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**Ques 19: Prove that 5-2**√3** is an irrational number.****Sol:** Let 5-2√3 is a rational number

∴ 5-2√3= p/q where p and q are co-prime integers and q ≠ 0.

Since, p and q are integers.

∴ p/2q is a rational number

i.e., is a rational number.

⇒√3 is a rational number.

But this contradicts the fact that √3 is an irrational number.

So, our assumption that (5-2√3) is a rational number is not correct.

∴ (5-2√3) is an irrational number.**Ques 20: Prove that (5+3**√**2)** **is an irrational number.****Sol: **Let (5+3√2) is a rational number.

∴ (5+3√2) = a/b [where ‘a’ and ‘b’ are co-prime integers and b ≠ 0]

⇒

⇒

⇒

‘a’ and ‘b’ are integers,

∴ is a rational number.

⇒ √2 is a rational number.

But this contradicts the fact that √2 is an irrational number.

∴ Our assumption that (5+3√2) is a rational is incorrect.

⇒ (5+3√2) is an irrational number.**Ques 21: Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form ‘3m’ or ‘3m + 1’ for some integer ‘m’.****Sol: **Let x be a positive integer

∴ x can be of the form 3p, (3p + 1) or (3p + 2)

When x = 3p, we have

x^{2} = (3p)^{2}

⇒ x^{2} = 9p^{2}

⇒ x^{2} = 3 (3p^{2})

⇒ x^{2} = 3m [Here 3p^{2} = m]

When x = (3p + 1), we have

x^{2 }= (3p + 1)^{2}

⇒ x^{2} = 9p^{2} + 6p + 1

⇒ x^{2} = 3p (3p + 2) + 1

= 3m + 1,

Where m =p (3p +2)

When x = 3p +2 , we have

x^{2 }= (3p + 2)^{2}

= 9p^{2} + 12p + 4

= 9p^{2} + 12p + 3 + 1

= 3 (3p^{2} + 4p + 1) + 1

= 3m + 1, where

m = 3p^{2} + 4p + 1

Thus, x^{2} is of the form 3m or 3m + 1.**Ques 22: Show that 2+√3 is an irrational number.Sol: **Let 2+√3

such that p and q are co-prime integers and q ≠ 0.

Since, p and q are integers.

∴ is rational

⇒ √3 is rational.

But this contradicts the fact that √3 is an irrational.

∴ Our assumption that (2+√3) is a rational is incorrect.

Thus, (2+√3) is an irrational.**Ques 23: Show that one and only one of n, n + 2 and n + 4 is divisible by 3. ****Sol: **Let us divide n by 3.

Let us get ‘q’ as quotient and ‘r’ remainder.

∴ n = 3 × q + r,where 0 ≤ r < 3

i.e., r = 0, 1, 2

when r = 0, then n = 3q ... (1)

when r = 1, then n = 3q + 1 ... (2)

when r = 2, then n = 3q + 2 ... (3)

From (1), n is divisible by 3

From (2), n = 3q + 1

Adding 2 to both sides, we have

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

⇒ n + 2 = 3q + 3

⇒ n + 2 = 3 (q + 1)

3 (q + 1) is divisible by 3,

∴ n + 2 is divisible by 3.

From (3),

n = 3q + 2

Adding 4 to both sides,

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

⇒ n + 4 = 3q + 6 = 3 (q + 2)

3 (q + 2) is divisible by 3.

∴ n + 4 is divisible by 3.

At one time, r has only one value out of 0, 1, 2. Hence, only one of n, n + 2, n + 4 is divisible by 3.

OR

Let q be an integer such that 3q, (3q + 1) or (3q + 2) is a positive integer. Let us consider the following cases :

Case-I: When n = 3q

3q ÷ 3, gives 0 as remainder

∴ n = 3q is divisible by 3.

Next n = 3q ⇒ n + 2 = 3q + 2

(3q + 2) ÷ 3, gives 2 as remainder.

⇒ n + 2 = (3q + 2) is not divisible by 3.

Again, n = 3q ⇒ n + 4 = 3q + 4 = (3q + 3) + 1

∴ [(3q + 3) + 1] ÷ 3, gives 1 as remainder.

⇒ n + 4 = (3q + 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

Since (3q + 1) ÷ 3, gives remainder as 1.

∴ n = (3q + 1) is not divisible by 3.

Next, n + 2 = (3q + 1) + 2 = (3q + 3) + 0

[(3q + 3) + 0] ÷ 3, gives remainder as 0

⇒ n + 2 = (3q + 1) + 2 is divisible by 3.

Again, n = (3q + 1) ⇒ n + 4 = (3q + 1) + 4 = (3q + 3) + 2

[(3q + 3) + 2] ÷ 3, gives remainder as 2

⇒ n + 4 = (3q + 1) + 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

Since (3q + 2) ÷ 3, gives remainder as 2

∴ n = 3q + 2 is not divisible by 3

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

[(3q + 3) + 1] ÷ 3, gives remainder as 1.

⇒ n + 2 = (3q + 2) + 2 is not divisible by 3

Again, n = 3q + 2 ⇒ n + 4 = (3q + 2) + 4 = (3q + 6) + 0

[(3q + 6) + 0] ÷ 3, gives remainder as 0

⇒ n + 4 = (3q + 2) + 4 is divisible by 3

Thus, (n + 4) is divisible by 3, but n and (n + 2) are not divisible by 3.**Ques 24: Show that (2+√5 )is an irrational number.Sol: **Let

∴ (2+√5) = p/q , such that p and q are co-prime integers and q ≠ 0

p and q are integers.

∴ is a rational.

⇒ √5 is a rational.

But, this contradicts the fact that √5 is an irrational.

∴ Our supposition that (2+√5) is rational is incorrect.

Thus,(2+√5)is an irrational.

We have 96 = 56 × 1 + 40 and

56 = 40 × 1 + 16

Similarly 40 = 16 × 2 + 8

and 16 = 8 × 2 + 0

Remainder is zero

∴ HCF of 56 and 40 is 8.

Now, 404 = 8 × 50 + 4

and 8 = 4 × 2 + 0

∵ Remainder is zero

∴ HCF of 404 and 8, is 4

Thus, the HCF of 56, 96 and 404 is 4.

∴ 3-√5 = p/q

⇒

⇒

Since, p and q are integers,

∴ is a rational number.

⇒ √5 is a rational number.

But this contradicts the fact that √5 is an irrational number.

∴ Our assumption that (3-√5) is a rational number’ is incorrect.

⇒ (3-√5) is an irrational number.**Ques 27: Prove that (5+ √2) is irrational.****Sol:** √2 = a/b where ‘a’ and ‘b’ are co-prime integers and b ≠ 0

⇒

⇒

since ‘a’ and ‘b’ are integers,

∴ is a rational.

⇒ **√**2 is a rational.

But this contradicts the fact that **√**2 is an irrational.

∴ Our assumption that (5+ √2) is a rational number is incorrect.

Thus, (5+ √2) is an irrational number.**Ques 28: Prove that 2**√**3-7** **is an irrational.****Sol: **Let 2√3-7** **is rational.

⇒

∵p and q are integers.

∴ is rational.

⇒ √3 is rational.

But we know that √3 is irrational,

∴ Our assumption that (2√3-7) is rational is wrong.

Hence 2√3-7 is irrational.

**Ques 29: If ‘n’ in an integer, then show that n ^{2} – 1 is divisible by 8.Sol:** Le q be an integer, then

4q + 1 or 4q + 3

Now,

Case-I: When

n = 4q + 1

∴ n

⇒ n

= 16q

= 8q (2q + 1), which is divisible by 8.

⇒ n

Case-II: When

n = 4q + 3

∴ n^{2} = (4q + 3)^{2} = 16q^{2} + 24q + 9

⇒ n^{2} – 1= 16q^{2} + 24q + 9 – 1 = 16q^{2} + 24q + 8

= 8(2q^{2} + 3q + 1) which is divisible by 8.

⇒ n^{2} – 1 is divisible by 8.

Thus, (n^{2} – 1) is divisible by 8.**Ques 30: Prove that if x and y are both odd positive integers, then x ^{2} + y^{2} is even but not divisible by 4.**

Again let for some integers m and n are such that

x = 2m + 1 and y = 2n + 1

∴ x

= 4m

= 4(m

(m + n) + 2

= 4[(m

= 4q + 2

[Where q = (m

⇒ x

Thus, 4q + 2 is an even number that is not divisible by 4, i.e. it leaves remainder 2.

Hence, x

∴ An odd positive integer can be of the form 6m + 1, 6m + 3 or 6m + 5

Now, we have

(6m + 1)

= 6(6m

= 6q + 1, q is an integer

(6m + 3)

= 6(6m

= 6q + 3, q is an integer

(6m + 5)

= 6(6m

= 6q + 1, q is an integer

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Sol:

Case-I: When n = 5q

Since 5q ÷ 5, gives remainder as 0 ⇒ n is divisible by 5

Next, n = 5q ⇒ n + 4 = 5q + 4

⇒ (n + 4) is not divisible by 5

and (5q + 4) ÷ 5, gives remainder as 4

Again, n = 5q ⇒ n + 8 = 5q + 8 = (5q + 5) + 3

⇒ (n + 8) is not divisible by 5

and [(5q + 5) + 3] ÷ 5, gives remainder as 3

Similarly,

n = 5q ⇒ n + 12 = 5q + 12 = (5q + 10) + 2

⇒ (n + 12) is not divisible by 5

and [(5q + 10) + 2] ÷ 5, gives remainder as 2

n = 5q ⇒ n + 16 = 5q + 16 = (5q + 15) + 1

⇒ (n + 16) is not divisible by 5

and [(5q + 15) + 1] ÷ 5, gives remainder as 1

Thus, n is divisible by 5, but (n + 4), (n + 8), (n + 12) and (n + 16) are not divisible by 5.

Case-II: When n = (5q + 1)

Here, (5q + 1) ÷ 5, gives remainder as 1 ⇒ n is not divisible by 5

Next n = (5q + 1)

⇒ n + 4 = (5q + 1) + 4 = (5q + 5) + 0

⇒ (n + 4) is divisible by 5

and [(5q + 5) + 0] ÷ 5, gives remainder as 0

Similarly,

n = (5q + 1) ⇒ n + 8 = (5q + 1) + 8

= (5q + 5) + 4

⇒ (n + 8) is not divisible by 5

and [(5q + 5) + 4] ÷ 5 gives remainder as 4

n = (5q + 1) ⇒ n + 12 = (5q + 1) + 12

= (5q + 10) + 3

⇒ (n + 12) is not divisible by 5

and [(5q + 10) + 3] ÷ 5, gives remainder as 3

n = 5q + 1 ⇒ n + 16 = 5q + 1 + 16

= (5q + 15) + 2

⇒ (n + 16) is not divisible by 5

and [(5q + 15) + 2] ÷ 5, gives remainder as 2

Thus, (n + 4) is divisible by 5, but n, (n + 8), (n + 12) and (n + 16) are not divisible by 5.

Case-III: When n = (5q + 2)

Here, (5q + 2) ÷ 5, gives remainder as 2

⇒ n is not divisible by 5

Next n = 5q + 2 ⇒ n + 4 = 5q + 2 + 4

⇒ 5q + 6 = (5q + 5) + 1

⇒ (n + 4) is not divisible by 5

and [(5q + 5) + 1] ÷ 5, gives remainder as 1

Similarly,

n = 5q + 2 ⇒ n + 8 = 5q + 2 + 8 = 5q + 10

⇒ (n + 8) is divisible by 5

and (5q + 10) ÷ 5, gives remainder as 0

n = 5q + 2 ⇒ n + 12 = 5q + 2 + 12

= (5q + 10) + 4

⇒ (n + 12) is not divisible by 5

and [(5q + 10) + 4] ÷ 5, gives remainder as 4

n = 5q + 2 ⇒ n + 16 = 5q + 2 + 16

= (5q + 15) + 3

⇒ (n + 16) is not divisible by 5

and [(5q + 15) + 3] ÷ 5, gives remainder 3

Thus, (n + 8) is divisible by 5, but n, (n + 4) (n + 12) and (n + 16) are not divisible by 5.

Case-IV: When n = (5q + 3)

Here, (5q + 3) ÷ 5, gives remainder as 3 ⇒ n is not divisible by 5.

Next n = 5q + 3 ⇒ n + 4 = (5q + 3) + 4

= (5q + 5) + 2

⇒ (n + 4) is not divisible by 5

and [(5q + 5) + 2] ÷ 5, gives remainder as 2

Similarly,

n = 5q + 3 ⇒ n + 8 = 5q + 3 + 8

= (5q + 10) + 1

⇒ (n + 8) is not divisible by 5

and [(5q + 10) +1] ÷ 5, gives remainder as 1

n = 5q + 3 ⇒ n + 12 = 5q + 3 + 12

= (5q + 15) + 0

⇒ (n + 12) is divisible by 5

and [(5q + 15) + 0] ÷ 5, gives remainder as 0

n = 5q + 3 ⇒ n + 16 = 5q + 3 + 16

= (5q + 15) + 4

⇒ (n + 16) is not divisible by 5

and [(5q + 15) + 4] ÷ 5, gives remainder as 4

Thus, (n + 16) is divisible by 5 but n, (n + 4) , (n + 8) and (n + 12) are not divisible by 5.

Case-V: When n = (5q + 4)

Here, (5q + 4) ÷ 5, gives remainder as 4 ⇒ n is not divisible by 5

Next, n = (5q + 4) ⇒ n + 4 = 5q + 4 + 4

= (5q + 5) + 3

⇒ (n + 4) is not divisible by 5

and [(5q + 5) + 3] ÷ 5, gives remainder as 3

Similarly,

n = 5q + 4 ⇒ n + 8 = 5q + 4 + 8

= (5q + 10) + 2

⇒ (n + 8) is not divisible by 5

and [(5q + 10) + 2] ÷ 5, gives remainder as 2

n = 5q + 4 ⇒ n + 12 = 5q + 4 + 12

= (5q + 15) + 1

⇒ (n + 12) is not divisible by 5

and [(5q + 15) + 1] ÷ 5, gives remainder as 1

n = 5q + 4 ⇒ n + 16 = 5q + 4 + 16

= (5q + 20) + 0

⇒ (n + 16) is divisible by 5

and [(5q + 20) + 0] ÷ 5, gives remainder 0

Thus, (n + 16) is divisible by 5, but n, n + 4, n + 8 and n + 12 are not divisible by 5.**Ques 33: Show that there is no positive integer ‘p’ for which ****is rational.Sol: **If possible let there be a positive integer p for which = a/b is equal to a rational i.e. where a and b are positive integers.

Now

Also,

Since a, b are integers

are rationals

⇒ (p + 1) and (p – 1) are perfect squares of positive integers, which is not possible (because any two perfect squares differ at least by 3). Hence, there is no positive integer p for which is rational.**Ques 34: Prove that is irrational, where p and q are primes.****Sol: **Let be rational

Let it be equal to ‘r’

i.e.

Squaring both sides, we have

⇒

⇒ ...(i)

Since, p, q are both rationals

Also, r^{2} is rational (∵ r is rational)

∴ RHS of (i) is a rational number

⇒ LHS of (i) should be rational i.e.**√**q should be rational.

But **√**q is irrational (∵ p is prime).

∴ We have arrived at a contradiction.

Thus, our supposition is wrong.

Hence, **√**p+√q is irrational.**Ques 35: Prove that ****√2+√3**** is irrational.Sol: **Let us suppose that

Also, let

∴

Squaring both sides, we have

⇒

⇒

⇒

i.e.√3 = a rational number

But it is a contradiction (√3 is irrational)

Hence, **√**2+√3 is irrational.

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