Real Numbers

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exists unique integers q and r which satisfies the condition a = bq + r where 0≤ r ≤ b.

Let the number be x.

By Euclid’s division lemma -

x = bq + r ….(1)

where q is the quotient,

r is the remainder

and b is the divisor.

According to the question, b = 61, r = 32, q = 27.

Putting the values in equation (1) -

∴ x = 61(27) + 32

⇒ x = 1679.

Hence, the given number is 1679.

By Euclid’s division lemma -

x = bq + r ….(1)

where q is the quotient,

r is the remainder

and b is the divisor.

According to the question, x = 1365, r = 32, q = 31, b = ?.

Putting the values in equation(1) -

∴ 1365 = b(31) + 32

⇒ 31b = 1333

⇒ b = 43

i.

∵ here 405 < 2520,

∴ b = 405 and a = 2520.

By Euclid’s division lemma -

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting it in equation (1) -

⇒ 2520 = 405(6) + 90.

Here 90 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 405; b = 90;

⇒ 405 = 90(4) + 45.

Here 45 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

a = 90; b = 45;

⇒ 90 = 45(2) + 0.

∵ remainder is zero.

∴ HCF is 45.

ii.

∵ here 504 < 1188,

∴ b = 504 and a = 1188.

By Euclid’s division lemma -

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting the values in equation (1) -

⇒ 1188 = 504(2) + 180.

Here 180 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 504; b = 180;

⇒ 504 = 180(2) + 144

Here 144 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now,a = 180; b = 144;

⇒ 180 = 144(1) + 36.

Here 36 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now,a = 144; b = 36;

⇒ 144 = 36(4) + 0.

∵ remainder is zero.

∴ HCF is 36.

iii.

∵ here 960 < 1575,

∴ b = 960 and a = 1575.

a = bq + r ….(1)

where q is the quotient, r is the remainder and b is the divisor.

Putting the values in equation (1) -

⇒ 1575 = 960(1) + 615.

Here 615 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 960; b = 615;

⇒ 960 = 615(1) + 345.

Here 345 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 615; b = 345;

⇒ 615 = 345(1) + 270.

Here 270 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 345; b = 270;

⇒ 345 = 270(1) + 75.

Here 75 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 270; b = 75;

⇒ 270 = 75(3) + 45.

Here 45 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 75; b = 45;

⇒ 75 = 45(1) + 30.

Here 30 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

Now, a = 45; b = 30;

⇒ 45 = 30(1) + 15.

Here 15 is the remainder, which is not zero. Again applying the Euclid’s divison lemma -

a = 30; b = 15;

⇒ 30 = 15(2) + 0.

∵ remainder is zero.

∴ HCF is 15.

Using Euclid’s divison lemma -

a = bq + r;

We want to show that any even integer is of the form '2q' and any odd integer is of the form'2q + 1'

So, we take the two integers a and 2.

When we divide a with 2, the possible values of remainder are 0 and 1, which means

a = 2q + 0 or a = 2q + 1

Also, 2q is always an even integer for any integral value of q and so 2q + 1 will always be an odd integer. (∵ 'even integer + 1' is always an odd integer)

∴ when a is a positive even integer it will always be of the form 2q

And when a is positive odd integer it will always be of the form 2q + 1.

∴ for every positive integer it is either odd or even.

Let take a as any positive integer and b = 6. a > b

Then using Euclid’s algorithm, we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will be 6q + 0, 6q + 1 , 6q + 2,6q + 3,6q + 4,6q + 5.

6q + 0 → 6 is divisible by 2 so it is an even number.

6q + 1 → 6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.

6q + 2 → 6 is divisible by 2 and 2 is also divisible by 2 so it is an even number.

6q + 3 → 6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number.

6q + 4 → 6 is divisible by 2 and 4 is also divisible by 2 it is an even number.

6q + 5 → 6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number.

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

Let a be any odd positive integer and b = 4. By division lemma there exist integer q and r such that

a = 4 q + r, where 0 ≤ r ≤ 4

so a = 4q or, a = 4q + 1 or, a = 4q + 2 or, a = 4q + 3

4q + 1 → 4 is divisible by 2 but 1 is not divisible by 2, so it is an odd number

4q + 2 → 4 is divisible by 2 and 2 is also divisible by 2, so it is an even number

4q + 3 → 4 is divisible by 2 but 3 is not divisible by 2, so it is an odd number

4q + 4 → 4 is divisible by 2 and 4 is also divisible by 2, so it is an even number

∴ any odd integer is of the form 4q + 1 or, 4q + 3.

i. HCF = 12, LCM = 252

Prime factorization of given numbers -

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

So HCF = product of common factors = 2 × 2 × 3 = 12

And LCM = product of prime factors with highest powers = 2² × 3² × 7 = 252

Verification -

HCF × LCM = product of numbers

⇒ LHS = 12 × 252 = 3024

⇒ RHS = 36 × 84 = 3024

∵ LHS = RHS(Hence verified)

ii. HCF = 1, LCM = 713

Prime factorization of given numbers -

23 = 23 × 1

31 = 31 × 1

So HCF = product of common factors = 1

And LCM = product of prime factors with highest powers = 23 × 31 = 713

Verification -

HCF × LCM = product of numbers

⇒ LHS = 1 × 713 = 713

⇒ RHS = 23 × 31 = 713

∵ LHS = RHS(Hence verified)

iii. HCF = 4, LCM = 9696

Prime factorization of given numbers -

96 = 2 × 2 × 2 × 2 × 2 × 3

404 = 2 × 2 × 101

So HCF = product of common factors = 2 × 2 = 4

And LCM = product of prime factors with highest powers = 2⁵ × 3 × 101 = 9696.

Verification -

HCF × LCM = product of numbers

⇒ LHS = 4 × 9696 = 38784

⇒ RHS = 96 × 404 = 38784

∵ LHS = RHS(Hence verified)

iv. HCF = 18, LCM = 1584

Prime factorization of given numbers -

144 = 2 × 2 × 2 × 2 × 3 × 3

198 = 2 × 3 × 3 × 11

So HCF = product of common factors = 2 × 3 × 3 = 18

And LCM = product of prime factors with highest powers = 2⁴ × 3² × 11 = 1584

Verification -

HCF × LCM = product of numbers

⇒ LHS = 18 × 1584 = 28512

⇒ RHS = 144 × 198 = 28512

∵ LHS = RHS(Hence verified)

v. HCF = 36, LCM = 11880

Prime factorization of given numbers -

396 = 2 × 2 × 3 × 3 × 11

1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5

So HCF = product of common factors = 2 × 2 × 3 × 3 = 36

And LCM = product of prime factors with highest powers = 2³ × 3³ × 5 × 11 = 11880

Verification -

HCF × LCM = product of numbers

⇒ LHS = 36 × 11880 = 427680

⇒ RHS = 396 × 1080 = 427680

∵ LHS = RHS(Hence verified)

vi. HCF = 128, LCM = 14976

Prime factorization of given numbers -

1152 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

1664 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 13

So HCF = product of common factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128

And LCM = product of prime factors with highest powers = 2⁷ × 3² × 13 = 14976

Verification -

HCF × LCM = product of numbers

⇒ LHS = 128 × 14976 = 1916928

⇒ RHS = 1152 × 1664 = 1916928

∵ LHS = RHS(Hence verified)

i. HCF = 1, LCM = 1800

Prime factorization of given numbers -

8 = 2 × 2 × 2 × 1

9 = 3 × 3 × 1

25 = 5 × 5 × 1

So HCF = product of common factors = 1

And LCM = product of prime factors with highest powers = 2³ × 3² × 5² = 1800

ii. HCF = 3, LCM = 420

Prime factorization of given numbers -

12 = 2 × 2 × 3

15 = 3 × 5

21 = 3 × 7

So HCF = product of common factors = 3

And LCM = product of prime factors with highest powers = 2² × 3 × 5 × 7 = 420

iii. HCF = 1, LCM = 11339

Prime factorization of given numbers -

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

So HCF = 1

And LCM = product of prime factors with highest powers = 17 × 23 × 29 = 11339

iv. HCF = 4, LCM = 360

Prime factorization of given numbers -

36 = 2 × 2 × 3 × 3

24 = 2 × 2 × 2 × 3

40 = 2 × 2 × 2 × 5

So HCF = product of common factors = 2 × 2 = 4

And LCM = product of prime factors with highest powers = 2³ × 3² × 5 = 360

v. HCF = 6, LCM = 2160

Prime factorization of given numbers -

30 = 2 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

So HCF = product of common factors = 2 × 3 = 6

And LCM = product of prime factors with highest powers = 2⁴ × 3³ × 5 = 2160

vi. HCF = 1, LCM = 1260

Prime factorization of given numbers -

21 = 3 × 7

28 = 2 × 2 × 7

36 = 2 × 2 × 3 × 3

45 = 3 × 3 × 5

So HCF = product of common factors = 1

And LCM = product of prime factors with highest powers = 2² × 3² × 5 × 7 = 1260

We know that LCM × HCF = product of numbers.

Let the other number be x.

So, 161 × x = 23 × 1449

⇒ x = 207

We know that LCM × HCF = product of numbers.

Let the other number be x.

So, 725 × x = 145 × 2175

⇒ x = 435

We know that LCM × HCF = product of numbers.

Let the LCM of numbers to be x.

So, 18 × x = 12960

⇒ x = 720

No, ∵ HCF does not divided LCM exactly

Using Euclid’s division lemma -

Take a = 760 and b = 18.

a = bq + r. where q is the quotient, r is the remainder and b is the divisor.

If HCF divides LCM completely, r = 0.

Putting the values -

Here 760 = 18(42) + 4

⇒ r = 4

∵ r is not equal to zero

∴ HCF does not divides LCM completely.

So this is not possible for two numbers to have HCF = 18 and LCM = 760.

(i) HCF for 69 and 92 is 23

So dividing both of them by 23 -

69 = 23 × 3

92 = 23 × 4

We get numerator as 3 and denominator as 4.

∴ simplest fraction is .

(ii) HCF for 473 and 645 is 43

So dividing both of them by 43 -

473 = 43 × 11

645 = 43 × 15

We get numerator as 11 and denominator as 15

∴ simplest fraction is .

(iii) HCF for 368 and 496 is 16

So, dividing both of them by 16 -

368 = 16 × 23

496 = 16 × 31

We get numerator as 23 and denominator as 31.

∴ simplest fraction is

(iv) HCF for 1095 and 1168 is 73

So, dividing both of them by 73 -

1095 = 73 × 15

1168 = 73 × 16

We get numerator as 15 and denominator as 16.

∴ simplest fraction is .

∵ we are getting 6 as remainder.

Subtracting 6 from the dividend will give us the exact division.

So, let’s subtract the remainder and find the HCF of numbers (largest number which divides both the number)

We need to find HCF of 438 - 6 = 432 and 606 - 6 = 600.

Prime factorization of numbers -

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

600 = 2 × 2 × 2 × 3 × 5 × 5

HCF of given numbers = 2 × 2 × 2 × 3 = 24.

∵ we are getting 5 and 7 as remainder.

Subtracting 5 and 7 from the dividends will give us the exact division.

So, let’s subtract the remainder and find the HCF of numbers (largest number which divides both the number) -

We need to find HCF of 320 - 5 = 315 and 457 - 7 = 450.

Prime factorization of numbers -

315 = 3 × 3 × 5 × 7

450 = 2 × 3 × 3 × 5 × 5

HCF of given numbers = 3 × 3 × 5 = 45.

Find the LCM of the numbers (the least number which all the given numbers divides) -

Prime factorization of numbers -

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

LCM of given numbers = product of prime factors with highest powers = 2³ × 5 × 7 × 13 = 3640.

∵ required remainder is 7.

We need to add 7 to the LCM of numbers.

∴ the least number that leaves remainder 7 will be 3640 + 7 = 3647.

Subtract the remainders from the given numbers:

If you look at the negative remainders -
Remainder from 28 = 28 - 8 is - 20.
Remainder from 32 = 32 - 12 is - 20

Find the LCM of the numbers (the least number which all the given numbers divide)

Prime factorization of numbers -

28 = 2 × 2 × 7

32 = 2 × 2 × 2 × 2 × 2

LCM of given numbers = product of prime factors with highest powers = 2⁵ × 7 = 224.

Smallest no, which leaves remainder 8 and 12 when divided by 28 and 32

= LCM - 20 = 224 - 20 = 204

Find the LCM of the numbers (the least number which all the given numbers divide) -

Prime factorization of numbers -

468 = 2 × 2 × 3 × 3 × 13

520 = 2 × 2 × 2 × 5 × 13

LCM of given numbers = product of prime factors with highest powers = 2³ × 3² × 5 × 13 = 4680

∵ the number is increased by 17 to get perfect division.

We need to subtract 17 to the LCM of numbers.

∴ the least number when added by 17 gives exact division will be 4680 - 17 = 4663.

Find the LCM of the numbers (the least number which all the given numbers divide) -

Prime factorization of numbers -

15 = 3 × 5

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

LCM of given numbers = product of prime factors with highest powers = 2³ × 3² × 5

= 360

We know that the greatest four-digit number is 9999.

So, the greatest value closest to 9999 which 360 can divide can be found by Euclid’s division lemma -

Putting the values in the equation a = bq + r -

9999 = 360(27) + 279

So, highest four - digit number 360 can completely divide = 9999 - remainder

= 9999 - 279

= 9720.

The number of room will be minimum if each room accommodates maximum number of participants.

∵ in each room the same number of participants are to be seated and all of them must be of the same subject.

Therefore, the number of participants in each room must be the HCF of 60, 84 and 108

So HCF of 60, 84 and 108 -

Prime factors of numbers are -

60 = 2 × 2 × 3 × 5

84 = 2 × 2 × 3 × 7

108 = 2 × 2 × 3 × 3 × 3

∴ HCF = 2 × 2 × 3 = 12

Therefore, in each room 12 participants can be seated.

Total number of students = 60 + 84 + 108 = 252

So minimum number of room required to accommodate all students = 252/12 = 21

The number of stacks will be minimum if each stack accommodates maximum number of books.

∵ height of each stack is the same and all of them are of the same subject.

Therefore, the number of books in each stack must be the HCF of 336,240 and 96.

So HCF of 336,240 and 96 -

Prime factors of numbers are -

336 = 2 × 2 × 2 × 2 × 3 × 7

240 = 2 × 2 × 2 × 2 × 3 × 5

96 = 2 × 2 × 2 × 2 × 2 × 3

∴ HCF = 2 × 2 × 2 × 2 × 3 = 48

Therefore, in each stack 48 books can be placed.

Total number of books = 336 + 240 + 96 = 672

So minimum number of stacks required to accommodate all books = 672/48 = 14

∵ length of each plank is the same and we are supposed to find plank of greatest possible length.

Therefore, the greatest possible length of each plank must be the HCF of 42, 49 and 63.

So HCF of 42, 49 and 63 -

Prime factors of numbers are -

42 = 2 × 3 × 7

49 = 7 × 7

63 = 3 × 3 × 7

∴ HCF = 7

Therefore, maximum plank length of each plank = 7m

Total available length of plank = 42 + 49 + 63 = 154m

So number of planks of this maximum possible length = 154/7 = 22.

We know 1m = 100cm.

So,

7m = 700cm,

3m 85cm = 300 + 85 = 385cm,

and 12m 95cm = 1200 + 95 = 1295cm.

The greatest possible length that can measure all the three given lengths will be HCF of the lengths -

So HCF of 700, 385 and 1295 -

Prime factors of numbers are -

700 = 2 × 2 × 5 × 5 × 7

385 = 5 × 7 × 11

1295 = 5 × 7 × 37

∴ HCF = 5 × 7 = 35.

So the greatest possible length that can be used = 35cm.

∵ each student gets the same number of pens and pencils.

Maximum number of students among which it can be distributed will be equal to HCF of number of pencils and number of pens.

So HCF of 1001 and 910 -

Prime factors of the numbers are -

1001 = 7 × 11 × 13

910 = 2 × 5 × 7 × 13

∴ HCF = 7 × 13 = 91.

We know 1m = 100cm.

So,

15m 17cm = 1500 + 17 = 1517cm

And 9m 2cm = 900 + 2 = 902 cm

Least number of square tiles required to pave the ceiling of this area if largest tiles are used.

Length of largest tile = H.C.F. of 1517 cm and 902 cm

Prime factors of 1517 and 902 -

1517 = 41 × 37

902 = 2 × 11 × 41

So HCF = 41

∴ Area of tiles of largest length and width = (41 × 41) cm²

And area of room = length × width = (1517 × 902) cm²

So minimum number of tiles required = (1517 × 902) cm²/ (41 × 41) cm² = 814.

Least length of cloth that can be measured using this rod will be LCM of the three lengths.

So LCM of 64, 80 and 96 -

Prime factors of the given numbers are -

64 = 2 × 2 × 2 × 2 × 2 × 2

80 = 2 × 2 × 2 × 2 × 5

96 = 2 × 2 × 2 × 2 × 2 × 3

So LCM = product of prime factors with highest powers = 2⁶ × 3 × 5 = 960

∴ least length of cloth that can be measured using the rods = 960 cm = 9.6m

The devices will beep simultaneously at the LCM of intervals of beeps

So LCM of 60 and 62 -

Prime factors of the numbers are -

60 = 2 × 2 × 3 × 5

62 = 2 × 31

So LCM = product of prime factors with highest powers = 2² × 3 × 5 × 31 = 1860 seconds

∴ devices will beep simultaneously after 1860 seconds.

We know 1 min = 60 seconds,

So 1860 seconds = 1860/60 minutes = 31 minutes.

∴ the next time they will beep simultaneously at 10:31 hrs.

The traffic lights will change at the LCM of intervals of all the three lights.

So LCM of 48, 72 and 108 -

Prime factors of the numbers are -

48 = 2 × 2 × 2 × 2 × 3

72 = 2 × 2 × 2 × 3 × 3

108 = 2 × 2 × 3 × 3 × 3

So LCM = product of prime factors with highest powers = 2⁴ × 3³ = 432

∴ traffic light will change after 432 seconds.

We know 1 min = 60 seconds,

So 432 seconds = 432/60 minutes = 7 minutes and 12 seconds.

∴ the next time lights will simultaneously change at 8:7:12 hrs.

All the clocks will toll together at the LCM of their intervals.

So LCM of the intervals = 2,4,6,8,10,12 = 120.

∴ all clocks will toll together after every 120 minutes.

Given interval = 30 hrs = 30 × 60 minutes = 1800 minutes {∵ 1hr = 60 minutes}

∴ number of times clock will tol = 1800/120 = 15.

plus one time when they commenced together

So the answer will be 16 times.

Let’s start from the bottom -

The factors of last numbers are 11 and 5.

So last number is 11 × 5 = 55

Now factors of the second number from bottom is 55 and 3

So second last number is 55 × 3 = 165

Also factors for the third number from bottom is 165 and 2.

So third last number is 165 × 2 = 330

Similarly factors for the first number is 330 and 2.

So first number is 330 × 2 = 660

(i)

∵ denominator is of the form, 2ⁿ × 5^m,

where n = 3 and m = 2.

∴ it is terminating in nature.

ii.

∵ denominator is of the form, 2ⁿ × 5^m,

125 = 5 × 5 × 5.

where n = 0 and m = 3.

∴ it is terminating in nature.

iii.

∵ denominator is of the form, 2ⁿ × 5^m,

800 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

where n = 5 and m = 2.

∴ it is terminating in nature.

iv.

∵ denominator is of the form, 2ⁿ × 5^m,

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

where n = 6 and m = 2.

∴ it is terminating in nature.

v.

∵ denominator is of the form, 2ⁿ × 5^m,

320 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

where n = 6 and m = 1.

∴ it is terminating in nature.

vi.

∵ denominator is of the form, 2ⁿ × 5^m,

3125 = 5 × 5 × 5 × 5 × 5

where n = 0 and m = 5.

∴ it is terminating in nature.

(i) Denominator is 2³ × 3 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(ii) Denominator is 2² × 3³ × 5 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(iii) Denominator is 2² × 5³ × 7² which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(iv) Denominator is 35 = 5 × 7 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(v) Denominator is 210 = 2 × 3 × 5 × 7 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(vi) Denominator is 147 = 3 × 7 × 7 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(vii) Denominator is 343 = 7 × 7 × 7 which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

(viii) Denominator is 455 = 5 × 91, which is not in the form 2ⁿ × 5^m

∴ the fraction will not be a terminating decimal.

i) Let x = 0. - (i)

Multiply 10 on both sides -

10x = 8. - (ii)

Subtract the equations (i) from (ii) -

9x = 8

So x =

ii) Let x = 2. - (i)

Multiply 10 on both sides -

10x = 24. - (ii)

Subtract the equations (i) from (ii) -

9x = 22

So x =

iii) Let x = 0. - (i)

Multiply 100 on both sides -

100x = 24. - (ii)

Subtract the equations (i) from (ii) -

99x = 24

So x = =

iv) Let x = 0.1 - (i)

Multiply 10 on both sides -

10x = 1. - (ii)

Multiply 10 on both sides -

100x = 12. - (iii)

Subtract the equations (ii) from (iii) -

90x = 11

So x =

v) Let x = 2.2 - (i)

Multiply 10 on both sides -

10x = 22. - (ii)

Multiply 10 on both sides -

100x = 224. - (iii)

Subtract the equations (ii) from (iii) -

90x = 202

So x = =

vi) Let x = 0. - (i)

Multiply 1000 on both sides -

1000x = 365. - (ii)

Subtract the equations (i) from (ii) -

999x = 365

So x =

A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non - zero denominator q.

(ii) irrational numbers

An irrational number is a number that cannot be expressed as a fraction p/q for any integers p and q. Irrational numbers have decimal expansions that neither terminate nor become periodic

(iii) real numbers.

Real numbers are numbers that can be found on the number line. This includes both the rational and irrational numbers.

A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non - zero denominator q. While, an irrational number is a number that cannot be expressed as a fraction p/q for any integers p and q. Irrational numbers have decimal expansions that neither terminate nor become periodic.

{π,3., 5.636363 …, 2.040040004 …, , 1.535335333 …, 3.121221222 …, are irrational}

i. rational ii. rational

iii. irrational iv. rational

v. rational vi. irrational

vii. irrational viii. irrational

ix. irrational x. irrational

(i) let's assume that √6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

⇒ a/b = √6

So let's take square of both the sides -

⇒ (a/b)(a/b) = (√6)( √6)
⇒ a²/b² = 6
⇒ a² = 6b²

Last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least 6) is even. So a² must be even. But any odd number times itself is odd, so if a² is even, then a is even.

Since a is even, there is some integer c such that:

⇒ 2c = a.

Now let's replace a with 2c:

⇒ a² = 6b²
⇒ (2c)² = (2)(3)b²
⇒ 2c² = 3b²

But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.

Thus, proved that √6 is irrational.

(ii) assume that 2 - √3 is rational
2 - √3 = a/b , where a and b are integers .
⇒ - √3 = a/b - 2
⇒ √3 = 2 - a/b
⇒ √3 = 2b/b - a/b
⇒ √3 = (2b – a) / b
we know that a, b and 2 are integers and they are also rational {i.e RHS is rational}
therefore √3 will be rational.
but we know that √3 is irrational.
there is a contradiction
so, 2 - √3 is an irrational number

(iii) assume that 3 + √2 is rational
3 + √2 = a/b , where a and b are integers .
⇒ √2 = a/b - 3
⇒ √2 = (a - 3b)/b
we know that a, b and 3 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 3 + √2 is an irrational number

(iv) assume that 2 + √5 is rational
2 + √5 = a/b , where a and b are integers .
⇒ √5 = a/b - 2
⇒ √5 = a/b - 2b/b
⇒ √5 = (a - 2b)/ b
we know that a, b and 2 are integers and they are also rational {i.e RHS is rational}
therefore √5 will be rational.
but we know that √5 is irrational.
there is a contradiction
so, 2 + √5 is an irrational number

(v) assume that 5 + 3√2 is rational
5 + 3√2 = a/b , where a and b are integers .
⇒ 3√2 = a/b - 5
⇒ 3√2 = a/b - 5b/b
⇒ √2 = (a - 5b)/3b
we know that a, b, 3 and 5 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 5 + 3√2 is an irrational number.

(vi) assume that 3√7 is rational
⇒ 3√7 = a/b , where a and b are integers .
⇒ √7 = a/3b
we know that a, b and 3 are integers and they are also rational {i.e RHS is rational}
therefore √7 will be rational.
but we know that √7 is irrational.
there is a contradiction
so, 3√7 is an irrational number

(vii) {Rationalising}

Now, let us assume that 3√5 / 5 is a rational number

⇒ 3√5/5 = p/q (where, p & q are integers, q not equal to 0)

⇒ 3√5 = 5p/q

⇒ √5 = 5p/3q

Here, LHS is an irrational number whereas RHS is a rational number.

So by contradiction 3√5 / 5 is an irrational number

(viii) assume that 2 - 3√5 is rational
2 - 3√5 = a/b , where a and b are integers .
⇒ - 3√5 = a/b - 2
⇒ 3√5 = 2 - a/b
⇒ 3√5 = 2b/b - a/b
⇒ √5 = (2b – a) / 3b
we know that a, b, 2 and 3 are integers and they are also rational {i.e RHS is rational}
therefore √5 will be rational.
but we know that √5 is irrational.
there is a contradiction
so, 2 - 3√5 is an irrational number

(ix) Multiplying both sides by (√5 - √3).
(√5 - √3) (√5 + √3) = 5 - 3 = 2

⇒ 2 = p/q × (√5 - √3)
⇒ (√5 - √3) = 2q/p,

∴ √5 - √3 is rational = 2q/p
⇒ √5 + √3 = p/q
⇒ √5 - √3 = 2q/p

Adding the equations -

, which is a rational number.

But we know that √5 is IRRATIONAL.

Therefore, the assumption is wrong and √3 + √5 is irrational.

Assume to be rational. So we can write it in the form of a/b where a and b are co - prime.

So , a/b =

And ∴ b/a = √3

Since b/a is rational but √3 is irrational.

By contradiction is irrational.

.

Let’s take (2 + ) and (2 ‒ ).

These are irrational numbers.

Their sum = (2 + ) + (2 ‒ ) = 4 {which is rational}.

ii) Let’s take (3 + ) and (3 ‒ ).

These are irrational numbers.

Their product = (3 + ) × (3 ‒ ) = 9 + 3√2 – 3√2 – (√2)²

= 9 – 4 = 5 {which is rational}

i. True ii. True

iii. False iv. False

v. True vi. True

(iii) Let’s take (2 + ) and (2 ‒ ).

These are irrational numbers.

Their sum = (2 + ) + (2 ‒ ) = 4 {which is rational}.

∴ it is false.

(iv) Let’s take (3 + ) and (3 ‒ ).

These are irrational numbers.

Their product = (3 + ) × (3 ‒ ) = 9 + 3√2 – 3√2 – (√2)²

= 9 – 4 = 5 {which is rational}

∴ it is false.

Assume that 2√3 - 1 is rational
2√3 = a/b , where a and b are integers .
⇒ 2√3 = a/b + 1
⇒ 2√3 = a/b + b/b
⇒ √3 = (a + b) / 2b
we know that a, b, and 2 are integers and they are also rational {i.e RHS is rational}
therefore √3 will be rational.
but we know that √3 is irrational.
there is a contradiction
so, 2√3 - 1 is an irrational number.

5√2 is an irrational number and subtraction of a rational number and an irrational number is an irrational number.

∴, 4 - 5√2 is irrational, where 5√2 is an irrational number while 4 is a rational number.

Alternative - Assume that 4 - 5√2 is rational
4 - 5√2 = a/b , where a and b are integers .
⇒ - 5√2 = a/b - 4
⇒ 5√2 = 4 - a/b
⇒ 5√2 = 4b/b - a/b
⇒ √2 = (4b – a) / 5b
we know that a, b, 4 and 5 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 4 - 5√2 is an irrational number

2√3 is an irrational number and subtraction of a rational number and an irrational number is an irrational number.

∴, 5 - 2√3 is irrational, where 2√3 is an irrational number while 5 is a rational number.

Alternative - Assume that 5 - 2√3 is rational
5 - 2√3 = a/b , where a and b are integers .
⇒ - 2√3 = a/b - 5
⇒ 2√3 = 5 - a/b
⇒ 2√3 = 5b/b - a/b
⇒ √3 = (5b – a) / 2b
we know that a, b, 2 and 5 are integers and they are also rational {i.e RHS is rational}
therefore √3 will be rational.
but we know that √3 is irrational.
there is a contradiction
so, 5 - 2√3 is an irrational number

assume that 5√2 is rational

where a and b are integers .

we know that a, b and 5 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 5√2 is an irrational number

∴ it is proved that 5√2 is irrational.

Now, let us assume that 2√7 / 7 is a rational number

⇒ 2√7/7 = p/q (where, p & q are integers, q not equal to 0)

⇒ 2√7 = 7p/q

⇒ √7 = 7p/2q

Here, LHS is an irrational number whereas RHS is a rational number.

So by contradiction 2√7 / 7 is an irrational number

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exists unique integers q and r which satisfies the condition a = bq + r where 0≤ r ≤ b.

The fundamental theorem of arithmetic, also called the unique factorization theorem or the unique - prime - factorization theorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors.

Factorization of 360 = 2 × 2 × 2 × 3 × 3 × 5 = (2³× 3²× 5)

Two prime numbers don’t have a common factor other than 1. So HCF = 1.

∵ HCF of two prime numbers is 1 and HCF × LCF = product of numbers.

So LCM of numbers = product of numbers = a × b = ab.

∵ HCF × LCM = product of numbers

∴ 25 × LCM = 1050

And LCM = 1050/25 = 42.

A whole number that can be divided evenly by numbers other than 1 or itself is a composite number. We can also say, “A non - prime number is a composite number”.

Two relatively prime numbers don’t have a common factor other than 1. So HCF = 1.

b = (2^m× 5^m), where m and n are some non - negative integers.

2√45 = 2√(3*3*5)
= 2*3√5
=6√5

Also 3√20 = 3√2*2*5
= 3*2√5
= 6√5

So numerator = 6√5 + 6√5 = 12√5

And denominator is 2√5.

∴ expression becomes = 6

.

(2” × 5”) = 10ⁿ {using rules of exponent}

And 10ⁿ either ends up with 1 for n = 0 and for n>0 and integer, ends up with zero.

Also for negative value of n it ends with fractions not ending with 5.

∴ this will never end with 5.

No, ∵ HCF does not divided LCM exactly

Using Euclid’s division lemma -

Take a = 520 and b = 25.

a = bq + r. where q is the quotient, r is the remainder and b is the divisor.

If HCF divides LCM completely, r = 0.

Here 520 = 25(20) + 20

So, r = 20

∵ r is not equal to zero

∴ HCF does not divides LCM completely.

So this is not possible for two numbers to have HCF = 25 and LCM = 520.

Let’s take (2 + ) and (2 ‒ ).

These are irrational numbers.

Their sum = (2 + ) + (2 ‒ ) = 4 {which is rational}.

Let’s take (3 + ) and (3 ‒ ).

These are irrational numbers.

Their product = (3 + ) × (3 ‒ ) = 9 + 3√2 – 3√2 – (√2)²

= 9 – 4 = 5 {which is rational}

Two relatively prime numbers don’t have a common factor other than 1.

i.e, HCF = 1 and we know HCF × LCM = product of numbers.

So LCM = product of numbers = a × b = ab.

No, ∵ HCF should divide LCM exactly.

Using Euclid’s division lemma -

Take a = 1200 and b = 500.

a = bq + r. where q is the quotient, r is the remainder and b is the divisor.

If HCF divides LCM completely, r = 0.

Here 1200 = 500(2) + 200

r = 200

∵ r is not equal to zero.

∴ HCF does not divides LCM completely.

So this is not possible for two numbers to have HCF = 500 and LCM = 1200.

Let x = 0. - (i)

Multiply 10 on both sides -

10x = 4. - (ii)

Subtract the equations (i) from (ii) -

⇒ 9x = 4

So x =

Let x = 0. - (i)

Multiply 100 on both sides -

100x = 23. - (ii)

Subtract the equations (i) from (ii) -

⇒ 99x = 23

So x =

Irrational numbers are non - terminating non - recurring decimals.

Thus, 0.15015001500015 … is an irrational number.

assume that is rational
⇒ = a/b, where a and b are integers.
⇒ √2 = 3a/b
we know that a, b and 3 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, is an irrational number.

∵, √3 = 1.732…

So, we may take 1.8 as the required rational number between √3 and 2.

Thus, the required rational number is 1.8 or

∵, 3. is a non - terminating repeating decimal.

∴, is a rational number.

Co - prime numbers are the numbers which have only one common divisor = 1 or their HCF = 1.

Here for option


(A) 14 = 2 × 7
35 = 5 × 7

∴HCF(14,25) = 7.

(B)18 = 2 × 3 × 3
25 =5 × 5


∴ HCF(18,25) = 1.

(C) 31 = 1 × 31

93 = 3 × 31


∴ HCF(31,93) = 31.

(D) 32 = 2 × 2 × 2 × 2 × 2

62 = 2 × 31

∴ HCF(32,62) = 2.

So answers will be (B).

a = (2²× 3³× 5⁴)

b = (2³× 3²× 5)

Here HCF (a, b) = product of common factors = 2² × 3² × 5 = 4 × 9 × 5 = 180.

HCF is product of common factors,

Here HCF = 2² × 3 × 5 = 4 × 3 × 5 = 60.

LCM of the given numbers = product of prime factors with highest powers = 2⁴× 5 × 7 × 3 = 1680.

We know that product of numbers = LCM × HCF.

Let the other number be x.

So, x × 54 = 27 × 162

Or x = 81.

We know that product of numbers = LCM × HCF

So, 1600 = 5 × LCM

And ∴ LCM = 320.

The largest number that can divide both the numbers will be the HCF of the two numbers -

So prime factors of the numbers –

1152 = 2⁷ × 3²

1664 = 2⁷ × 13.

∴ HCF of number = 2⁷ = 128.

∵ we are getting 5 and 8 as remainder.

Subtracting 5 and 8 from the dividends will give us the exact division.

So let’s subtract the remainder and find the HCF of numbers (largest number which divides both the number) -

We need to find HCF of 70 - 5 = 65 and 125 - 8 = 117.

Prime factorization of numbers -

65 = 5 × 13

117 = 3 × 3 × 13

HCF of given numbers = 13.

∵ we are getting 5 as remainder.

Subtracting 5 from the dividend will give us the exact division.

So let’s subtract the remainder and find the HCF of numbers (largest number which divides both the number) -

We need to find HCF of 245 - 5 = 240 and 1029 - 5 = 1024.

Prime factorization of numbers -

240 = 2 × 2 × 2 × 2 × 3 × 5

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

HCF of given numbers = 2 × 2 × 2 × 2 = 16.

Prime factors of the numbers –

1095 = 3 × 5 × 73

1168 = 2⁴ × 73

Dividing them -

We get numerator as 3 × 5 = 15 and denominator as 2⁴ = 16.

The theorem itself.

Given: A number when divided by 143 leaves 31 as remainder.

To find: the remainder when the same number is divided by 13.

Solution:

Let the number be a.

By Euclid’s division lemma -

Putting the values of divisor as 143 and remainder as 31.

a = 143(q) + 31, where q is the quotient when divided by 143.

a = {13(11)}q + 31

a = 13(11)q + 13(2) + 5

a = 13(11q + 2) + 5 ..... (i)

Again when it is divided by 13, let the remainder be r.

So, a = 13(p) + r ......(ii)

where p is the quotient when divided by 13.

Comparing (i) and (ii) -

11q + 2 = p and r = 5.

So remainder = 5.

An irrational number is non - terminating non-repeating decimal and can’t be expressed in p/q form. So here option D is neither terminating non-repeating decimal.

∵ π is non - terminating non-periodic in nature and does not satisfy the rational number definition. It is irrational.

∵ the given number is non - terminating but repeating decimal. ∴ it’s a rational number.

∵ the given number is neither terminating nor repeating decimal, it’s an irrational number.

∵ the number is non - terminating but repeating decimal as after 2 places of decimal it repeats. ∴ it’s a rational number.

For the terminating decimal the denominator must be in the form 2ⁿ × 5^m where m and n are non - negative integers.

For option C denominator is 625 which can be written as 2⁰ × 5⁴.

∴ option C is the answer.

Denominator = 2² × 5 = 2 × 10.

So, expression is equal to = = 1.85

∴ it terminates after two decimal places.

The expression = 11.8024, which terminates after 4 places of decimal.

∵ the given number is terminating decimal.

∴ it’s a rational number.

Least prime factor being 3 or 5 means that 2 is not a factor.

So the numbers are both odd.

a + b will be even no. as a is odd and b is odd {∵ odd + odd = even}

Least prime factor of an even no is 2.

is a non - terminating non - repeating decimal. So, it’s an irrational number.

Assume to be rational. So we can write it in the form of a/b where a and b are co - prime.

So , a/b =

And ∴ b/a = √2

Since b/a is rational but √2 is irrational.

By contradiction is irrational.

Sum of a rational and an irrational number is an irrational number.

Here 2 is rational and is irrational. So their sum is irrational.

Least number that is divisible by all the natural numbers from 1 to 10 will be LCM of the numbers -

LCM of numbers from 1 to 10 = product of prime factors with highest powers = 2³ × 3² × 5 × 7 = 2520.

Here the denominator 150 = 2 × 3 × 5² which is not of the form 2ⁿ × 5^m.Hence the given fraction is non - terminating in nature.

Also (multiply and divide by 2)

For a terminating decimal, the denominator must be in the form 2ⁿ × 5^m where m and n are non negative integers.

For option B denominator is 80 which can be written as 2⁴ × 5.

∴ option B is the answer.

Using Euclid’s division lemma –

n = 9(q) + 7, where q is the quotient when divided by 9.

3n - 1 = 3(9q + 7) - 1
⇒ 3n - 1 = 27q + 21 - 1
⇒ 3n - 1 = 27q + 20
Now, we have to set the above equation in a way, such that
3n - 1 = 9q' + r', where 0 ≤ r < 9

⇒ 3n - 1 = 27q + 18 + 2
⇒ 3n - 1 = 9(3q + 2) + 2

According to Euclid's divison lemma, we have
r = 2, when 3n - 1 is divided by 9

0. = and 0. = .

∴,0. + 0. =

For a number to end with zero, it must contain 5 as one of it’s prime factor.

In case of 4ⁿ, 5 can’t be a prime factor. So it can’t end with zero.

Let the other number be x.

We know product of numbers = LCM × HCF.

So, x × 81 = 27 × 162

And x = 54.

The denominator 30 can be written as 2 × 3 × 5, which is not in the form 2ⁿ × 5^m.

∴ it is not a terminating decimal.

Prime factors of the numbers are -

148 = 2 × 2 × 37

185 = 5 × 37

So HCF = 37

Dividing the numbers we get numerator as 4 and denominator as 5.

, , π, 0.232332333…

An irrational number is non - terminating, non - repeating and cannot be represented in a fractional form. So the decimals which are non - terminating and non - repeating are irrationals.

√3 is an irrational number and addition of a rational number and an irrational number is an irrational number.

∴, 2 + √3 is irrational, where √3 is an irrational number while 2 is a rational number.

Alternative - Assume that 2 + √3 is rational
, where a and b are integers.


we know that a, b and 2 are integers, therefor a - 2b is also an integer
therefore √3 will be rational.
but we know that √3 is irrational.
there is a contradiction
so, 2 + √3 is an irrational number

12 = 2 × 2 × 3

15 = 3 × 5

18 = 2 × 3 × 3

27 = 3 × 3 × 3

So HCF = 3

And LCM = product of prime factors with highest powers = 2² × 3³ × 5 = 540

Let’s take (2 + ) and (2 ‒ ).

These are irrational numbers.

Their sum = (2 + ) + (2 ‒ ) = 4 {which is rational}.

4620 = 2 × 2 × 3 × 5 × 7 × 11 = 2²× 3 × 5 × 7 × 11

Prime factors of numbers are -

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5

HCF = 2 × 2 × 2 × 3 × 3 = 72

HCF of numerators (8,10,16) = 2 and HCF of denominators (9,27,81) = 9.

LCM of numerators(8,10,16) = 80 and LCM of denominators(9,27,81) = 81.

So HCF of fraction == .

and LCM of fraction == .

∵ we are getting 6 and 8 as remainder.

Subtracting 6 and 8 from the dividends will give us the exact division.

So let’s subtract the remainder and find the HCF of numbers (largest number which divides both the number) -

We need to find HCF of 546 - 6 = 540 and 764 - 8 = 756.

Prime factorization of numbers -

540 = 2 × 2 × 3 × 3 × 3 × 5

756 = 2 × 2 × 3 × 3 × 3 × 7

HCF of given numbers = 2 × 2 × 3 × 3 × 3 = 108

Let us assume that √3 is a rational number.

That is, we can find integers a and b (≠ 0) such that √3 = (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 co - prime.

So, √3b = a

⇒ 3b² = a² (Squaring on both sides)

Therefore, a² is divisible by 3.

∴ ‘a’ is also divisible by 3.

So, we can write a = 3c for some integer c.

Equation (1) becomes,

3b² = (3c)²

3b² = 9c²

b² = 3c²

This means that b² is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are co - prime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.

So, we conclude that √3 is irrational.

Let a be any odd positive integer and b = 4.

By division lemma there exist integer q and r such that

a = 4 q + r, where 0 ≤ r ≤ 4

so a = 4q or, a = 4q + 1 or, a = 4q + 2 or, a = 4q + 3

4q + 1 4 is divisible by 2 but 1 is not divisible by 2,so it is an odd number

4q + 2 4 is divisible by 2 and 2 is also divisible by 2,so it is an even number

4q + 3 4 is divisible by 2 but 3 is not divisible by 2,so it is an odd number

4q + 4 4 is divisible by 2 and 4 is also divisible by 2,so it is an even number

∴, any odd integer is of the form 4q + 1 or, 4q + 3.

We applied Euclid Division algorithm on n and 3.

a = bq + r on putting a = n and b = 3

n = 3q + r, 0 < r < 3

So,

n = 3q (I)

n = 3q + 1 (II)

n = 3q + 2 (III)

Case - I: When n = 3q

In this case, we have

n = 3q, which is divisible by 3

Now, n = 3q

n + 2 = 3q + 2

n + 2 leaves remainder 2 when divided by 3

Again, n = 3q

n + 4 = 3q + 4 = 3(q + 1) + 1

n + 4 leaves remainder 1 when divided by 3

n + 4 is not divisible by 3.

Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3.

Case - II: when n = 3q + 1

In this case, we have

n = 3q + 1,

n leaves remainder 1 when divided by 3.

n is divisible by 3

Now, n = 3q + 1

n + 2 = (3q + 1) + 2 = 3(q + 1)

n + 2 is divisible by 3.

Again, n = 3q + 1

n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2

n + 4 leaves remainder 2 when divided by 3

n + 4 is not divisible by 3.

Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3.

Case - III: When n + 3q + 2

In this case, we have

n = 3q + 2

n leaves remainder 2 when divided by 3.

n is not divisible by 3.

Now, n = 3q + 2

n + 2 = 3q + 2 + 2 = 3(q + 1) + 1

n + 2 leaves remainder 1 when divided by 3

n + 2 is not divisible by 3.

Again, n = 3q + 2

n + 4 = 3q + 2 + 4 = 3(q + 2)

n + 4 is divisible by 3.

Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3.

3√2 is an irrational number and subtraction of a rational number and an irrational number is an irrational number.

∴ 4 + 3√2 is irrational, where 3√2 is an irrational number while 4 is a rational number.

Alternative - Assume that 4 + 3√2 is rational
4 + 3√2 = a/b, where a and b are integers.
⇒ 3√2 = a/b - 4
⇒ 3√2 = a/b – 4b/b
⇒ √2 = (a – 4b) / 3b
we know that a, b, 3 and 4 are integers and they are also rational {i.e RHS is rational}
therefore √2 will be rational.
but we know that √2 is irrational.
there is a contradiction
so, 4 + 3√2 is an irrational number