Real Numbers NCERT Solutions

Unit 1 - Real Numbers

Unit 1 Real Numbers Exercise 1.1, 1.2, 1.3, 1.4 Solutions

Exercise 1.1: Solutions of Questions on Page Number: 7

Q1: Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

Answer :

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 x 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain 135 = 90 x 1 + 45

We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain 90 = 2 x 45 + 0

Since the remainder is zero, the process stops. Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii)196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 x 195 + 0

Since the remainder is zero, the process stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196.

(iii)867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 867 = 255 x 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 x 2 + 51

We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain 102 = 51 x 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

Q2: Show that any positive odd integer is of the form, or where q is some integer.

Answer :

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd, Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid’s algorithm to find the HCF.

616 = 32 x 19 + 8

32 = 8 x 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides, we get,

x² = (3q)² = 9q² = 3 × 3q²

Let 3q² = m

Therefore, x²= 3m ……………………..(1)

x²= (3q + 1)²= (3q)²+1²+2×3q×1 = 9q² + 1 +6q = 3(3q²+2q) +1

Substitute, 3q²+2q = m, to get,

x²= 3m + 1 ……………………………. (2)

x²= (3q + 2)²= (3q)²+2²+2×3q×2 = 9q²+ 4 + 12q = 3 (3q²+ 4q + 1)+1

Again, substitute, 3q²+4q+1 = m, to get,

x²= 3m + 1…………………………… (3)

Hence, from equations 1, 2 and 3, we can say that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

x = 3q

x = 3q + 1

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i): When r = 0, then,

x²= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³

Case (ii): When r = 1, then,

x³ = (3q+1)³ = (3q)³+1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

Taking 9 as common factor, we get,

x³= 9(3q³+3q²+q)+1

Putting = m, we get,

Putting (3q³+3q²⁺q) = m, we get ,

x³ = 9m+1

Case (iii): When r = 2, then,

x³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8

Taking 9 as common factor, we get,

x³=9(3q³+6q²+4q)+8

Putting (3q³+6q²+4q) = m, we get ,

x³ = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Excercise 1.2: Solutions of Questions on Page number: 11

1. Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solutions:

140 = 2 × 2 × 5 × 7 × 1 = 2²×5×7

56 = 2 × 2 × 13 × 3 × 1 = 2²× 13 × 3

3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3²×5²×17

5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solutions:

(i) 26 and 91

Expressing 26 and 91 as product of its prime factors, we get,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, product of 26 and 91 = 26 × 91 = 2366

And Product of LCM and HCF = 182 × 13 = 2366

Hence, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92

Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Verification

Now, product of 510 and 92 = 510 × 92 = 46920

And Product of LCM and HCF = 23460 × 2 = 46920

Hence, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54

Expressing 336 and 54 as product of its prime factors, we get,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM(336, 54) = = 3024

And HCF(336, 54) = 2×3 = 6

Verification

Now, product of 336 and 54 = 336 × 54 = 18,144

And Product of LCM and HCF = 3024 × 6 = 18,144

Hence, LCM × HCF = product of the 336 and 54.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solutions:

(i) 12, 15 and 21

Writing the product of prime factors for all the three numbers, we get,

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Writing the product of prime factors for all the three numbers, we get,

17=17×1

23=23×1

29=29×1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Writing the product of prime factors for all the three numbers, we get,

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution: As we know that,

HCF×LCM=Product of the two given numbers

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: If the number 6ⁿ ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.

Prime factorization of 6ⁿ = (2×3)ⁿ

Therefore, the prime factorization of 6ⁿ doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, ⁶ⁿis not divisible by 5 and thus it proves that 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution: By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;

7 × 11 × 13 + 13

Taking 13 as common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore, LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3 Page: 14

1. Prove that √5 is irrational.

Solutions: Let us assume, that √5 is rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x

Squaring both the sides, we get,

(y√5)² = x²

⇒5y² = x²……………………………….. (1)

Thus, x² is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y² = (5k)²

⇒y² = 5k²

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

2. Prove that 3 + 2√5 + is irrational.

Solutions: Let us assume 3 + 2√5 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y

Rearranging, we get,

Since, x and y are integers, thus,

is a rational number.

Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.

So, we conclude that 3 + 2√5 is irrational.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Solutions:

(i) 1/√2

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

Exercise 1.4 Page: 17

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2³5²) (vii) 129/(2²5⁷7⁵) (viii) 6/15 (ix) 35/50 (x) 77/210

Solutions:

Note: If the denominator has only factors of 2 and 5 or in the form of 2^m ×5ⁿ then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

(i) 13/3125