Ques 21: Prove that 52√3 is an irrational number.
Sol:
Let 52√3 is a rational number
∴ 52√3= p/q where p and q are coprime 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 (52√3) is a rational number is not correct.
∴ (52√3) is an irrational number.
Ques 22: 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 coprime 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 23: 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 24: 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 :
CaseI: 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 3CaseII: 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 3CaseIII: 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 25: Show that (2+√5 )is an irrational number.
Sol:
Let (2+√5) is a rational number.
∴ (2+√5) = p/q , such that p and q are coprime 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.
Ques 26: Using Euclid’s division algorithm, find the HCF of 56, 96 and 404.
Sol:
Using the Euclid’s division algorithm for 56 and 96
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.
Ques 27: Prove that 3√5 is an irrational number.
Sol:
Let (3√5) is a rational number.
∴ 3√5 = p/q , such that p and q are coprime integers and q ≠ 0.⇒
⇒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 28: Prove that (5+ √2) is irrational.
Sol:
√2 = a/b where ‘a’ and ‘b’ are coprime 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 29: Prove that 2√37 is an irrational.
Sol:
Let 2√37 is rational.
⇒
∵p and q are integers.
∴ is rational.
⇒ √3 is rational.
But we know that √3 is irrational,
∴ Our assumption that (2√37) is rational is wrong.
Hence 2√37 is irrational.
Ques 30: 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,
CaseI: When
n = 4q + 1
∴ n^{2 }= (4q + 1)^{2} = 16q^{2} + 8q + 1
⇒ n^{2} – 1 = 16q^{2} + 8q + 1 – 1
= 16q^{2} + 8q
= 8q (2q + 1), which is divisible by 8.
⇒ n^{2} – 1 is divisible by 8.CaseII: 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 31: Prove that if x and y are both odd positive integers, then x^{2} + y^{2} is even but not divisible by 4.
Sol:
Let ‘q’ be an integer, then any odd positive integer is of the form 2q + 1
Again let for some integers m and n are such that
x = 2m + 1 and y = 2n + 1
∴ x^{2} + y^{2} = (2m + 1)^{2} + (2n + 1)^{2}
= 4m^{2} + 4m + 1 + 4n^{2} +4n + 1
= 4(m^{2} + n^{2}) + 4
(m + n) + 2
= 4[(m^{2} + n^{2}) + (m + n)] + 2
= 4q + 2
[Where q = (m^{2} + n^{2}) + (m + n)]
⇒ x^{2} + y^{2 }= 2[2q + 1], which an even number
Thus, 4q + 2 is an even number that is not divisible by 4, i.e. it leaves remainder 2.
Hence, x^{2} + y^{2} is even but not divisible by 4.
Ques 32: Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
Sol:
Let ‘m’ be an integer such that 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 can be any positive integer.
∴ An odd positive integer can be of the form 6m + 1, 6m + 3 or 6m + 5
Now, we have
(6m + 1)^{2 } = 36m^{2} + 12m + 1
= 6(6m^{2} + 2m) + 1
= 6q + 1, q is an integer
(6m + 3)^{2} = 36m^{2} + 36m + 9
= 6(6m^{2} + 6m + 1) + 3
= 6q + 3, q is an integer
(6m + 5)^{2 }= 36m^{2} + 60m + 25
= 6(6m^{2} + 10m + 4) + 1
= 6q + 1, q is an integer
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.
Ques 33: Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
Sol:
Let q be an integer such that 5q, (5q + 1) or (5q + 2) is a positive integer. Let us consider the following cases:
CaseI: 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.CaseII: 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.CaseIII: 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.CaseIV: 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.CaseV: 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 34: 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 integerare 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 35: 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.
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1. What are real numbers? 
2. How do we classify numbers as real numbers? 
3. What is the difference between rational and irrational numbers? 
4. How can real numbers be used in everyday life? 
5. Can all real numbers be represented on the number line? 

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