NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers — Complete Guide with Step-by-Step Solutions

Statement: For any two positive integers $a$ and $b$, there exist unique integers $q$ (quotient) and $r$ (remainder) such that:

This is simply the mathematical statement of what happens when you divide one number by another. For example, when you divide $17$ by $5$: $17 = 5 \times 3 + 2$. Here $a = 17$, $b = 5$, $q = 3$, $r = 2$, and indeed $0 \le 2 < 5$.

The key insight is the uniqueness of $q$ and $r$. Given any pair $(a, b)$, there is exactly one pair $(q, r)$ satisfying the conditions. This uniqueness is what makes Euclid's Division Algorithm work reliably every time.

The algorithm uses the lemma repeatedly to find the HCF of two numbers:

Step 1: Apply the division lemma to $a$ and $b$ (where $a > b$): $a = bq_1 + r_1$.
Step 2: If $r_1 = 0$, then HCF $= b$. Stop.
Step 3: If $r_1 \neq 0$, apply the lemma to $b$ and $r_1$: $b = r_1 q_2 + r_2$.
Step 4: Continue this process until the remainder becomes $0$. The last non-zero remainder (i.e., the last divisor) is the HCF.

Why does this work? At each step, $\text{HCF}(a, b) = \text{HCF}(b, r_1) = \text{HCF}(r_1, r_2) = \ldots$ The remainders keep decreasing ($r_1 > r_2 > r_3 > \ldots \ge 0$), so the process must terminate. When the remainder hits $0$, the previous remainder is the HCF.

The beauty of this algorithm is that it works for any pair of positive integers, no matter how large, without needing to find prime factorisations.

Statement: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

For example:
- $12 = 2^2 \times 3$
- $180 = 2^2 \times 3^2 \times 5$
- $17 = 17$ (prime numbers are their own factorisation)

This theorem is the foundation for finding HCF and LCM using prime factorisation:
- HCF = product of the smallest power of each common prime factor
- LCM = product of the greatest power of each prime factor that appears in either number

Important relationship:

This identity is extremely useful for verification and for finding one quantity when the other three are known.

A number that cannot be written in the form $\frac{p}{q}$ (where $p$ and $q$ are integers with $q \neq 0$) is called irrational. Examples include $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\pi$, and $e$.

The standard proof technique for showing a number is irrational is proof by contradiction (also called reductio ad absurdum):
1. Assume the number is rational (i.e., it can be written as $\frac{p}{q}$ in lowest terms).
2. Derive a logical consequence that contradicts the assumption.
3. Conclude that the original assumption was wrong, so the number must be irrational.

A critical lemma used in these proofs: **If a prime $p$ divides $a^2$, then $p$ divides $a$.** This follows from the Fundamental Theorem of Arithmetic.

The decimal expansion of a rational number $\frac{p}{q}$ (in lowest terms) is:
- Terminating if and only if $q$ is of the form $2^m \times 5^n$ (where $m, n$ are non-negative integers).
- Non-terminating repeating if $q$ has any prime factor other than $2$ or $5$.

To convert a terminating decimal to a fraction, multiply numerator and denominator to make the denominator a power of $10$. For example:

This concept connects to the Fundamental Theorem because the primality of the denominator's factors determines the decimal behaviour.

Solution:
We need HCF(135, 225). Since $225 > 135$, we start with $a = 225$ and $b = 135$.

Step 1: Apply the division lemma:

Step 2: Now apply to $135$ and $90$:

Step 3: Apply to $90$ and $45$:

The last non-zero remainder (equivalently, the last divisor) is $45$.

Answer: $\text{HCF}(135, 225) = 45$

Verification: $135 = 45 \times 3$ and $225 = 45 \times 5$. Since $\gcd(3, 5) = 1$, confirmed that $45$ is indeed the HCF. $\checkmark$

Solution:
Since $38220 > 196$, we start with $a = 38220$ and $b = 196$.

Step 1: Apply the division lemma:

The remainder is $0$ in the very first step! This means $196$ divides $38220$ exactly.

Answer: $\text{HCF}(196, 38220) = 196$

Verification: $38220 \div 196 = 195$ exactly, so $196$ is indeed a factor of $38220$. And since $196$ divides itself, the HCF is $196$. $\checkmark$

Key takeaway: When the first division gives remainder $0$, the smaller number is itself the HCF. This happens whenever one number is a multiple of the other.

Solution:
Since $867 > 255$, start with $a = 867$ and $b = 255$.

Step 1:

Step 2:

Step 3:

Answer: $\text{HCF}(867, 255) = 51$

Verification: $867 = 51 \times 17$ and $255 = 51 \times 5$. Since $\gcd(17, 5) = 1$, confirmed. $\checkmark$

Solution (Board Exam Style):
Since $12576 > 4052$:

Step 1: $12576 = 4052 \times 3 + 420$
Step 2: $4052 = 420 \times 9 + 272$
Step 3: $420 = 272 \times 1 + 148$
Step 4: $272 = 148 \times 1 + 124$
Step 5: $148 = 124 \times 1 + 24$
Step 6: $124 = 24 \times 5 + 4$
Step 7: $24 = 4 \times 6 + 0$

Answer: $\text{HCF}(4052, 12576) = 4$

Note: This problem requires more steps, which is typical for larger numbers. The algorithm always terminates because the remainders form a strictly decreasing sequence: $420 > 272 > 148 > 124 > 24 > 4 > 0$.

Solution:
Let $a$ be any positive integer. By Euclid's Division Lemma with divisor $b = 2$:

So $r = 0$ or $r = 1$.
- If $r = 0$: $a = 2q$ — this is even.
- If $r = 1$: $a = 2q + 1$ — this is odd.

Therefore, every positive even integer is of the form $2q$, and every positive odd integer is of the form $2q + 1$, for some non-negative integer $q$. $\square$

Solution:
Let $a$ be any positive integer. Applying Euclid's Division Lemma with $b = 4$:

So $r \in \{0, 1, 2, 3\}$, giving us four forms:
- $a = 4q$ (even)
- $a = 4q + 1$ (odd)
- $a = 4q + 2 = 2(2q + 1)$ (even)
- $a = 4q + 3$ (odd)

The odd cases are $4q + 1$ and $4q + 3$. Therefore, any positive odd integer is of the form $4q + 1$ or $4q + 3$. $\square$

Board exam tip: This type of problem can appear as a 2-mark or 3-mark question. The key is to apply Euclid's Lemma with the appropriate divisor and then identify which remainders give odd values.

Solution:
Let $a$ be any positive integer. By Euclid's Division Lemma with $b = 3$:

Case 1: $a = 3q \implies a^2 = 9q^2 = 3(3q^2) = 3m$ where $m = 3q^2$.

Case 2: $a = 3q + 1 \implies a^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$ where $m = 3q^2 + 2q$.

Case 3: $a = 3q + 2 \implies a^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$ where $m = 3q^2 + 4q + 1$.

In all cases, $a^2$ is either of the form $3m$ or $3m + 1$. $\square$

Observation: The square of any integer is never of the form $3m + 2$. This is a useful result for irrationality proofs involving $\sqrt{3}$.

Solution:

(i) $140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$

(ii) $156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$

(iii) $3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 5 \times 85 = 3^2 \times 5 \times 5 \times 17$

(iv) $5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$

(v) $7429 = 17 \times 437 = 17 \times 19 \times 23$

Tip for board exams: Use the factor tree method or repeated division by primes. Start with the smallest prime that divides the number.

Solution:
Prime factorisations:

HCF = product of common prime factors with lowest powers $= 13$

LCM = product of all prime factors with highest powers $= 2 \times 7 \times 13 = 182$

Verification: $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$
$a \times b = 26 \times 91 = 2366$ $\checkmark$

Solution:
Prime factorisations:

HCF $= 2$ (the only common prime factor)

LCM $= 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$

Verification: $2 \times 23460 = 46920 = 510 \times 92$ $\checkmark$

Solution:
Prime factorisations:

HCF $= 2^1 \times 3^1 = 6$ (smallest powers of common primes $2$ and $3$)

LCM $= 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$ (highest powers of all primes)

Verification: $6 \times 3024 = 18144 = 336 \times 54$ $\checkmark$

Solution:
Prime factorisations:

HCF $= 3$ (the only prime common to all three, with minimum power $1$)

LCM $= 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

Note: For three numbers, the relationship $\text{HCF} \times \text{LCM} = a \times b \times c$ does not hold in general. That identity only works for two numbers. You can verify: $3 \times 420 = 1260$, but $12 \times 15 \times 21 = 3780$.

Solution:
Using the fundamental relationship for two numbers:

Answer: $\text{LCM}(306, 657) = 22338$

Solution:
For a number to end with digit $0$, it must be divisible by $10 = 2 \times 5$. In other words, both $2$ and $5$ must appear in its prime factorisation.

Now, $6^n = (2 \times 3)^n = 2^n \times 3^n$.

The prime factorisation of $6^n$ contains only the primes $2$ and $3$. The prime $5$ never appears.

By the Fundamental Theorem of Arithmetic, since the prime factorisation is unique, $5$ cannot be a factor of $6^n$.

Therefore, $6^n$ cannot end with the digit $0$ for any natural number $n$. $\square$

Solution:

(i) $7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 = 13 \times 2 \times 3 \times 13 = 2 \times 3 \times 13^2$

Since this number has factors other than $1$ and itself, it is composite.

(ii) $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5040 + 5 = 5(1008 + 1) = 5 \times 1009$

Since $5045 = 5 \times 1009$ has factors other than $1$ and itself, it is composite.

Key insight: Taking out the common factor is the trick. In (i), $13$ is common; in (ii), $5$ is common.

Solution:
We need to find the maximum length of planks that can be laid along both the $60$ m side and the $50$ m side without cutting. This means the plank length must divide both $60$ and $50$ — i.e., we need $\text{HCF}(60, 50)$.

Answer: The maximum length of each plank is $10$ m.

Number of planks needed: along the $60$ m side $= 6$, along the $50$ m side $= 5$, total $= 2(6 + 5) = 22$ planks (for the full perimeter).

Solution:
Assume, for the sake of contradiction, that $\sqrt{5}$ is rational.

Then we can write $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are co-prime integers (i.e., $\gcd(p, q) = 1$) and $q \neq 0$.

Squaring both sides:

This tells us $5$ divides $p^2$. Since $5$ is prime, by our lemma, $5$ divides $p$.

Let $p = 5k$ for some integer $k$. Substituting in (1):

From (2), $5$ divides $q^2$, so $5$ divides $q$.

But now **both $p$ and $q$ are divisible by $5$**, which contradicts our assumption that $p$ and $q$ are co-prime.

Therefore, our assumption is wrong, and $\sqrt{5}$ is irrational. $\square$

Template observation: This exact proof works for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{7}$, or any $\sqrt{p}$ where $p$ is prime — just replace $5$ with the relevant prime.

Solution:
Assume, for contradiction, that $3 + 2\sqrt{5}$ is rational.

Then $3 + 2\sqrt{5} = \frac{a}{b}$ for some co-prime integers $a$ and $b$ with $b \neq 0$.

Rearranging:

Now, since $a$ and $b$ are integers, the right side $\frac{a - 3b}{2b}$ is a rational number.

But we have already proved (in Problem 1) that $\sqrt{5}$ is irrational.

A rational number cannot equal an irrational number. This is a contradiction.

Therefore, $3 + 2\sqrt{5}$ is irrational. $\square$

Key technique: This is the "isolation" method — isolate the irrational part ($\sqrt{5}$) on one side and show the other side is rational, giving a contradiction.

Solution:
Assume, for contradiction, that $\frac{1}{\sqrt{2}}$ is rational.

Then $\frac{1}{\sqrt{2}} = \frac{a}{b}$ for some co-prime integers $a, b$ with $b \neq 0$.

Rearranging:

Since $a$ and $b$ are integers with $a \neq 0$, the right side is rational.

But $\sqrt{2}$ is irrational (can be proved exactly like $\sqrt{5}$ in Problem 1, replacing $5$ with $2$).

This contradiction means $\frac{1}{\sqrt{2}}$ is irrational. $\square$

Solution (Full proof — this is the most commonly asked version in board exams):

Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$, where $\gcd(p, q) = 1$.

Squaring: $p^2 = 2q^2 \quad \cdots (1)$

So $2$ divides $p^2 \implies 2$ divides $p$. Let $p = 2m$.

Substituting in (1): $4m^2 = 2q^2 \implies q^2 = 2m^2$

So $2$ divides $q^2 \implies 2$ divides $q$.

But both $p$ and $q$ are even, contradicting $\gcd(p, q) = 1$.

Therefore, $\sqrt{2}$ is irrational. $\square$

Solution:
Assume $\sqrt{3} = \frac{p}{q}$ with $\gcd(p, q) = 1$.

$p^2 = 3q^2 \implies 3 | p^2 \implies 3 | p$. Let $p = 3m$.

$9m^2 = 3q^2 \implies q^2 = 3m^2 \implies 3 | q^2 \implies 3 | q$.

Both divisible by $3$, contradicting co-primality. Hence $\sqrt{3}$ is irrational. $\square$

Solution:
Assume $5 + 3\sqrt{2} = \frac{a}{b}$ with $\gcd(a, b) = 1$.

The right side is rational, but $\sqrt{2}$ is irrational. Contradiction.

Therefore, $5 + 3\sqrt{2}$ is irrational. $\square$

The Rule: $\frac{p}{q}$ (in lowest terms) has a terminating decimal $\iff$ $q = 2^m \times 5^n$ (for non-negative integers $m, n$).

(i) $\frac{13}{3125}$: Here $3125 = 5^5 = 2^0 \times 5^5$. Since the denominator is of the form $2^m \times 5^n$, the decimal is terminating.

(ii) $\frac{17}{8}$: Here $8 = 2^3 = 2^3 \times 5^0$. Terminating.

(iii) $\frac{64}{455}$: First reduce: $\gcd(64, 455) = 1$ (since $64 = 2^6$ and $455 = 5 \times 7 \times 13$). The denominator has prime factors $5, 7, 13$. Since $7$ and $13$ are present (not just $2$s and $5$s), the decimal is non-terminating repeating.

(iv) $\frac{15}{1600}$: Simplify: $\frac{15}{1600} = \frac{3}{320}$. Now $320 = 2^6 \times 5$. Since $320 = 2^6 \times 5^1$, the decimal is terminating.

(v) $\frac{29}{343}$: Here $343 = 7^3$. The denominator contains a prime factor ($7$) other than $2$ or $5$. So the decimal is non-terminating repeating.

(vi) $\frac{23}{2^3 \times 5^2}$: The denominator is already in the form $2^m \times 5^n$ with $m = 3$, $n = 2$. Terminating.

(vii) $\frac{129}{2^2 \times 5^7 \times 7^5}$: The denominator has factor $7^5$. Non-terminating repeating.

(viii) $\frac{6}{15}$: Simplify first! $\frac{6}{15} = \frac{2}{5}$. Denominator $= 5 = 2^0 \times 5^1$. Terminating.

Critical reminder: Always simplify the fraction to lowest terms first before checking the denominator. If you skip this step, you might incorrectly classify fractions like $\frac{6}{15}$ (which simplifies to $\frac{2}{5}$, terminating) or $\frac{14}{21}$ (which simplifies to $\frac{2}{3}$, non-terminating).

Solution:
For each terminating fraction, we convert the denominator to a power of $10$ by multiplying both numerator and denominator by the appropriate factor of $2$ or $5$:

(i) $\frac{13}{3125} = \frac{13 \times 32}{100000} = \frac{416}{100000} = 0.00416$

(ii) $\frac{17}{8} = \frac{17 \times 125}{1000} = \frac{2125}{1000} = 2.125$

(iv) $\frac{15}{1600} = \frac{3}{320} = \frac{3 \times 3125}{320 \times 3125} = \frac{9375}{1000000} = 0.009375$

(vi) $\frac{23}{200} = \frac{23 \times 5}{1000} = \frac{115}{1000} = 0.115$

(viii) $\frac{6}{15} = \frac{2}{5} = \frac{4}{10} = 0.4$

Method: To make the denominator a power of $10 = 2 \times 5$, check which prime is "missing" or has a smaller exponent. If denominator is $2^a \times 5^b$, multiply by $2^{\max(a,b)-a} \times 5^{\max(a,b)-b}$ to get $10^{\max(a,b)}$.

Solution:

(i) $0.\overline{6} = 0.666\ldots$

Let $x = 0.666\ldots$
$10x = 6.666\ldots$
$10x - x = 6$
$9x = 6$
$x = \frac{6}{9} = \frac{2}{3}$

(ii) $0.4\overline{7} = 0.4777\ldots$

Let $x = 0.4777\ldots$
$10x = 4.777\ldots$
$100x = 47.777\ldots$
$100x - 10x = 47.777\ldots - 4.777\ldots = 43$
$90x = 43$
$x = \frac{43}{90}$

(iii) $0.\overline{001} = 0.001001001\ldots$

Let $x = 0.001001\ldots$
$1000x = 1.001001\ldots$
$1000x - x = 1$
$999x = 1$
$x = \frac{1}{999}$

Technique: Multiply by $10^k$ where $k$ is the number of repeating digits, then subtract to eliminate the repeating part.

Problem: Find HCF and LCM of $8, 9, 25$.

Solution:
$8 = 2^3$, $9 = 3^2$, $25 = 5^2$

These three numbers share no common prime factor.

$\text{HCF} = 1$ (co-prime numbers)

$\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$

Key insight: When three numbers are pairwise co-prime, their HCF is $1$ and their LCM is the product of all three.

Problem: Show that $n^2 - 1$ is divisible by $8$ for any odd positive integer $n$.

Solution:
Any odd positive integer can be written as $n = 2k + 1$ for some non-negative integer $k$.

Now, among any two consecutive integers $k$ and $k+1$, one must be even. So $k(k+1)$ is always even, say $k(k+1) = 2m$.

Therefore $n^2 - 1 = 4 \times 2m = 8m$, which is divisible by $8$. $\square$

Problem: Prove that the product of three consecutive positive integers is divisible by $6$.

Solution:
Let the three consecutive integers be $n, n+1, n+2$.

Divisibility by 2: Among any two consecutive integers, one is even. So $n(n+1)(n+2)$ has at least one even factor, making the product divisible by $2$.

Divisibility by 3: By Euclid's Lemma with $b = 3$, any integer is of the form $3q$, $3q+1$, or $3q+2$. Among three consecutive integers, exactly one is divisible by $3$.

Since the product is divisible by both $2$ and $3$, and $\gcd(2, 3) = 1$, the product is divisible by $2 \times 3 = 6$. $\square$