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Irrational Numbers & Decimal Expansions

🎓 Class 10 Mathematics CBSE Theory Ch 1 — Real Numbers ⏱ ~35 min
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This MCQ module is based on: Irrational Numbers & Decimal Expansions

This mathematics assessment will be based on: Irrational Numbers & Decimal Expansions
Targeting Class 10 level in Number Theory, with Intermediate difficulty.

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1.3 Revisiting Irrational Numbers

In Class IX, you were introduced to irrational numbers? and many of their properties. You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) and, in general, \(\sqrt{p}\) is irrational, where \(p\) is a prime. One of the theorems we use in our proof is the Fundamental Theorem of Arithmetic.

Recall
A number \(s\) is called irrational if it cannot be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Some examples of irrational numbers with which you are already familiar are:

\(\sqrt{2},\; \sqrt{3},\; \sqrt{15},\; \pi,\; -\dfrac{\sqrt{2}}{\sqrt{3}},\; 0.10110111011110\ldots\), etc.

Before we prove that \(\sqrt{2}\) is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Theorem 1.2
Let \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.
Proof of Theorem 1.2
1
Let the prime factorisation of \(a\) be as follows:
\(a = p_1 \cdot p_2 \cdot \ldots \cdot p_n\), where \(p_1, p_2, \ldots, p_n\) are primes, not necessarily distinct.
2
Therefore, \(a^2 = (p_1 \cdot p_2 \cdot \ldots \cdot p_n)(p_1 \cdot p_2 \cdot \ldots \cdot p_n) = p_1^2 \cdot p_2^2 \cdot \ldots \cdot p_n^2\).
3
Now, we are given that \(p\) divides \(a^2\). Therefore, from the Fundamental Theorem of Arithmetic, it follows that \(p\) is one of the prime factors of \(a^2\).
4
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of \(a^2\) are \(p_1, p_2, \ldots, p_n\). So \(p\) is one of \(p_1, p_2, \ldots, p_n\).
5
Now, since \(a = p_1 \cdot p_2 \cdot \ldots \cdot p_n\), and \(p\) is one of \(p_1, p_2, \ldots, p_n\), it follows that \(p\) divides \(a\). ■

Proof that \(\sqrt{2}\) is Irrational

Theorem 1.3
\(\sqrt{2}\) is irrational.

The proof is based on a technique called proof by contradiction?. (This technique is discussed in some detail in Appendix 1.)

Proof of Theorem 1.3: \(\sqrt{2}\) is irrational
1
Assume the opposite. Let us assume, to the contrary, that \(\sqrt{2}\) is rational.
2
Write as a fraction. So, we can find integers \(r\) and \(s\) (\(s \neq 0\)) such that \(\sqrt{2} = \frac{r}{s}\). Suppose \(r\) and \(s\) have a common factor other than 1. Then, we divide by the common factor to get \(\sqrt{2} = \frac{a}{b}\), where \(a\) and \(b\) are coprime (no common factors other than 1).
3
Rearrange. So, \(b\sqrt{2} = a\).
4
Square both sides. Squaring on both sides and rearranging, we get \(2b^2 = a^2\). Therefore, 2 divides \(a^2\).
5
Apply Theorem 1.2. Now, by Theorem 1.2, it follows that 2 divides \(a\).
6
Write \(a = 2c\). So, we can write \(a = 2c\) for some integer \(c\).
7
Substitute back. Substituting for \(a\), we get \(2b^2 = 4c^2\), that is, \(b^2 = 2c^2\).
8
Deduce 2 divides \(b\). This means that 2 divides \(b^2\), and so 2 divides \(b\) (again using Theorem 1.2 with \(p = 2\)).
9
Contradiction! Therefore, \(a\) and \(b\) have at least 2 as a common factor. But this contradicts the fact that \(a\) and \(b\) are coprime (they have no common factors other than 1).
10
Conclusion. This contradiction has arisen because of our incorrect assumption that \(\sqrt{2}\) is rational. So, we conclude that \(\sqrt{2}\) is irrational. ■

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

Prove that \(\sqrt{3}\) is irrational.
Step 1: Let us assume, to the contrary, that \(\sqrt{3}\) is rational.

Step 2: That is, we can find integers \(a\) and \(b\) (\(b \neq 0\)) such that \(\sqrt{3} = \frac{a}{b}\). Suppose \(a\) and \(b\) have a common factor other than 1, then we can divide by the common factor, and assume that \(a\) and \(b\) are coprime.

Step 3: So, \(b\sqrt{3} = a\).

Step 4: Squaring on both sides, we get \(3b^2 = a^2\).

Step 5: Therefore, \(a^2\) is divisible by 3, and by Theorem 1.2, it follows that \(a\) is also divisible by 3.

Step 6: So, we can write \(a = 3c\) for some integer \(c\).

Step 7: Substituting for \(a\), we get \(3b^2 = 9c^2\), that is, \(b^2 = 3c^2\).

Step 8: This means that \(b^2\) is divisible by 3, and so \(b\) is also divisible by 3 (using Theorem 1.2 with \(p = 3\)).

Step 9: Therefore, \(a\) and \(b\) have at least 3 as a common factor. But this contradicts the fact that \(a\) and \(b\) are coprime.

Conclusion: This contradiction has arisen because of our incorrect assumption that \(\sqrt{3}\) is rational. So, we conclude that \(\sqrt{3}\) is irrational. ■

Example 6: Show that \(5 - \sqrt{3}\) is irrational.

Prove that \(5 - \sqrt{3}\) is irrational.
Step 1: Let us assume, to the contrary, that \(5 - \sqrt{3}\) is rational.

Step 2: That is, we can find coprime \(a\) and \(b\) (\(b \neq 0\)) such that \(5 - \sqrt{3} = \frac{a}{b}\).

Step 3: Therefore, \(5 - \frac{a}{b} = \sqrt{3}\).

Step 4: Rearranging, we get \(\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}\).

Step 5: Since \(a\) and \(b\) are integers, \(\frac{5b - a}{b}\) is rational, and so \(\sqrt{3}\) is rational.

Step 6: But this contradicts the fact that \(\sqrt{3}\) is irrational (proved in Example 5 above).

Conclusion: This contradiction has arisen because of our incorrect assumption that \(5 - \sqrt{3}\) is rational. So, we conclude that \(5 - \sqrt{3}\) is irrational. ■
General Principle
In Class IX, we mentioned that:
  • the sum or difference of a rational and an irrational number is irrational, and
  • the product and quotient of a non-zero rational and an irrational number is irrational.

Example 7: Show that \(3\sqrt{2}\) is irrational.

Prove that \(3\sqrt{2}\) is irrational.
Step 1: Let us assume, to the contrary, that \(3\sqrt{2}\) is rational.

Step 2: That is, we can find coprime \(a\) and \(b\) (\(b \neq 0\)) such that \(3\sqrt{2} = \frac{a}{b}\).

Step 3: Rearranging, we get \(\sqrt{2} = \frac{a}{3b}\).

Step 4: Since 3, \(a\) and \(b\) are integers, \(\frac{a}{3b}\) is rational, and so \(\sqrt{2}\) is rational.

Step 5: But this contradicts the fact that \(\sqrt{2}\) is irrational (Theorem 1.3).

Conclusion: This contradiction has arisen because of our incorrect assumption that \(3\sqrt{2}\) is rational. So, we conclude that \(3\sqrt{2}\) is irrational. ■
Think about it: Is \(\sqrt{2} + \sqrt{3}\) rational or irrational? It is irrational. If it were rational, say \(r\), then \(\sqrt{3} = r - \sqrt{2}\). Squaring: \(3 = r^2 - 2r\sqrt{2} + 2\), so \(\sqrt{2} = \frac{r^2 - 1}{2r}\), which is rational -- a contradiction.
Interactive: Number Line Explorer for Surds
See where \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) and other square roots lie on the number line between integers
Click a button above to see the position on the number line.
Activity: Proof by Contradiction Practice
L4 Analyse
Challenge: Can you use the same proof structure to show that \(\sqrt{7}\) is irrational?
  1. Write down the assumption: Assume \(\sqrt{7}\) is rational, so \(\sqrt{7} = \frac{a}{b}\) where \(a, b\) are coprime integers, \(b \neq 0\).
  2. Square both sides to get \(7b^2 = a^2\).
  3. Conclude that 7 divides \(a^2\), and by Theorem 1.2, 7 divides \(a\). Write \(a = 7c\).
  4. Substitute back to get \(b^2 = 7c^2\), showing 7 divides \(b\) as well.
  5. Identify the contradiction with the coprime assumption.

Solution: Following the five steps above, we find that both \(a\) and \(b\) are divisible by 7, contradicting the fact they are coprime. Therefore our assumption is wrong, and \(\sqrt{7}\) is irrational. The key is that the proof works for any prime \(p\) -- Theorem 1.2 ensures that if \(p \mid a^2\) then \(p \mid a\).

Competency-Based Questions

Scenario: A student named Riya is working on a geometry project. She measures the diagonal of a unit square and gets \(\sqrt{2}\) cm. Her friend says, "That is just 1.414, a terminating decimal -- so it is rational." Riya disagrees.
Q1. Is Riya's friend correct? Identify the flaw in the friend's reasoning.
L4 Analyse
Answer: Riya's friend is incorrect. The value 1.414 is only an approximation of \(\sqrt{2}\). The actual decimal expansion of \(\sqrt{2} = 1.41421356\ldots\) is non-terminating and non-repeating. A number is rational only if its decimal expansion terminates or repeats. Since \(\sqrt{2}\) does neither, it is irrational.
Q2. Which of the following is irrational?
L2 Understand
  • (a) \(\sqrt{4}\)
  • (b) \(\sqrt{9}\)
  • (c) \(\sqrt{8}\)
  • (d) \(\sqrt{25}\)
Answer: (c) \(\sqrt{8}\). \(\sqrt{4} = 2\), \(\sqrt{9} = 3\), \(\sqrt{25} = 5\) are all rational. But \(\sqrt{8} = 2\sqrt{2}\), and since \(\sqrt{2}\) is irrational, \(2\sqrt{2}\) is also irrational (product of non-zero rational with irrational is irrational).
Q3. If the side of a square is rational, can its diagonal ever be rational? Justify.
L5 Evaluate
Answer: If the side is \(s\) (rational), the diagonal is \(s\sqrt{2}\). Since \(\sqrt{2}\) is irrational and \(s \neq 0\) is rational, their product \(s\sqrt{2}\) is irrational. So the diagonal of a square with rational sides is always irrational. The only exception would be \(s = 0\), but a square with zero side length is degenerate.
Q4. Create a proof (analogous to the proof of Theorem 1.3) to show that \(\sqrt{11}\) is irrational. Outline the key steps.
L6 Create
Answer: (1) Assume \(\sqrt{11}\) is rational, i.e., \(\sqrt{11} = \frac{a}{b}\) with \(a, b\) coprime. (2) Square: \(11b^2 = a^2\). (3) So 11 divides \(a^2\). Since 11 is prime, by Theorem 1.2, 11 divides \(a\). Write \(a = 11c\). (4) Substitute: \(11b^2 = 121c^2 \Rightarrow b^2 = 11c^2\). So 11 divides \(b\). (5) Both \(a\) and \(b\) divisible by 11 contradicts coprimality. Hence \(\sqrt{11}\) is irrational. ■

Assertion--Reason Questions

Assertion (A): \(\sqrt{2}\) is irrational.
Reason (R): If \(p\) is a prime and \(p\) divides \(a^2\), then \(p\) divides \(a\).
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) -- Both are true. The proof that \(\sqrt{2}\) is irrational relies directly on Theorem 1.2 (the Reason). R is used in Steps 5 and 8 of the proof. R explains A.
Assertion (A): \(5 - \sqrt{3}\) is irrational.
Reason (R): The difference of a rational number and an irrational number is always irrational.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) -- \(5\) is rational and \(\sqrt{3}\) is irrational. Their difference \(5 - \sqrt{3}\) is irrational. R correctly states the general property, and it directly explains why A is true.
Assertion (A): \(\sqrt{4} + \sqrt{9}\) is irrational.
Reason (R): The sum of two square roots is always irrational.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R false.
Answer: (d) -- A is false: \(\sqrt{4} + \sqrt{9} = 2 + 3 = 5\), which is rational. R is also false: the sum of two square roots is not always irrational (as this counter-example shows). Both A and R are false.

Frequently Asked Questions — Real Numbers

What is Irrational Numbers & Decimal Expansions in NCERT Class 10 Mathematics?

Irrational Numbers & Decimal Expansions is a key concept covered in NCERT Class 10 Mathematics, Chapter 1: Real Numbers. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Irrational Numbers & Decimal Expansions step by step?

To solve problems on Irrational Numbers & Decimal Expansions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 1: Real Numbers?

The essential formulas of Chapter 1 (Real Numbers) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Irrational Numbers & Decimal Expansions important for the Class 10 board exam?

Irrational Numbers & Decimal Expansions is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Irrational Numbers & Decimal Expansions?

Common mistakes in Irrational Numbers & Decimal Expansions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Irrational Numbers & Decimal Expansions?

End-of-chapter NCERT exercises for Irrational Numbers & Decimal Expansions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.

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