Short Answer Questions: Real Numbers - 2

# Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2

Ques 21: 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 22: Prove that (5+32) 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 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
x2 = (3p)2
⇒ x2 = 9p2
⇒ x2  = 3 (3p2)
⇒ x2 = 3m [Here 3p2 = m]
When x  = (3p + 1),  we have
x=  (3p + 1)2
⇒ x2 = 9p2 + 6p + 1
⇒ x2 = 3p (3p + 2) + 1
= 3m + 1,
Where m =p (3p +2)
When x   = 3p +2 ,  we have
x= (3p + 2)2
= 9p2 + 12p + 4
= 9p2 + 12p + 3 + 1
= 3 (3p2 + 4p + 1) + 1
= 3m + 1,  where
m  = 3p2 + 4p + 1
Thus, x2 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
(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 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 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.

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 co-prime 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 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 29: Prove that 23-7 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 30: If ‘n’ in an integer, then show that n2 – 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= (4q + 1)2 = 16q2 + 8q + 1
⇒ n2 – 1 = 16q2 + 8q + 1 – 1
= 16q2 + 8q
= 8q (2q + 1), which is divisible by 8.
⇒ n2 – 1 is divisible by 8.

Case-II: When
n = 4q + 3
∴ n2  =  (4q + 3)2 = 16q2 + 24q + 9
⇒ n2 – 1=  16q2 + 24q + 9 – 1 = 16q2 + 24q + 8
= 8(2q2 + 3q + 1) which is divisible by 8.
⇒ n2 – 1 is divisible by 8.
Thus, (n2 – 1) is divisible by 8.

Ques 31: Prove that if x and y are both odd positive integers, then x2 + y2 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
∴ x2 + y2 = (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 +4n + 1
= 4(m2 + n2) + 4
(m + n) + 2
= 4[(m2 + n2) + (m + n)] + 2
= 4q + 2
[Where q = (m2 + n2) + (m + n)]
⇒  x2 + y= 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, x2 + y2 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) =  36m2 + 12m + 1
= 6(6m2 + 2m) + 1
= 6q + 1, q is an integer
(6m + 3)2  =  36m2 + 36m + 9
= 6(6m2 + 6m + 1) + 3
= 6q + 3, q is an integer
(6m + 5)= 36m2 + 60m + 25
= 6(6m2 + 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:
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 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 integer

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 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, r2 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.

The document Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2

 1. What are real numbers?
Ans. Real numbers are a set of numbers that include both rational and irrational numbers. They can be represented on the number line and are used to describe quantities and measurements in the real world.
 2. How do we classify numbers as real numbers?
Ans. Numbers can be classified as real numbers if they can be expressed as a decimal or a fraction. This includes whole numbers, integers, rational numbers, and irrational numbers.
 3. What is the difference between rational and irrational numbers?
Ans. Rational numbers can be expressed as a ratio of two integers, while irrational numbers cannot be expressed as a fraction. Irrational numbers have non-repeating and non-terminating decimal representations.
 4. How can real numbers be used in everyday life?
Ans. Real numbers are used in various everyday situations, such as measuring distances, temperatures, and time. They are also used in financial calculations, scientific research, and engineering designs.
 5. Can all real numbers be represented on the number line?
Ans. No, not all real numbers can be represented on the number line. For example, irrational numbers like √2 or π are not exactly representable on a number line because their decimal representations are non-repeating and non-terminating. However, they can be approximated on the number line.

## Mathematics (Maths) Class 10

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