Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Short Answer Questions: Real Numbers - 2

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

Ques 1: 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.

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

Since, p and q are integers.
∴  p/2q is a rational number
i.e., Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2  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 2: 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

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

‘a’ and ‘b’ are integers,
∴  Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 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 3: Prove that 1/√3 is irrational. 

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

Ques 4: Consider the numbers 4n , where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero.

Sol

If the number 4n , for any n, were to end with the digit zero, then it would be divisible by 5. 

That is, the prime factorisation of 4n would contain the prime 5.

This is not possible because 4n = (2)2n ; so the only prime in the factorisation of 4n is 2. 

So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n .

So, there is no natural number n for which 4n ends with the digit zero. 

Ques 5: Prove that Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2is irrational, where p and q are primes.

Sol: 

Let Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2be rational
Let it be equal to ‘r’
i.e. Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2
Squaring both sides, we have

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

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2  ...(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.

Ques 6: 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
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2
p and q are integers.
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2  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 7: 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.

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

Since, p and q are integers,
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 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 8: 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 9: 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.

Ques10:  Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method. 

Sol :

 We have : 6 = 2 × 3, 72 = 23 × 32 , 120 = 23 × 3 × 5 

Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.

 So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 23 , 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. 

So, LCM (6, 72, 120) = 23 × 32 × 51 = 360 

Ques 11: Show that there is no positive integer ‘p’ for which Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2is rational.

Sol: 

If possible let there be a positive integer p for which Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 = a/b is equal to a rational i.e. where a and b are positive integers.

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

Now

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2
Also,

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2
Since a, b are integer

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 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 Class 10 Maths Chapter 1 Question Answers - Real Numbers - 2 is rational.

Ques 12: 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 13: 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.

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

1. What are real numbers and how are they classified?
Ans. Real numbers are all the numbers on the number line, including rational numbers (like integers and fractions) and irrational numbers (like √2 and π). They are classified into groups such as natural numbers, whole numbers, integers, rational numbers, and irrational numbers.
2. How do you perform basic operations with real numbers?
Ans. Basic operations with real numbers include addition, subtraction, multiplication, and division. To perform these operations, follow standard arithmetic rules, ensuring to pay attention to the signs (positive or negative) and the order of operations (PEMDAS/BODMAS).
3. What are the properties of real numbers?
Ans. The properties of real numbers include the commutative property (order doesn't matter for addition and multiplication), associative property (grouping doesn't matter), distributive property (a(b + c) = ab + ac), identity property (adding 0 or multiplying by 1), and inverse property (every number has an additive inverse and multiplicative inverse).
4. How can real numbers be represented on a number line?
Ans. Real numbers can be represented on a number line by marking points that correspond to each number. The distance from the origin (0) to any point represents the value of that number, with positive numbers extending to the right and negative numbers to the left.
5. What is the significance of irrational numbers in real numbers?
Ans. Irrational numbers are significant because they fill in the gaps between rational numbers on the number line. They cannot be expressed as simple fractions, and examples include numbers like √2 and π, which are crucial in various mathematical contexts, such as geometry and calculus.
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