CBSE Class 10 Maths Standard · Chapter 1

NCERT Solutions: Real Numbers.

Exercise-wise worked methods for the Real Numbers chapter. Use these explanations beside the official NCERT textbook, then practise the same method yourself.

Open Official NCERT Textbook

Chapter coverage

Fundamental Theorem of Arithmetic
Irrational number proofs
Prime-factor reasoning

Exercise-wise solution guide

1NCERT Chapter 1Fundamental Theorem of Arithmetic

Prime factorisation of 156

Write the prime factorisation of $156$ and identify every prime factor used.

Answer

$156 = 2^{2} \times 3 \times 13$ .

Worked method

1. $156 = 2 \times 78 = 2 \times 2 \times 39$ .
2. $39 = 3 \times 13$ , and both $3$ and $13$ are prime.
3.Therefore $156 = 2^{2} \times 3 \times 13$ .

Key idea: Every composite number has a unique prime factorisation apart from the order of factors.

2NCERT Chapter 1Fundamental Theorem of Arithmetic

Showing a number is composite

Show that $7 \times 11 \times 13 + 13$ is composite.

Answer

$7 \times 11 \times 13 + 13 = 13 \times 78$ , so it is composite.

Worked method

1.Take the common factor $13$ outside.
2. $7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1)$ .
3. $13 (77 + 1) = 13 \times 78$ , a product of two integers greater than $1$ .

Key idea: To prove a number is composite, express it as a product of two whole numbers greater than $1$.

3NCERT Chapter 1Fundamental Theorem of Arithmetic

Prime factors in an LCM

Use prime factorisation to find the LCM of $140$ and $792$ .

Answer

$LCM = 27720$ .

Worked method

1. $140 = 2^{2} \times 5 \times 7$ and $792 = 2^{3} \times 3^{2} \times 11$ .
2.Take every prime with its greatest power.
3. $LCM = 2^{3} \times 3^{2} \times 5 \times 7 \times 11 = 27720$ .

Key idea: For an LCM, retain every prime factor at its greatest required power.

4NCERT Chapter 1Fundamental Theorem of Arithmetic

When can $6^n$ end in zero?

Can $6^{n}$ end with the digit $0$ for a natural number $n$ ? Explain using prime factors.

Answer

No, $6^{n}$ cannot end in $0$ .

Worked method

1.A number ending in $0$ must have $10 = 2 \times 5$ as a factor.
2. $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$ .
3.There is no factor $5$ , so $6^{n}$ cannot be divisible by $10$ .

Key idea: A final zero needs both factors $2$ and $5$.

5NCERT Chapter 1Irrational numbers

Why $\sqrt{5}$ is irrational

Prove that $5$ is irrational.

Answer

$5$ is irrational.

Worked method

1.Assume $5 = \frac{p}{q}$ in lowest form.
2.Then $p^{2} = 5 q^{2}$ , so $p$ is divisible by $5$ .
3.Writing $p = 5 k$ shows $q$ is also divisible by $5$ .
4.This contradicts $\frac{p}{q}$ being in lowest form.

Key idea: Assume a lowest-form fraction and prove numerator and denominator share a factor.

6NCERT Chapter 1Irrational numbers

Why $3+\sqrt{5}$ is irrational

Show that $3 + 5$ is irrational.

Answer

$3 + 5$ is irrational.

Worked method

1.Assume $3 + 5$ is rational.
2.Subtract rational number $3$ .
3.Then $5$ would be rational, a contradiction.

Key idea: A rational number added to a non-zero irrational number stays irrational.

7NCERT Chapter 1Irrational numbers

Why $5-\sqrt{3}$ is irrational

Show that $5 - 3$ is irrational.

Answer

$5 - 3$ is irrational.

Worked method

1.Assume $5 - 3$ is rational.
2.Then $3 = 5 - (5 - 3)$ would be rational.
3.This contradicts that $3$ is irrational.

Key idea: A rational number minus an irrational number cannot become rational.

8NCERT Chapter 1Irrational numbers

Why $\sqrt{2}+\sqrt{3}$ is irrational

Show that $2 + 3$ is irrational.

Answer

$2 + 3$ is irrational.

Worked method

1.Assume $2 + 3 = r$ for rational $r$ .
2.Squaring gives $5 + 26 = r^{2}$ .
3.Thus $6 = \frac{r ^{2} - 5}{2}$ would be rational.
4.But $6$ is irrational, so the assumption is false.

Key idea: For sums of square roots, square once and isolate another irrational square root.

Real Numbers NCERT Solutions FAQ

Are these NCERT Real Numbers solutions for CBSE Class 10 Maths Standard?

Yes. The methods cover the current Chapter 1 focus: prime factorisation, the Fundamental Theorem of Arithmetic and irrational numbers.

Should I copy the solution directly?

First attempt the NCERT exercise in your textbook. Then use this page to compare the method, correct one step at a time and rewrite the final answer yourself.

Why is the official NCERT textbook linked here?

The official textbook is the source for the exact exercise wording. This page provides original explanation and worked method guidance, so students can study alongside their own NCERT copy.

Real Numbers Practice Real Numbers Important Questions

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Showing a number is composite

Show that

7 \times 11 \times 13 + 13

is composite.

Answer

$7 \times 11 \times 13 + 13 = 13 \times 78$ , so it is composite.

Worked method

1.Take the common factor $13$ outside.
2. $7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1)$ .
3. $13 (77 + 1) = 13 \times 78$ , a product of two integers greater than $1$ .

Key idea: To prove a number is composite, express it as a product of two whole numbers greater than $1$.

When can $6^n$ end in zero?

Can

6^{n}

end with the digit

0

for a natural number

n

? Explain using prime factors.

Answer

No, $6^{n}$ cannot end in $0$ .

Worked method

1.A number ending in $0$ must have $10 = 2 \times 5$ as a factor.
2. $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$ .
3.There is no factor $5$ , so $6^{n}$ cannot be divisible by $10$ .

Key idea: A final zero needs both factors $2$ and $5$.

Why $\sqrt{5}$ is irrational

Prove that

5

is irrational.

Answer

$5$ is irrational.

Worked method

1.Assume $5 = \frac{p}{q}$ in lowest form.
2.Then $p^{2} = 5 q^{2}$ , so $p$ is divisible by $5$ .
3.Writing $p = 5 k$ shows $q$ is also divisible by $5$ .
4.This contradicts $\frac{p}{q}$ being in lowest form.

Key idea: Assume a lowest-form fraction and prove numerator and denominator share a factor.

Why $\sqrt{2}+\sqrt{3}$ is irrational

Show that

2 + 3

is irrational.

Answer

$2 + 3$ is irrational.

Worked method

1.Assume $2 + 3 = r$ for rational $r$ .
2.Squaring gives $5 + 26 = r^{2}$ .
3.Thus $6 = \frac{r ^{2} - 5}{2}$ would be rational.
4.But $6$ is irrational, so the assumption is false.

Key idea: For sums of square roots, square once and isolate another irrational square root.

Real Numbers NCERT Solutions FAQ

Are these NCERT Real Numbers solutions for CBSE Class 10 Maths Standard?

Yes. The methods cover the current Chapter 1 focus: prime factorisation, the Fundamental Theorem of Arithmetic and irrational numbers.

Should I copy the solution directly?

First attempt the NCERT exercise in your textbook. Then use this page to compare the method, correct one step at a time and rewrite the final answer yourself.

Why is the official NCERT textbook linked here?

The official textbook is the source for the exact exercise wording. This page provides original explanation and worked method guidance, so students can study alongside their own NCERT copy.