Below are clear, step-by-step solutions to all NCERT Exercise questions from Chapter 1 (Real Numbers). Copy these straight into your notebook — each question is solved and explained.
Exercise 1.1
Q1. Use Euclid’s division algorithm to find the H.C.F. of the following pairs of numbers :
(a) 135 and 225
(b) 196 and 38220
(c) 867 and 255
(a) HCF of 135 and 225
Apply Euclid’s algorithm (divide and take remainders):
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Last non-zero remainder = 45.
HCF = 45
(b) HCF of 196 and 38220
Divide 38220 by 196:
38220 = 196 × 195 + 0
Remainder is 0 in one step → divisor 196 is the HCF.
HCF = 196
(c) HCF of 867 and 255
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Last non-zero remainder = 51.
HCF = 51
Exercise 1.2
Q1. Express HCF(306,657) as 306m + 657n for integers m, n.
First find HCF using Euclid’s algorithm:
657 = 306 × 2 + 45
306 = 45 × 6 + 36
45 = 36 × 1 + 9
36 = 9 × 4 + 0
So HCF = 9.
Now express 9 as linear combination of 306 and 657 by back-substitution.
From the steps:
(1) 45 = 657 − 306×2
(2) 36 = 306 − 45×6
(3) 9 = 45 − 36
Substitute (2) into (3):
9 = 45 − (306 − 45×6) = 45 − 306 + 45×6 = 45×7 − 306
Now substitute (1) for 45:
9 = (657 − 306×2)×7 − 306
= 657×7 − 306×14 − 306
= 657×7 − 306×15
Thus 9 = 306(−15) + 657(7)
So one solution: m = −15, n = 7.
Q2. Show that √5 is irrational.
Proof by contradiction:
Assume √5 is rational. Then √5 = p/q where p, q are integers with gcd(p,q)=1.
Squaring gives:5q2=p25 q^2 = p^25q2=p2
So p² is divisible by 5 ⇒ p is divisible by 5 ⇒ p = 5k.
Substitute:5q2=25k2⇒q2=5k25 q^2 = 25 k^2 \Rightarrow q^2 = 5 k^25q2=25k2⇒q2=5k2
So q² divisible by 5 ⇒ q divisible by 5.
Thus p and q both divisible by 5 → contradicts gcd(p,q)=1.
Hence assumption false → √5 is irrational.
Exercise 1.3
This exercise deals with decimal expansions, terminating/non-terminating and conversion between decimals and fractions.
Q1. Express each of the following as p/q in simplest form and say whether the decimal representation is terminating or non-terminating repeating.
(i) 2/5
(ii) 1/3
(iii) 3/8
(iv) 4/125
Solutions:
(i) 25\frac{2}{5}52. Denominator = 5 = 515^151 (only 2 and 5 allowed) ⇒ terminating. Decimal = 0.4
(ii) 13\frac{1}{3}31. Denominator 3 (prime other than 2,5) ⇒ non-terminating repeating. Decimal = 0.3‾\overline{3}3
(iii) 38\frac{3}{8}83. Denominator 8 = 232^323 ⇒ terminating. Decimal = 0.375
(iv) 4125\frac{4}{125}1254. Denominator 125 = 535^353 ⇒ terminating. Decimal = 0.032
Q2. Convert the following terminating decimals to p/q in simplest form:
(i) 0.375
(ii) 1.5
(iii) 6.125
Solutions:
(i) 0.375=3751000=380.375 = \frac{375}{1000} = \frac{3}{8}0.375=1000375=83 after dividing numerator & denominator by 125.
(ii) 1.5=1510=321.5 = \frac{15}{10} = \frac{3}{2}1.5=1015=23.
(iii) 6.125=61251000=4986.125 = \frac{6125}{1000} = \frac{49}{8}6.125=10006125=849 after dividing by 125.
Q3. Express the following as p/q and say whether the decimal representation terminates or not:
(i) 0.6 → 610=35\frac{6}{10} = \frac{3}{5}106=53. Denominator 5 ⇒ terminating.
(ii) 0.47 → 47100\frac{47}{100}10047. Denominator 100 = 22522^2 5^22252 ⇒ terminating.
(iii) 1.732 → 17321000=433250\frac{1732}{1000} = \frac{433}{250}10001732=250433. Denominator 250 = 2×532×5^32×53 ⇒ terminating.
Q4. Write the decimal expansion of 1/11.
Long division or known fact:111=0.09‾=0.090909…\frac{1}{11} = 0.\overline{09} = 0.090909\ldots111=0.09=0.090909…
Non-terminating repeating.
Exercise 1.4
This exercise is about converting recurring decimals to fractions.
Q1. Convert to fractions:
(i) 0.3‾0.\overline{3}0.3
(ii) 0.27‾0.2\overline{7}0.27 (i.e. 0.27777…)
(iii) 0.09‾0.\overline{09}0.09
(iv) 1.321‾1.3\overline{21}1.321
Solutions (standard method):
(i) x=0.3‾x = 0.\overline{3}x=0.3
10x = 3.\overline{3}
10x − x = 3 ⇒ 9x = 3 ⇒ x = 3/9 = 1/3.
(ii) x=0.27‾=0.27777…x = 0.2\overline{7} = 0.27777\ldotsx=0.27=0.27777…
Non-repeating part length m = 1 (digit 2), repeating block length k = 1 (digit 7).
Multiply by 10m+k=102=10010^{m+k}=10^2=10010m+k=102=100: 100x=27.7‾100x = 27.\overline{7}100x=27.7
Multiply by 10m=1010^m = 1010m=10: 10x=2.7‾10x = 2.\overline{7}10x=2.7
Subtract: 90x=2590x = 2590x=25 ⇒ x=25/90=5/18x = 25/90 = 5/18x=25/90=5/18.
Answer: 5/18
(iii) x=0.09‾=0.090909…x = 0.\overline{09} = 0.090909\ldotsx=0.09=0.090909…
k = 2. Multiply by 100:
100x = 9.\overline{09}
100x − x = 9 ⇒ 99x = 9 ⇒ x = 9/99 = 1/11.
(iv) x=1.321‾=1.3212121…x = 1.3\overline{21} = 1.3212121\ldotsx=1.321=1.3212121…
m = 1 (digit 3), k = 2 (digits 21).
Multiply 10m+k=103=100010^{m+k}=10^3=100010m+k=103=1000: 1000x=1321.21‾1000x = 1321.\overline{21}1000x=1321.21
Multiply 10m=1010^m = 1010m=10: 10x=13.21‾10x = 13.\overline{21}10x=13.21
Subtract: 990x=1308990x = 1308990x=1308 ⇒ x=1308/990x = 1308/990x=1308/990. Simplify:
Divide numerator & denominator by 6: 1308/6=218, 990/6=1651308/6 = 218,\; 990/6 = 1651308/6=218,990/6=165 ⇒ 218/165218/165218/165. gcd(218,165)=1, so final is 218/165.
Q2. Express 0.08\overline{3} (i.e. 0.08333…) as a fraction.
Let x=0.083‾x = 0.08\overline{3}x=0.083. Here m = 2 (0.08) and k = 1 (3).
10m+k=103=1000:1000x=83.3‾10^{m+k} = 10^3 = 1000: 1000x = 83.\overline{3}10m+k=103=1000:1000x=83.3
10m=100:100x=8.3‾10^m = 100: 100x = 8.\overline{3}10m=100:100x=8.3
Subtract: 900x=75900x = 75900x=75 ⇒ x=75/900=1/12x = 75/900 = 1/12x=75/900=1/12.
Answer: 1/12
Q3. Convert 0.142857‾0.\overline{142857}0.142857 to a fraction.
Let x=0.142857‾x = 0.\overline{142857}x=0.142857. k=6.
106x=142857.142857‾10^6 x = 142857.\overline{142857}106x=142857.142857
So (106−1)x=142857(10^6 – 1)x = 142857(106−1)x=142857 ⇒ 999999x=142857999999 x = 142857999999x=142857
Divide: x=142857/999999=1/7x = 142857/999999 = 1/7x=142857/999999=1/7 (known simplification).
Exercise 1.5
This exercise contains proofs of irrationality and related theoretical questions.
Q1. Prove that √7 is irrational.
Same method as earlier proofs.
Assume √7 = p/q in lowest terms. Squaring: 7q2=p27q^2 = p^27q2=p2. ⇒ p² divisible by 7 ⇒ p divisible by 7 ⇒ p = 7k. Substitute:
7q2=49k2⇒q2=7k27q^2 = 49k^2 ⇒ q^2 = 7k^27q2=49k2⇒q2=7k2 ⇒ q divisible by 7.
Thus p and q share factor 7, contradicting gcd(p,q)=1. Hence √7 is irrational.
Q2. Prove that √p is irrational for any prime p.
Assume √p = a/b in lowest terms. Square: pb2=a2p b^2 = a^2pb2=a2. So a² divisible by p ⇒ a divisible by p ⇒ set a = p c. Substitute => pb2=p2c2⇒b2=pc2p b^2 = p^2 c^2 ⇒ b^2 = p c^2pb2=p2c2⇒b2=pc2 ⇒ b divisible by p. Contradiction. Hence √p irrational.
Q3. Show that the decimal expansion of a rational number either terminates or repeats.
When performing long division of p by q (q>0) there are only finitely many possible remainders: 0,1,2,…,q−1.
- If at some stage remainder becomes 0 → division terminates (terminating decimal).
- If remainder never becomes 0, at some step a remainder must repeat (pigeonhole principle since only q−1 non-zero remainders possible). Once a remainder repeats, the sequence of digits from then on will repeat → decimal is repeating. Hence any rational number has either terminating or repeating decimal.
Q4. Prove that between any two distinct real numbers there exists an irrational number.
Sketch (standard construction): Let a < b. Because rationals are dense we can choose a rational r such that a/2<r<b/2a/\sqrt{2} < r < b/\sqrt{2}a/2<r<b/2. Then r2r\sqrt{2}r2 is between a and b (multiply inequality by √2). Since r is rational and √2 is irrational, r2r\sqrt{2}r2 is irrational. So there exists an irrational between a and b. Thus irrationals are dense in ℝ.
(Alternative constructive proofs exist; key idea: multiply/divide by irrational.)
Q5. Prove that p1/np^{1/n}p1/n is irrational for prime p and integer n > 1.
Assume p1/n=a/bp^{1/n} = a/bp1/n=a/b in lowest terms. Raise to n: p=an/bnp = a^n / b^np=an/bn ⇒ an=pbna^n = p b^nan=pbn. So left side divisible by p ⇒ a divisible by p ⇒ set a = p k ⇒ substitute: pbn=pnkn⇒bn=pn−1knp b^n = p^n k^n ⇒ b^n = p^{n-1} k^npbn=pnkn⇒bn=pn−1kn ⇒ b divisible by p. Contradiction to gcd(a,b)=1. Hence p1/np^{1/n}p1/n is irrational.
EXTRA: Frequently asked short problems (quick answers)
- HCF × LCM = Product (for two numbers). Use after finding HCF to compute LCM or vice versa.
- Convert 0.232323… to fraction.
Let x = 0.\overline{23}. 100x − x = 23 ⇒ 99x = 23 ⇒ x = 23/99. - Is 13/125 terminating?
125 = 535^353 ⇒ denominator only 2 & 5? yes (only 5) ⇒ terminating. - Express 0.090909… as fraction. = 1/11 (we did above).
SUMMARY & HOW TO USE THESE SOLUTIONS
- For HCF problems: always show the division steps clearly (each line
a = b × q + r) and identify the last non-zero remainder. - For expressing HCF as linear combination: do back-substitution (we showed for 306 & 657).
- For irrationality proofs: assume rational p/q in lowest terms, square (or raise to necessary power), deduce both p and q have a common prime factor → contradiction.
- For decimals: reduce denominator to check for only 2s and 5s (terminating) or other primes (repeating). For converting repeating decimals, identify lengths m (non-repeating) and k (repeating) and use 10m+kx−10mx10^{m+k}x – 10^m x10m+kx−10mx.