:

• write numbers in prime factor form,
• find the HCF and LCM of numbers,
• study divisibility,
• compare numbers more easily,
• understand many later ideas in algebra and number theory.

It is like saying that prime numbers are the “building blocks” of all composite numbers.

5. How to Find Prime Factorization

To factor a composite number into primes, keep dividing it by prime numbers until only prime factors remain.

You can use:

• repeated division, or
• a factor tree.

While factoring, remember:

• Keep breaking composite factors further.
• Stop only when every factor is prime.
• Write repeated prime factors using exponents if needed.

6. Worked Example 1: Prime Factorization of a Small Number

Find the prime factorization of 24.

Step 1: Start dividing by the smallest prime number, 2.

$$24 = 2 \times 12$$

Step 2: Factor 12.

$$12 = 2 \times 6$$

Step 3: Factor 6.

$$6 = 2 \times 3$$

Now 2 and 3 are prime, so we stop.

Therefore,

$$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$$

Answer: The prime factorization of 24 is \(2^3 \times 3\).

7. Worked Example 2: Same Number, Different Factoring Paths

Show that 36 has a unique prime factorization even if we begin in different ways.

Method 1:

$$36 = 6 \times 6$$

Now factor each 6:

$$6 = 2 \times 3$$

So,

$$36 = (2 \times 3)(2 \times 3) = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$$

Method 2:

$$36 = 4 \times 9$$

Now factor each part:

$$4 = 2 \times 2, \quad 9 = 3 \times 3$$

So,

$$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$$

Both methods give the same prime factors.

Answer: The unique prime factorization of 36 is \(2^2 \times 3^2\).

8. Worked Example 3: Using Prime Factorization to Study Divisibility

Check whether 84 is divisible by 18 using prime factorization.

Step 1: Prime factorize 84.

$$84 = 2 \times 42 = 2 \times 2 \times 21 = 2^2 \times 3 \times 7$$

Step 2: Prime factorize 18.

$$18 = 2 \times 9 = 2 \times 3^2$$

Step 3: Compare the prime factors.

For 84 to be divisible by 18, the prime factorization of 84 must contain all the prime factors of 18.

But

$$18 = 2 \times 3^2$$

and

$$84 = 2^2 \times 3 \times 7$$

84 has only one factor of 3, but 18 needs two factors of 3.

So 84 is not divisible by 18.

Answer: 84 is not divisible by 18.

9. Worked Example 4: Using the Theorem in Exponential Form

Write the prime factorization of 540.

Step 1: Break the number into easier factors.

$$540 = 54 \times 10$$

Step 2: Factor each part.

$$54 = 2 \times 27 = 2 \times 3^3$$

$$10 = 2 \times 5$$

Step 3: Multiply all prime factors together.

$$540 = (2 \times 3^3)(2 \times 5)$$

$$540 = 2^2 \times 3^3 \times 5$$

Answer: The prime factorization of 540 is \(2^2 \times 3^3 \times 5\).

10. Important Notes and Common Mistakes

• Do not stop too early. If a factor is composite, factor it again. For example, in \(24 = 3 \times 8\), you must still factor 8 into \(2 \times 2 \times 2\).

• Use only prime factors in the final answer. A correct final form for 18 is \(2 \times 3^2\), not \(6 \times 3\).

• Order does not matter. \(2 \times 3 \times 2\) and \(2^2 \times 3\) represent the same prime factorization.

• 1 is not a prime number. Do not include 1 in prime factorization.

11. How the Theorem Helps with HCF and LCM

Prime factorization is often used to find the HCF and LCM of numbers.

For example, take 12 and 18.

$$12 = 2^2 \times 3$$

$$18 = 2 \times 3^2$$

The HCF uses common prime factors with the smallest powers:

$$\text{HCF} = 2 \times 3 = 6$$

The LCM uses all prime factors with the greatest powers:

$$\text{LCM} = 2^2 \times 3^2 = 36$$

This method works because every composite number has a unique prime factorization.

12. A Simple Way to Remember the Theorem

You can remember the Fundamental Theorem of Arithmetic like this:

Every composite number has one and only one prime factorization, apart from the order of the factors.

Or even shorter:

Composite numbers are made of primes in a unique way.

Brief Summary

The Fundamental Theorem of Arithmetic says that every composite number can be written as a product of prime numbers, and this product is unique except for the order of the factors.

This theorem is the foundation of prime factorization. It helps us understand divisibility and is very useful in finding HCF, LCM, and studying the structure of numbers.

Whenever you factor a number completely into primes, you are using this theorem.

Prime Factorization and Divisibility Rules are important tools in number systems. They help us break numbers into smaller parts, test whether one number divides another, and find useful values like the Greatest Common Divisor (GCD) and Lowest Common Multiple (LCM).

In this lesson, you will learn how to:

• identify prime and composite numbers,
• write numbers as a product of prime factors,
• use divisibility rules to factor numbers faster,
• find the GCD and LCM using prime factorization,
• use the relation $$\text{LCM} \times \text{GCD} = a \times b$$ for two numbers.

1. Prime and Composite Numbers

A prime number is a number greater than 1 that has exactly two positive factors: 1 and itself.

Examples of prime numbers are: 2, 3, 5, 7, 11, and 13.

A composite number is a number greater than 1 that has more than two factors.

Examples of composite numbers are: 4, 6, 8, 9, 10, and 12.

The number 1 is neither prime nor composite.

2. Prime Factorization

Prime factorization means writing a composite number as a product of prime numbers only.

For example, the prime factorization of 12 is:

$$12 = 2 \times 2 \times 3 = 2^2 \times 3$$

Every composite number has a unique prime factorization, except for the order of the factors. This means the same prime factors will always appear, even if we find them in a different order.

One common method is the factor tree.

For example, to factor 60:

$$60 = 6 \times 10 = (2 \times 3) \times (2 \times 5) = 2^2 \times 3 \times 5$$

3. Divisibility Rules

Divisibility rules help you quickly decide whether a number can be divided by another number without doing full division. These rules are very useful when finding prime factors.

• Divisible by 2: The last digit is 0, 2, 4, 6, or 8.
• Divisible by 3: The sum of the digits is divisible by 3.
• Divisible by 5: The last digit is 0 or 5.
• Divisible by 9: The sum of the digits is divisible by 9.
• Divisible by 10: The last digit is 0.

For example, consider 126:

• It is divisible by 2 because it ends in 6.
• It is divisible by 3 because \(1+2+6=9\), and 9 is divisible by 3.
• It is divisible by 9 because \(1+2+6=9\), and 9 is divisible by 9.

These rules help us choose factors quickly when doing prime factorization.

4. Finding GCD Using Prime Factorization

The Greatest Common Divisor (GCD) of two numbers is the largest number that divides both of them exactly. It is also called the Highest Common Factor (HCF).

To find the GCD using prime factorization:

1. Write each number as a product of prime factors.
2. Find the prime factors common to both numbers.
3. Take the smallest power of each common prime.
4. Multiply them.

5. Finding LCM Using Prime Factorization

The Lowest Common Multiple (LCM) of two numbers is the smallest positive number that is a multiple of both numbers.

To find the LCM using prime factorization:

1. Write each number as a product of prime factors.
2. List all prime factors that appear in either number.
3. Take the greatest power of each prime.
4. Multiply them.

6. Important Relation Between LCM and GCD

For any two positive integers \(a\) and \(b\):

$$\text{LCM}(a,b) \times \text{GCD}(a,b) = a \times b$$

This relation is very useful. If you know the GCD, you can find the LCM, and vice versa.

Worked Example 1: Prime Factorization Using Divisibility Rules

Find the prime factorization of 180.

Step 1: Use divisibility rules.

• 180 ends in 0, so it is divisible by 10.
• Also, it is even, so it is divisible by 2.

Let us factor it step by step:

$$180 = 2 \times 90$$

$$90 = 2 \times 45$$

$$45 = 3 \times 15$$

$$15 = 3 \times 5$$

So,

$$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$$

Answer: The prime factorization of 180 is $$2^2 \times 3^2 \times 5$$

Worked Example 2: Find the GCD of 48 and 60

First, write each number as a product of prime factors.

$$48 = 2^4 \times 3$$

$$60 = 2^2 \times 3 \times 5$$

The common prime factors are 2 and 3.

• The smallest power of 2 is \(2^2\).
• The smallest power of 3 is \(3^1\).

So,

$$\text{GCD}(48,60) = 2^2 \times 3 = 4 \times 3 = 12$$

Answer: $$\text{GCD}(48,60)=12$$

Worked Example 3: Find the LCM of 48 and 60

Use the same prime factorizations:

$$48 = 2^4 \times 3$$

$$60 = 2^2 \times 3 \times 5$$

Take the greatest power of each prime factor that appears:

• For 2, use \(2^4\)
• For 3, use \(3^1\)
• For 5, use \(5^1\)

So,

$$\text{LCM}(48,60) = 2^4 \times 3 \times 5 = 16 \times 3 \times 5 = 240$$

Answer: $$\text{LCM}(48,60)=240$$

Worked Example 4: Use the Relation \(\text{LCM} \times \text{GCD} = a \times b\)

Two numbers are 18 and 24. Their GCD is 6. Find their LCM.

Use the formula:

$$\text{LCM} \times \text{GCD} = a \times b$$

Substitute the values:

$$\text{LCM} \times 6 = 18 \times 24$$

$$\text{LCM} \times 6 = 432$$

$$\text{LCM} = \frac{432}{6} = 72$$

Answer: The LCM of 18 and 24 is $$72$$

7. How GCD and LCM Are Connected to Prime Factors

Prime factorization makes the difference between GCD and LCM very clear:

• For GCD, choose only the common prime factors with the smaller powers.
• For LCM, choose all prime factors present, using the larger powers.

For example, if

$$a = 2^3 \times 3^2 \times 5$$

and

$$b = 2^2 \times 3 \times 7$$

Then:

$$\text{GCD}(a,b)=2^2 \times 3 = 12$$

$$\text{LCM}(a,b)=2^3 \times 3^2 \times 5 \times 7$$

8. Common Mistakes to Avoid

• Stopping too early: In prime factorization, keep factoring until every factor is prime.
• Mixing up GCD and LCM: GCD uses smallest powers of common primes; LCM uses greatest powers of all primes present.
• Forgetting a prime factor: Check your work carefully.
• Using the formula incorrectly: The relation $$\text{LCM} \times \text{GCD} = a \times b$$ works for two numbers.

9. Quick Strategy for Solving Problems

1. Use divisibility rules to break the number into factors quickly.
2. Write the prime factorization using exponents where possible.
3. For GCD, choose common primes with smaller exponents.
4. For LCM, choose all primes with larger exponents.
5. If needed, use $$\text{LCM} \times \text{GCD} = a \times b$$ to check your answer.

Summary

Prime factorization means expressing a number as a product of prime numbers. Divisibility rules help us do this faster by showing which numbers divide a given number exactly.

Once numbers are written in prime factor form, we can find the GCD by taking common prime factors with the smallest powers, and the LCM by taking all prime factors with the greatest powers. For two numbers \(a\) and \(b\), remember the important relation:

$$\text{LCM}(a,b) \times \text{GCD}(a,b) = a \times b$$

Mastering these ideas will help you solve many problems in number systems and real numbers.

Proof of Irrationality by Contradiction

In this lesson, we will learn how to prove that some numbers are irrational using a method called proof by contradiction.

You may already know that a rational number can be written as a fraction of two integers, like \(\frac{3}{4}\), \(-2\), or \(\frac{11}{7}\). An irrational number cannot be written in this form. Numbers like \(\sqrt{2}\), \(\sqrt{3}\), and \(\pi\) are irrational.

But how do we prove that a number such as \(\sqrt{2}\) is irrational? We use a logical method called contradiction.

Proof by contradiction means we start by assuming the opposite of what we want to prove. Then, if that assumption leads to something impossible or inconsistent, our assumption must be wrong.

So if we want to prove that \(\sqrt{2}\) is irrational, we begin by assuming that \(\sqrt{2}\) is rational.

Step 1: Recall what rational means.

If \(\sqrt{2}\) is rational, then it can be written as

where \(a\) and \(b\) are integers, \(b \ne 0\), and the fraction is in lowest terms. That means \(a\) and \(b\) have no common factor other than 1.

Why do we insist on lowest terms? Because if a fraction is not simplified, we can reduce it. Starting with lowest terms is important, because later we will show that both numbers must still have a common factor. That is the contradiction.

Step 2: Square both sides.

This tells us that \(a^2\) is even, because it equals \(2 \times b^2\).

Important fact: If \(a^2\) is even, then \(a\) must also be even.

This is because:

• the square of an even number is even, and
• the square of an odd number is odd.

So if \(a\) is even, we can write

for some integer \(k\).

Step 3: Substitute back.

Now \(b^2\) is even, so \(b\) must also be even.

So both \(a\) and \(b\) are even. That means both are divisible by 2.

But this is impossible, because we said at the start that \(\frac{a}{b}\) was in lowest terms.

We have reached a contradiction.

Therefore, our original assumption was false. So:

This is the basic structure of the proof.

1. Assume the square root is rational.
2. Write it as a fraction in lowest terms.
3. Square both sides.
4. Show the numerator is divisible by a prime number.
5. Then show the denominator is also divisible by the same prime number.
6. This contradicts the fact that the fraction was in lowest terms.
7. Therefore, the square root is irrational.

A useful fact about prime numbers

For proofs involving \(\sqrt{3}\), \(\sqrt{5}\), and other square roots of prime numbers, we use this idea:

If a prime number divides \(a^2\), then that prime number must divide \(a\).

For example:

• if 3 divides \(a^2\), then 3 divides \(a\),
• if 5 divides \(a^2\), then 5 divides \(a\).

This works because prime numbers have very strict factor behavior.

Worked Example 1: Prove that \(\sqrt{2}\) is irrational

Assume that \(\sqrt{2}\) is rational.

Then

where \(a\) and \(b\) are integers with no common factor, and \(b \ne 0\).

Square both sides:

So \(a^2\) is even, which means \(a\) is even.

Let \(a = 2k\).

Substitute into the equation:

So \(b^2\) is even, which means \(b\) is even.

Then both \(a\) and \(b\) are even, so they have a common factor 2.

This contradicts the fact that \(\frac{a}{b}\) is in lowest terms.

Therefore, \(\sqrt{2}\) is irrational.

Worked Example 2: Prove that \(\sqrt{3}\) is irrational

Assume that \(\sqrt{3}\) is rational.

Then we can write

where \(a\) and \(b\) are integers in lowest terms.

Square both sides:

This means \(a^2\) is divisible by 3. Therefore, \(a\) is also divisible by 3.

So let

for some integer \(k\).

Substitute into \(a^2 = 3b^2\):

Now \(b^2\) is divisible by 3, so \(b\) is divisible by 3.

That means both \(a\) and \(b\) are divisible by 3.

But then \(a\) and \(b\) have a common factor 3, which contradicts the fact that the fraction was in lowest terms.

Therefore,

Worked Example 3: Prove that \(\sqrt{5}\) is irrational

Assume that \(\sqrt{5}\) is rational.

Then

where \(a\) and \(b\) are integers with no common factor.

Square both sides:

So \(a^2\) is divisible by 5, which means \(a\) is divisible by 5.

Let

Substitute:

So \(b^2\) is divisible by 5, which means \(b\) is divisible by 5.

Then both \(a\) and \(b\) are divisible by 5.

This contradicts the claim that \(\frac{a}{b}\) is in lowest terms.

Therefore,

What all these examples have in common

• We assume the square root is rational.
• We write it as \(\frac{a}{b}\) in simplest form.
• After squaring, we get an equation like \(a^2 = pb^2\), where \(p\) is prime.
• That shows \(p\) divides \(a\).
• Substituting back shows \(p\) also divides \(b\).
• So \(a\) and \(b\) have a common factor \(p\), which is impossible.

General result

Using the same method, we can prove that if \(p\) is a prime number, then \(\sqrt{p}\) is irrational.

The proof follows exactly the same pattern:

1. Assume \(\sqrt{p} = \frac{a}{b}\) in lowest terms.
2. Square both sides to get

3. Since \(p\) divides \(a^2\), it must divide \(a\).
4. Write \(a = pk\), then substitute back.
5. This shows \(p\) also divides \(b\).
6. So \(a\) and \(b\) both have factor \(p\), a contradiction.

Therefore, \(\sqrt{p}\) is irrational for every prime number \(p\).

Common mistakes to avoid

• Forgetting to say the fraction is in lowest terms. This is essential for the contradiction.
• Skipping the assumption. In contradiction proofs, you must clearly state the opposite assumption first.
• Saying “if \(a^2\) is divisible by 3, then \(a\) might be divisible by 3.” Here it is stronger: \(a\) must be divisible by 3.
• Ending too early. You must explain why both \(a\) and \(b\) having the same factor causes a contradiction.

How to write a strong exam-style proof

A clear proof should include these parts:

1. Assume the number is rational.
2. Write it as \(\frac{a}{b}\), where \(a\) and \(b\) are integers with no common factor.
3. Square both sides.
4. Show the prime divides both \(a\) and \(b\).
5. State that this contradicts the fraction being in lowest terms.
6. Conclude that the number is irrational.

Quick check

If a student says, “\(\sqrt{2} = 1.4142135...\), so it is irrational because the decimal does not end,” that is not a complete proof.

It may suggest irrationality, but a formal proof needs logic that works with certainty. That is why contradiction is so important.

Brief Summary

To prove that \(\sqrt{2}\), \(\sqrt{3}\), or \(\sqrt{5}\) is irrational, we assume it is rational and write it as a fraction in lowest terms. After squaring, we show that the numerator and denominator must both be divisible by the same prime number. This is impossible if the fraction was already in lowest terms, so our assumption must be false. Therefore, the square root is irrational.

Decimal Expansions of Rational Numbers

When we write a fraction as a decimal, the decimal does not always end. Sometimes it stops after a few digits, and sometimes the digits continue forever in a repeating pattern. In this lesson, we will learn how to tell without dividing whether a rational number will have a terminating decimal or a non-terminating repeating decimal.

This idea is an important part of the number system. It connects fractions, prime factorization, and decimal representation in a very useful way.

Recall: A rational number is any number that can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\).

Examples of rational numbers are:

• \(\frac{1}{2}\)
• \(\frac{7}{8}\)
• \(\frac{3}{11}\)
• \(-\frac{5}{4}\)

When we convert these to decimals:

• \(\frac{1}{2} = 0.5\) which terminates
• \(\frac{7}{8} = 0.875\) which terminates
• \(\frac{3}{11} = 0.272727\ldots\) which repeats
• \(-\frac{5}{4} = -1.25\) which terminates

So rational numbers can have decimal expansions of two types:

• Terminating decimal: the decimal ends after a finite number of digits.
• Non-terminating recurring (repeating) decimal: the decimal goes on forever, but a block of digits repeats.

Main Question: How can we decide which type a rational number will have?

The answer depends on the prime factors of the denominator, after the fraction is written in its lowest form.

Important Rule: Let \(\frac{p}{q}\) be a rational number in lowest form, where \(q \ne 0\). Then:

• If the prime factorization of \(q\) has only 2s and/or 5s, then the decimal expansion terminates.
• If the prime factorization of \(q\) has any prime factor other than 2 or 5, then the decimal expansion is non-terminating repeating.

This rule works because our decimal system is based on 10, and

$$10 = 2 \times 5$$

Any terminating decimal can be written with denominator \(10\), \(100\), \(1000\), and so on. These denominators are powers of 10:

$$10^n = 2^n \times 5^n$$

So if a fraction can be changed into an equivalent fraction whose denominator is some power of 10, then its decimal will terminate. That is possible only when the denominator has no prime factors other than 2 and 5.

Step-by-step method

1. Write the fraction in lowest form.
2. Find the prime factorization of the denominator.
3. Check the prime factors:

• If they are only 2 and/or 5, the decimal terminates.
• If any other prime appears, the decimal repeats.

Let us now see this with worked examples.

Worked Example 1: Determine the decimal expansion of \(\frac{3}{8}\)

Step 1: Check if the fraction is in lowest form.

\(3\) and \(8\) have no common factor other than 1, so it is already in lowest form.

Step 2: Prime factorize the denominator.

$$8 = 2 \times 2 \times 2 = 2^3$$

Step 3: Look at the prime factors.

The denominator has only the prime factor 2.

Conclusion: The decimal expansion will terminate.

In fact,

$$\frac{3}{8} = 0.375$$

Worked Example 2: Determine the decimal expansion of \(\frac{7}{15}\)

Step 1: The fraction \(\frac{7}{15}\) is already in lowest form.

Step 2: Prime factorize the denominator.

$$15 = 3 \times 5$$

Step 3: Look at the prime factors.

The denominator contains \(3\), which is neither 2 nor 5.

Conclusion: The decimal expansion will be non-terminating repeating.

In fact,

$$\frac{7}{15} = 0.4666\ldots$$

The digit 6 repeats forever after the 4.

Worked Example 3: Determine the decimal expansion of \(\frac{18}{24}\)

This example is important because we must first reduce the fraction.

Step 1: Write in lowest form.

$$\frac{18}{24} = \frac{3}{4}$$

We divided numerator and denominator by 6.

Step 2: Prime factorize the denominator of the reduced fraction.

$$4 = 2^2$$

Step 3: Look at the prime factors.

The denominator has only 2 as a prime factor.

Conclusion: The decimal expansion terminates.

$$\frac{18}{24} = \frac{3}{4} = 0.75$$

Important Note: Always reduce the fraction first. If you do not, you may get confused by extra factors that cancel out.

Worked Example 4: Determine the decimal expansion of \(\frac{13}{60}\)

Step 1: Check lowest form.

\(13\) and \(60\) have no common factor other than 1, so the fraction is in lowest form.

Step 2: Prime factorize the denominator.

$$60 = 2^2 \times 3 \times 5$$

Step 3: Look at the prime factors.

The denominator contains \(3\), which is not 2 or 5.

Conclusion: The decimal expansion is non-terminating repeating.

Its decimal form will go on forever, but with a repeating pattern.

Why only 2 and 5?

A terminating decimal like \(0.125\) can be written as

$$0.125 = \frac{125}{1000}$$

Now,

$$1000 = 10^3 = 2^3 \times 5^3$$

So any terminating decimal becomes a fraction whose denominator is made only of 2s and 5s. That is why the denominator of a rational number in lowest form must have only 2 and/or 5 for the decimal to terminate.

Quick Check Table

• \(\frac{1}{4}\): \(4 = 2^2\) → terminating
• \(\frac{9}{20}\): \(20 = 2^2 \times 5\) → terminating
• \(\frac{2}{9}\): \(9 = 3^2\) → non-terminating repeating
• \(\frac{5}{12}\): \(12 = 2^2 \times 3\) → non-terminating repeating
• \(\frac{11}{25}\): \(25 = 5^2\) → terminating

Common mistakes to avoid

• Not reducing the fraction first: Always write the fraction in lowest form before checking the denominator.
• Looking at the numerator: The rule depends on the denominator, not the numerator.
• Thinking non-terminating means irrational: A repeating decimal is still rational. For example, \(\frac{1}{3} = 0.333\ldots\) is rational.
• Forgetting that 2 and 5 can both appear: A denominator like \(40 = 2^3 \times 5\) still gives a terminating decimal.

Try to think:

• \(\frac{5}{64}\): since \(64 = 2^6\), it terminates.
• \(\frac{4}{21}\): since \(21 = 3 \times 7\), it repeats.
• \(\frac{9}{50}\): since \(50 = 2 \times 5^2\), it terminates.

Summary

To decide whether the decimal expansion of a rational number terminates or repeats, first write the fraction in lowest form. Then prime factorize the denominator.

If the denominator has only 2 and/or 5 as prime factors, the decimal expansion is terminating. If the denominator has any prime factor other than 2 or 5, the decimal expansion is non-terminating repeating.

This rule helps you classify decimal expansions quickly and accurately without doing long division every time.

Operations on Surds and Radicals

In mathematics, a surd is a root that cannot be simplified into a whole number or a fraction. For example, \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{5}\) are surds because they are irrational numbers.

A radical expression is any expression that contains a root symbol, such as \(\sqrt{7}\), \(3\sqrt{2}\), or \(\sqrt{12}+\sqrt{3}\).

Learning how to work with surds is important because they appear often in algebra, geometry, and real numbers. In this lesson, you will learn how to simplify, add, subtract, and multiply surds correctly.

1. Simplifying surds

To simplify a surd, look for a perfect square factor inside the radical. Then use the rule:

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

This helps us take square factors out of the root.

For example, since \(12=4\times 3\), we can write:

$$\sqrt{12}=\sqrt{4\times 3}=\sqrt{4}\sqrt{3}=2\sqrt{3}$$

Here are some common perfect squares to remember:

• \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100\)

Some quick examples:

• \(\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}\)
• \(\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\)
• \(\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}\)

2. Like surds

Just like algebraic terms, surds can only be added or subtracted when they are like terms. This means they must have the same radical part after simplification.

For example:

• \(2\sqrt{3}\) and \(5\sqrt{3}\) are like surds
• \(\sqrt{2}\) and \(3\sqrt{5}\) are not like surds

3. Adding and subtracting surds

To add or subtract surds:

1. Simplify each surd if possible.
2. Combine only the like surds.

For example:

$$2\sqrt{5}+7\sqrt{5}=9\sqrt{5}$$

But:

$$\sqrt{2}+\sqrt{3}$$

cannot be combined because the surds are different.

4. Multiplying surds

When multiplying surds, multiply the numbers outside the roots and the numbers inside the roots separately.

Use the rule:

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

Example:

$$\sqrt{3}\times\sqrt{6}=\sqrt{18}=3\sqrt{2}$$

If there are coefficients, multiply them too:

$$2\sqrt{3}\times 4\sqrt{5}=8\sqrt{15}$$

5. Squaring a surd

When a surd is squared, the square root and the square cancel each other.

For example:

$$\left(\sqrt{7}\right)^2=7$$

And:

$$\left(3\sqrt{2}\right)^2=9\times 2=18$$

6. Important rules to remember

• \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\)
• \(\sqrt{ab}=\sqrt{a}\sqrt{b}\)
• Only like surds can be added or subtracted.
• Always simplify surds first before combining them.

Worked Example 1: Simplifying a surd

Simplify \(\sqrt{45}\).

Step 1: Find the largest perfect square factor of 45.

$$45=9\times 5$$

Step 2: Split the radical.

$$\sqrt{45}=\sqrt{9\times 5}=\sqrt{9}\sqrt{5}$$

Step 3: Simplify.

$$\sqrt{45}=3\sqrt{5}$$

Answer: \(3\sqrt{5}\)

Worked Example 2: Adding and subtracting surds

Simplify \(\sqrt{12}+\sqrt{27}-\sqrt{3}\).

Step 1: Simplify each surd.

$$\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$$

$$\sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}$$

So the expression becomes:

$$2\sqrt{3}+3\sqrt{3}-\sqrt{3}$$

Step 2: Combine like surds.

$$\left(2+3-1\right)\sqrt{3}=4\sqrt{3}$$

Answer: \(4\sqrt{3}\)

Worked Example 3: Multiplying surds

Simplify \((2\sqrt{6})(3\sqrt{15})\).

Step 1: Multiply the coefficients.

$$2\times 3=6$$

Step 2: Multiply the surds.

$$\sqrt{6}\times\sqrt{15}=\sqrt{90}$$

So:

$$6\sqrt{90}$$

Step 3: Simplify \(\sqrt{90}\).

$$\sqrt{90}=\sqrt{9\times 10}=3\sqrt{10}$$

Step 4: Multiply.

$$6\times 3\sqrt{10}=18\sqrt{10}$$

Answer: \(18\sqrt{10}\)

Worked Example 4: A mixed operation

Simplify \(3\sqrt{8}+2\sqrt{18}-\sqrt{50}\).

Step 1: Simplify each surd.

$$\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$$

$$\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}$$

$$\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$$

Step 2: Rewrite the expression.

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

$$6\sqrt{2}+6\sqrt{2}-5\sqrt{2}$$

Step 3: Combine like surds.

$$\left(6+6-5\right)\sqrt{2}=7\sqrt{2}$$

Answer: \(7\sqrt{2}\)

Common mistakes to avoid

• Do not add unlike surds. For example, \(\sqrt{2}+\sqrt{3}\neq\sqrt{5}\).
• Always simplify first. For example, \(\sqrt{8}+\sqrt{2}\) should become \(2\sqrt{2}+\sqrt{2}=3\sqrt{2}\).
• Do not forget to simplify after multiplying. For example, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\).

Quick check

• Simplify \(\sqrt{32}\)
• Simplify \(2\sqrt{7}+5\sqrt{7}\)
• Simplify \(\sqrt{20}-\sqrt{5}\)
• Simplify \((\sqrt{2})(\sqrt{18})\)

Answers

• \(\sqrt{32}=4\sqrt{2}\)
• \(2\sqrt{7}+5\sqrt{7}=7\sqrt{7}\)
• \(\sqrt{20}-\sqrt{5}=2\sqrt{5}-\sqrt{5}=\sqrt{5}\)
• \((\sqrt{2})(\sqrt{18})=\sqrt{36}=6\)

Summary

Surds are roots that cannot be written exactly as fractions or whole numbers. To work with them, first simplify by taking out perfect square factors. Then add or subtract only like surds, and multiply surds by multiplying inside the radicals and simplifying the result. With practice, these steps become straightforward and help you handle irrational numbers confidently.

Rationalizing Denominators means rewriting a fraction so that there is no irrational number in the denominator. In 10th Grade Maths, this usually means removing square roots from the bottom of a fraction.

This is useful because expressions are easier to compare, simplify, and use in later algebra when the denominator is rational. A rational denominator is a denominator that is a whole number, integer, or fraction without a surd.

For example, instead of writing \(\frac{3}{\sqrt{5}}\), we prefer to write an equivalent form with no square root in the denominator.

Main idea: multiply the fraction by a form of 1 that removes the surd from the denominator.

If the denominator is just one surd, such as \(\sqrt{a}\), multiply by the same surd:

This works because \(\sqrt{a} \cdot \sqrt{a} = a\), which is rational.

If the denominator is a binomial containing surds, such as \(a+\sqrt{b}\) or \(\sqrt{m}+\sqrt{n}\), multiply by its conjugate.

The conjugate of an expression is formed by changing the sign in the middle:

• The conjugate of \(a+b\) is \(a-b\)
• The conjugate of \(a-b\) is \(a+b\)
• The conjugate of \(\sqrt{3}+2\) is \(\sqrt{3}-2\)
• The conjugate of \(\sqrt{5}-\sqrt{2}\) is \(\sqrt{5}+\sqrt{2}\)

Why does this help? Because of the identity:

When we multiply conjugates, the middle terms cancel, and the denominator becomes rational.

There are two common cases:

1. Denominator has one surd — multiply by that surd.
2. Denominator has two terms with a surd — multiply by the conjugate.

Steps for rationalizing a denominator:

1. Look at the denominator carefully.
2. Decide whether to multiply by the same surd or by the conjugate.
3. Multiply both numerator and denominator by that expression.
4. Simplify the numerator and denominator.
5. Check that there is no surd left in the denominator.

Now let us look at worked examples from easy to harder.

Example 1: A denominator with one surd

Simplify \(\frac{7}{\sqrt{3}}\).

The denominator is just \(\sqrt{3}\), so multiply top and bottom by \(\sqrt{3}\).

Answer: \(\frac{7\sqrt{3}}{3}\)

The denominator is now rational, so the expression is rationalized.

Example 2: A denominator with a number times a surd

Simplify \(\frac{5}{2\sqrt{6}}\).

Multiply top and bottom by \(\sqrt{6}\).

Answer: \(\frac{5\sqrt{6}}{12}\)

Notice that we only needed to remove the square root part. The 2 stayed in the denominator.

Example 3: A binomial denominator using a conjugate

Simplify \(\frac{4}{\sqrt{5}+1}\).

The denominator has two terms, \(\sqrt{5}+1\), so we multiply by its conjugate, \(\sqrt{5}-1\).

Now simplify.

Use difference of squares in the denominator:

So,

Answer: \(\sqrt{5}-1\)

This example shows why conjugates are powerful: the denominator becomes a whole number very neatly.

Example 4: Both terms in the denominator are surds

Simplify \(\frac{3}{\sqrt{7}-\sqrt{2}}\).

The conjugate of \(\sqrt{7}-\sqrt{2}\) is \(\sqrt{7}+\sqrt{2}\).

Multiply:

Now simplify the denominator using difference of squares:

So the result is

Answer: \(\frac{3\sqrt{7}+3\sqrt{2}}{5}\)

Common mistakes to avoid:

• Multiplying only the denominator. You must multiply both numerator and denominator by the same expression.
• Using the wrong conjugate. If the denominator is \(\sqrt{3}+2\), the conjugate is \(\sqrt{3}-2\), not \(2-\sqrt{3}\) written carelessly without understanding the sign change.
• Forgetting the square root rule. Remember \(\sqrt{a}\cdot\sqrt{a}=a\).
• Expanding conjugates incorrectly. For example, \((a+b)(a-b)=a^2-b^2\), not \(a^2+b^2\).
• Stopping too early. After rationalizing, always simplify the final answer if possible.

Quick check:

• \(\frac{2}{\sqrt{11}} = \frac{2\sqrt{11}}{11}\)
• \(\frac{1}{3+\sqrt{2}}\) should be multiplied by \(\frac{3-\sqrt{2}}{3-\sqrt{2}}\)
• \((\sqrt{8}+1)(\sqrt{8}-1)=8-1=7\)

Summary

Rationalizing the denominator means rewriting a fraction so the denominator does not contain a surd. If the denominator has one surd, multiply by that surd. If the denominator has two terms, multiply by the conjugate so that the denominator becomes rational using \((a+b)(a-b)=a^2-b^2\).

With practice, you will quickly recognize which method to use. Always finish by simplifying your answer fully.