Real Numbers

Concept used :
To obtain the HCF of two positive integers, say c and d, with c > d,we follow
the steps below:
Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

(i) We know that,

= 225>135

Applying Euclid’s division algorithm:
(Dividend = Divisor × Quotient + Remainder)

225 = 135 ×1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90×1+45

Here remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence,

HCF of (135, 225) = 45

(ii)
We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0 (Dividend = Divisor × Quotient + Remainder)

Remainder = 0

Hence,

HCF of (196, 38220) = 196

(iii)
We know that,

867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102 (Dividend = Divisor × Quotient + Remainder)

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence,

(HCF 0f 867 and 255) = 51

To Prove: Any Positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5

Proof: To prove the statement by Euclid's lemma we have to consider divisor as 6 and then find out the possible remainders when divided by 6

By taking,’ a’ as any positive integer and b = 6.

Applying Euclid’s algorithm

a = 6 q + r
As divisor is 6 the remainder can take only 6 values from 0 to 5

Here, r = remainder = 0, 1, 2, 3, 4, 5 and q ≥ 0

So, total possible forms are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5

6q + 0 , (6 is divisible by 2, its an even number)

6q + 1, ( 6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, ( 6 and 4 both are divisible by 2, its an even number)

6q + 5 , (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 6q + 1, or 6q + 3, or 6q+5
Hence, Proved.

By using, Euclid’s division algorithm

616 = 32×19+8

Remainder ≠ 0

So, again Applying Euclid’s division algorithm

32 = 8×4+0

HCF of (616, 32) is 8.

So,

They can march in 8 columns each.

To Prove: Square of any number is of the form 3 m or 3 m +1
Proof: to prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3m and 3m + 1 as the form
So,
By taking, ’ a’ as any positive integer and b = 3.

Applying Euclid’s algorithm a = bq + r.

a = 3q + r

Here, r = remainder = 0, 1, 2 and q ≥ 0 as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.

So, putting all the possible values of the remainder in, a = 3q + r
a = 3q or 3q+1 or 3q+2

And now squaring all the values,
When a= 3q
Squaring both sides we get,

a² = (3q)²
a² = 9q²
a² =3 (3q²)
a² = 3 k₁
Where k₁ = 3q²
When a=3q+1
Squaring both sides we get,
a² = (3q + 1)²
a² = 9q² + 6q + 1
a² =3( 3q² + 2q )+ 1
a² = 3k₂ + 1
Where k₂= 3q² + 2q
When a = 3q+2
Squaring both sides we get,
a² = (3q + 2)²
a² = 9q² + 12q + 4

a² = 9q² + 12q + 3+1
a² = 3( 3q² + 4q + 1) +1

a² = 3k₃ + 1

Where k₁, k₂ and k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma:
a = bq + r So, we have the following cases:

Case I When a = 3q

In this case, we have

a³ = (3q)³ = 27q³ = 9(3q³ ) = 9m, where m = 3q³

Case II When a = 3q + 1

In this case, we have

a³ = (3q + 1)³

⇒27q³ + 27q² + 9q + 1

⇒9q(3q² + 3q + 1) + 1

⇒a³ = 9m + 1, where m = q(3q² + 3q + 1)

Case III When a = 3q + 2

In this case, we have

a³ = (3q + 1)³

⇒27q³ + 54q² + 36q + 8

⇒9q(3q² + 6q + 4) + 8

⇒a³ = 9m + 8, where m = q(3q² + 6q + 4)

Hence, a³ is the form of 9m or, 9m + 1 or, 9m + 8

(i) 140 = 2×2×5×7 = 2²×5×7

(ii) 156 = 2×2×3×13 = 2²×3×13

(iii) 3825 = 3×3×5×5×17 = 3²×5²×17

(iv) 5005 = 5×7×11×13

(v) 7429 = 17×19×23

(i) 26 = 2×13

91 = 7×13

HCF =13

LCM = 2×7×13 =182

Product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

Hence, product of two numbers = HCF×LCM

(ii) 510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460

HCF×LCM = 46920

Hence, product of two numbers = HCF×LCM

(iii) 336 = 2×2×2×2×3×7

336 = 2⁴×3×7

54 = 2×3×3×3

54 =2×3³

HCF = 2×3 = 6

LCM = 2⁴×3³×7 = 3024

Product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

Hence, product of two numbers = HCF×LCM

Prime factors of any number is the representation of a number as a product of prime numbers it is composed of
for example: Prime factors of 20 = 2 x 2 x 5
HCF = Highest common factor = The product of the factors that are common to the numbers
LCF = Least Common Factor = Product of all the factors of numbers without duplicating the factor
(i)
let us write prime factors of the given numbers
12 = 2×2×3 = 2²×3

15 = 3×5

21 = 3×7
Only 3 is common in all the three numbers, therefore

HCF = 3

LCM = 2²×3×5×7 = 420

(ii) Let us write the prime factors of the given numbers
17 = 1×17

23 = 1×23

29 = 1×29

HCF = 1

LCM = 17×23×29 = 11339

(iii) Let us write the prime factors of the given numbers
8=2×2×2

9 = 3×3

25 = 5×5

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

Given: HCF of (306, 657) = 9

We know that,

LCM × HCF = product of two numbers

LCM×HCF = 306×657

LCM =

LCM = 22338

We need to find can 6ⁿ end with zero
If any number has last digit 0,

Then, it should be divisible by 10
Factors of 10 = 2×5

So,

Value 6ⁿ should be divisible by 2 and 5

Prime factorisation of 6ⁿ = (2×3)ⁿ

Hence,

6ⁿ is divisible by 2 but not by 5.

It can not end with 0.

Composite numbers are those numbers, which can be written in the form of the product of two or more integers, and at least one of them should not be 1
(i)
7 × 11 × 13 + 13

Taking 13 as common, we get,

= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 13 × 6

Hence, it is a composite number.
Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as common , we get,

= 5×(7×6×4×3×2×1+1)

= 5× (1008+1)

= 5×1009

Hence, it is a composite number.

Both of them start from the same point and start moving in same direction
Sonia takes 18 minutes and Ravi takes 12 minutes to complete the circle.
After 12 minutes Ravi will be back to the starting point and Sonia must have covered (12/18) = 2/3 of the rounds
After 12 more minutes Ravi has completed 2 rounds and Sonia must have covered (24/18) = 4/3 of the rounds
After 12 more minutes Ravi has completed 3 rounds and Sonia must have completed (36/18) = 2 rounds
Hence after 36 minutes both will again meet at the starting point.
Alternate Method:
They will meet again after LCM of both values at starting point.

18 = 2×3×3

And

12 = 2×2×3

LCM of 12 and 18 = 2×2×3×3 = 36

Therefore,

Ravi and Sonia will meet together at the starting point after 36 minutes.

Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)² = a²

⇒ 5b² = a²

⇒ a²/5 = b²

Hence, 5 divides a²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

5b² = a²

Putting a = 5c,

⇒ 5b² = (5c)²

⇒ 5b² = 25c²

⇒ b² = 5c²

⸫ b²/5 = c²

Hence, 5 divides b²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational

To Prove: 3 + 2√5 is irrational
Proof:

Let is rational.

Therefore,

We can find two integers p & q where, (q ≠ 0) such that

Since p and q are integers, will also be rational and therefore, is rational.

We know that √5 is irrational but according to above statement it has to be rational

So, both the comments are contradictory,

Hence, the number should have been irrational to make the statement correct.

Therefore, is irrational

(i) Let is rational.

Therefore, we can find two integers p & q where, q ≠0 such that

is rational as p and q are integers.

Therefore, is rational which contradicts to the fact that is irrational.

Hence, our assumption is false and is irrational.

(ii) Let is rational.

Therefore, we can find two integers p & q where, q≠0 such that

for some integers p and q

is rational as p and q are integers.

Therefore, should be rational.

This contradicts the fact that is irrational.

Therefore our assumption that is rational is false.

Hence, is irrational.

(iii) Let be rational.

Therefore, we can find two integers p & q where q≠0, such that

Since p and q are integers, is also rational

Hence, should be rational, this contradicts the fact that is irrational.

So, our assumption is false and hence, is irrational.

(i)

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 5⁵

The denominator is of the form 5^m

Hence, the decimal expansion of is terminating.

(ii)

Factorize the denominator we get,

8 = 2 × 2 ×2 = 2³

The denominator is of the form 2^m

Hence, the decimal expansion of is terminating.

(iii)

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^m × 5ⁿ, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)

Factorize the denominator we get,

1600 = 2⁶×5²

The denominator is in the form 2^m × 5ⁿ

Hence, the decimal expansion of is terminating.

(v)

Factorize the denominator we get,

343 = 7³

Since the denominator is not in the form of 2^m × 5ⁿ, it has 7 as its factors.

So, the decimal expansion of non-terminating repeating.

(vi)

The denominator is in the form 2^m×5ⁿ

Hence, the decimal expansion of is terminating.

(vii)

Since, the denominator is not in the form of 2^m × 5ⁿ, as it has 7 in denominator.

So, the decimal expansion of is non-terminating repeating.

(viii)

The denominator is in the form 5ⁿ

Hence, the decimal expansion of is terminating.

(ix)

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2^m×5ⁿ

Hence, the decimal expansion of is terminating.

(x)

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2^m × 5ⁿ, as it has 3 in denominator.

So, the decimal expansion of non-terminating repeating.

(i)

(ii)

(iv)

(vi)

(viii)

(ix)

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form and q is of the form 2^m×5ⁿ

That is, the prime factor of q will be 2 or 5 or both.

(ii) 0.120120012000120000...

The decimal expansion is neither terminating nor recurring.

Therefore, the given number is an irrational number.

(iii)

Since the decimal expansion is non-terminating but recurring, the given number is a rational number of the form and q is not of the form 2^m×5ⁿ that is, the prime factors of q will also have a factor other than 2 or 5.