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

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

Q1: The decimal representation of 6/1250 will terminate after how many places of decimal?

1. Simplify the fraction to its lowest form.

2. Now we can either convert the lowest fraction into decimals or a quicker way to do this is to convert the denominator to a multiple of 10 so that it is easier to convert it into decimals.
3. So we multiply with and divide by  2to convert the denominator to 104. 
24 2

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

Therefore, this representation will terminate after 4 decimal places.

Question for Short Answer Questions: Real Numbers
Try yourself:Q.2. If HCF of a and b is 12 and product of these numbers is 1800. Then what is LCM of these numbers?
View Solution
 

Q2: If ‘p’ is prime, prove that √p is irrational.

Let p be rational in the simplest form a/b, where p is prime.
∴ a and b are integers having no common factor other than 1 and b ≠ 0.
⇒ Now, p = a/b
⇒ Squaring both sides, we have
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 
⇒ pb= a2      ...(1)
 Since pb2 is divisible by p, a2 is also divisible by p.
⇒ a is also divisible by p    ...(2)
Let a = pc for some integer c.
⇒ Substituting a = pc in (1), we have
pb2 = (pc)2
⇒ pb= p2c2
⇒ b2  = pc2
pc2 is divisible by p,
∴ b2 is divisible by p
⇒ b is divisible by p    ...(3)
From (2) and (3),
⇒ p is a common factor of ‘a’ and ‘b’. But this contradicts our assumption that a and b are co-prime.
∴ Our assumption that p is rational is wrong. Thus, p is irrational if p is prime.

Q3: Find the HCF of 18 and 24 using prime factorisation.

Using factor tree method, we have:
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
∴ 18 = 2 × 3 × 3 = 2 × 32
24 = 2 × 2 × 2 × 3 = 23 × 3
HCF = Product of common prime factors with lowest powers.
⇒ HCF (18, 24) = 3 × 2  = 6


Q4: Find the LCM of 10, 30 and 120.

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1            Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1             Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1           
∴ 10 = 2 × 5 = 21 × 51
30 = 2 × 3 × 5 = 21 × 31 × 51
120 = 2 × 2 × 2 × 3 × 5 = 23 × 31 × 51
LCM = Product of each prime factor with highest powers
⇒ LCM of 10, 30 and 120 = 23 × 3 × 5 = 120.

Q5: ExpressClass 10 Maths Chapter 1 Question Answers - Real Numbers - 1 as a rational number in the simplest form.

Let x = Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 = 0.666 .....    ...(1)
∴ 10x = 0.666 ..... × 10
10 x = 6.666 .......(2)
Subtracting (1) from (2), we have:
⇒ 10x - x = 6.666 ..... - 0.666 .....
⇒ 9x = 6
⇒ x = 6/9 = 2/3  
Thus,Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1= 2/3


Q6: Express Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 as a rational number in the simplest form.

Let    x  = Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1   = 1.161616 ........(1)
∴ 100x = (1.161616 .....) × 100
⇒ 100x = 116.161616 ........(2)
Subtracting (1) from (2), we have:
100x - x = [116.161616 .....] – [1.161616 .....]
⇒ 99x = 115
⇒ x = 115/99  
Thus,  Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 =115/99

Q7: Use Euclid's division algorithm to find HCF of 870 and 225.

We have 870 = 3 × 225 + 195
⇒ 225 = 1 × 195 + 30
⇒ 195 = 6 × 30 + 15
⇒ 30 = 2 × 15 + 0
∴ HCF (870, 225) is 15.

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

Q8: Find the LCM and HCF of 1296 and 5040 by prime factorisation method.

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  and     Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 
∴ 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
= 24 × 32 × 5 × 7  and
⇒ 1296 =  2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 24 × 34
∴ LCM = Product of each prime factor with highest powers
=  24 × 34 × 5 × 7
= 16 × 81 × 5 × 7 = 45360
HCF = Product of common prime factors with lowest powers
= 24 × 32
= 16 × 9 = 144

Q9: Prove that √3 is irrational.

Let 3 be rational in the simplest form of P/q.
i.e., p and q are integers having no common factor other than 1 and q ≠ 0.
Now,   3 = p/q  
Squaring both sides, we have
⇒ (√3)2 = (p/q)2
⇒ 3 = p2/q2   
⇒ 3q2 = p............(1)
Since 3q2 is divisible by 3
∴ p2 is also divisible by 3
⇒ p is divisible by 3 ..........(2)
Let p = 3c for some integer ‘c’.
Substituting p = 3c in (1), we have:
⇒ 3q = (3c)2
⇒ 3q2 = 9c2
⇒ q2  = 3c2
3cis divisible by 3
∴ q2 is divisible by 3
⇒ q is divisible by 3   ...(3)
From (2) and (3)
3 is a common factor of ‘p’ and ‘q’. But this contradicts our assumption that p and q are having no common factor other than 1.
∴ Our assumption that 3 is rational is wrong.
Thus, 3 is an irrational.

Q10: Show that  3√2 is irrational.

Let 3√2 be a rational number
∴ p/q = 3√2 where p and q are prime to each other and q ≠ 0.
∴ p/3q = √2   ...(1)
Since, p is integer and 3q is also integer (3q≠ 0).
∴ p/3q is a rational number.
From (1), √2 is a rational number.
But this contradicts the fact that √2 is irrational. Therefore, our assumption that 3√2 is rational is incorrect.
Hence, 3√2 is irrational.

Q11: Show that 2 - √3 is an irrational number.

Let  2 - √3 is rational.
∴ It can be expressed as p/q where p and q are integers (prime to each other) such that q ≠ 0.
∴  2 - √3   =  p/q  
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1     ...(1)

∵ p is an integer}
∴ q is an integer}
⇒ p/q is a rational number.
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 is a rational number.   ...(2)
From (1) and (2), √3 is a rational number. This contradicts the fact that √3 is an irrational number.
∴ Our assumption that (2 -√3) is a rational number is not correct. Thus, (2 - √3) is irrational.

Q12: Using Euclid’s division algorithm, find the HCF of 56, 88 and 404. 

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


Q13: Express Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 in the decimal form.

We have

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

[multiplying and dividing by 5]

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

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

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

Q14: Express Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  in the p/q form.

Let    x    =   Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 
or x = 5.4178178178 .....
∴ 10x = 54.178178178 ........(1)
Also 1000 (10x) = 54178.178178178 .....
⇒ 10000x    =    54178.178178178 ........(2)
Subtracting (1) from (2), we have:
⇒ 10000x - 10x    
= [54178.178178.....] - [54.178178 .....]
⇒ 9990x = 54124

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Thus, Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1

Q15: State whether Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  is a rational number or not.

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 = 1.23333..... is a non-terminating repeating decimal.
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 is a rational number.
3/4 is in the form of p/q, where q ≠ 0  [Here 4 ≠ 0]
∴ 3/4 is a rational number.
Since the sum of two rational numbers is a rational number.
Therefore, Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  is a rational number.

Q16: The LCM of two numbers is 45 times their HCF. If one of the numbers is 225 and sum of their LCM and HCF is 1150, find the other number.

One of the numbers = 225
Let the other number = x
Also LCM = 45 (HCF)                 ...(1)
And LCM + HCF =1150
⇒ (45 HCF) + HCF = 1150
⇒ 46 HCF = 1150
⇒ HCF = 1150/46 = 25
From (1),
LCM = 45 × 25
∴ LCM × HCF = (45 × 25) × 25
Now,  LCM × HCF = Product of the numbers
∴  x × 225 = (45 × 25) × 25
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1

= 125
Thus, the required number is 125.

Q17: Three different containers contain 496 litres, 403 litres and 713 litres of a mixture. What is the capacity of the biggest container that can measure all the different quantities exactly?

For the capacity of the biggest container, we have to find the HCF.

HCF: By Long Division method

First find the HCF of two numbers, 496 and 403
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
The HCF of 496 and 403 = 31
Now find the HCF of 31 and 713
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
HCF of 713 and 31 is 31
So, the maximum capacity is 31 liters.


Q18: Prove that 3+√2 is an irrational number.  

Let 3 + √2 is a rational number.
∴ 3 + √2  = a/b such that ‘a’ and ‘b’ are co-prime integers and b ≠ 0.
We have

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

 Since a and b are integers,
∴  Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 is rational.
⇒  Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  √2 is a rational. This contradicts the fact that √2 is irrational.
∴ Our assumption that 3 + √2 is rational is not correct.
⇒ (3 + √2) is an irrational number.

Q19: 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 - 1

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

Q20: 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 - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1

‘a’ and ‘b’ are integers,
∴  Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 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.

Q21: 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.

Q22: Show that one and only one of n, n + 2 and n + 4 is divisible by 3. 

Sol: 

Let n be any integer. When dividing n by 3, there are three possible remainders, 0, 1, or 2. So, n can be written in one of the following forms:

  1. n = 3qn=3q (remainder 0)
  2. n=3q+1 (remainder 1)
  3. n=3q+2 (remainder 2)

Now, we will check the divisibility of nn, n+2, and n+4 in each case:

Case 1: n = 3q
  • n = 3q, so nn is divisible by 3.
  • n + 2 = 3q + 2n+2 = 3q+2, which leaves a remainder of 2 when divided by 3, so n+2 is not divisible by 3.
  • n+4 = 3q+4 = 3(q+1)+1, which leaves a remainder of 1 when divided by 3, so n+4 is not divisible by 3.

Thus, in this case, nnn is divisible by 3, but n+2n + 2n+2 and n+4n + 4n+4 are not.

Case 2: n = 3q+1
  • n=3q+1, which leaves a remainder of 1 when divided by 3, so nn is not divisible by 3.
  • n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1)n+2 = 3q+1+2 = 3q+3 = 3(q+1), so n + 2n+2 is divisible by 3.
  • n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2, which leaves a remainder of 2 when divided by 3, so n + 4n+4 is not divisible by 3.

Thus, in this case, n+2 is divisible by 3, but n and n + 4n+4 are not.

Case 3: n = 3q + 2
  • n = 3q + 2n=3q+2, which leaves a remainder of 2 when divided by 3, so n is not divisible by 3.
  • n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1n+2 = 3q+2+2 = 3q+4 = 3(q+1)+1, which leaves a remainder of 1 when divided by 3, so n + 2n+2 is not divisible by 3.
  • n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2)n+4 = 3q+2+4 = 3q+6 = 3(q+2), so n + 4n+4 is divisible by 3.

Thus, in this case, n + 4n+4 is divisible by 3, but nn and n+2 are not.

Conclusion:

In each of the three cases, exactly one of nn, n + 2n+2, or n + 4n+4 is divisible by 3, and the other two are not. Therefore, one and only one of nn, n+2, or n + 4n+4 is divisible by 3.

Q23: 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 - 1
p and q are integers.
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  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.

Q24: 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 - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1

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

Q25: Prove that (5+ √2) is irrational.

Sol: 

√2 = a/b where ‘a’ and ‘b’ are co-prime integers and b ≠ 0

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

since ‘a’ and ‘b’ are integers,
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 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.

Q26: Prove that 23-7 is an irrational.

Sol: 

Let  2√3-7 is rational.

Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
∵p and q are integers.
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1  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.

Q27: If ‘n’ in an odd  integer, then show that n2 – 1 is divisible by 8.

Sol: 

Le q be an integer, then
4q + 1 or 4q + 3
Now,

 When n is odd
  • Any odd integer can be written as n=2k+1n = 2k + 1n=2k+1, where kkk is an integer.
  • Then: n2=(2k+1)2=4k2+4k+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1           n21=(4k2+4k+1)1=4k(k+1)n^2 - 1 = (4k^2 + 4k + 1) - 1 = 4k(k + 1)
  • Since k(k + 1)k(k+1) is always an even number (the product of two consecutive integers), we can write: n21=8for some integer for some integer m.

Q28: Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

Sol:

Let x = 2m + 1x=2m+1 and y = 2n + 1y=2n+1, where m and nn are integers (since x and y are odd integers).

Now:

x= (2m+1)= 4m2+4m+1       y^2 = (2n + 1)^2 = 4n^2 + 4n + 1y2 =(2n+1)= 4n+ 4n+1

Adding these:

x+ y2 = (4m2+4m+1) + (4n2+4n+1) = 4(m2+n2+m+n) + 2

Thus, x+ y= 4q+2, where q = m^2 + n^2 + m + nq=m2+n2+m+n.

Since x2+y2 is of the form 4q+24q + 2, it is an even number but not divisible by 4 (because it leaves a remainder of 2 when divided by 4).

Therefore, x2+y2x^2 + y^2 is even but not divisible by 4.

Q29: 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.

Q30: 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 n=5k+rn = 5k + r, where kk is an integer and r is the remainder when nn is divided by 5. The possible values of rr are 0, 1, 2, 3, or 4.

We will check each value of rr and see which one leads to a multiple of 5.

Case 1: r = 0
  • n=5k, so nn is divisible by 5.
  • n+4=5k+4n + 4 = 5k + 4, not divisible by 5.
  • n+8=5k+8=5(k+1)+3n + 8 = 5k + 8 = 5(k + 1) + 3, not divisible by 5.
  • n+12=5k+12=5(k+2)+2n + 12 = 5k + 12 = 5(k + 2) + 2, not divisible by 5.
  • n+16=5k+16=5(k+3)+1n + 16 = 5k + 16 = 5(k + 3) + 1, not divisible by 5.

Thus, only nn is divisible by 5.

Case 2: r = 1
  • n=5k+1n = 5k + 1, not divisible by 5.
  • n+4=5k+5n + 4 = 5k + 5, divisible by 5.
  • n+8=5k+9=5(k+1)+4n + 8 = 5k + 9 = 5(k + 1) + 4, not divisible by 5.
  • n+12=5k+13=5(k+2)+3n + 12 = 5k + 13 = 5(k + 2) + 3, not divisible by 5.
  • n+16=5k+17=5(k+3)+2n + 16 = 5k + 17 = 5(k + 3) + 2, not divisible by 5.

Thus, only n+4 is divisible by 5.

Case 3: r = 2
  • n=5k+2n = 5k + 2, not divisible by 5.
  • n+4=5k+6n + 4 = 5k + 6, not divisible by 5.
  • n+8=5k+10n + 8 = 5k + 10, divisible by 5.
  • n+12=5k+14=5(k+2)+4n + 12 = 5k + 14 = 5(k + 2) + 4, not divisible by 5.
  • n+16=5k+18=5(k+3)+3, not divisible by 5.

Thus, only n+8n + 8n+8 is divisible by 5.

Case 4: r = 3
  • n=5k+3n = 5k + 3, not divisible by 5.
  • n+4=5k+7=5(k+1)+2n + 4 = 5k + 7 = 5(k + 1) + 2, not divisible by 5.
  • n+8=5k+11=5(k+2)+1n + 8 = 5k + 11 = 5(k + 2) + 1, not divisible by 5.
  • n+12=5k+15n + 12 = 5k + 15, divisible by 5.
  • n+16=5k+19=5(k+3)+4n + 16 = 5k + 19 = 5(k + 3) + 4, not divisible by 5.

Thus, only n+12 is divisible by 5.

Case 5: r = 4
  • n=5k+4n = 5k + 4, not divisible by 5.
  • n+4=5k+8=5(k+1)+3n + 4 = 5k + 8 = 5(k + 1) + 3, not divisible by 5.
  • n+8=5k+12=5(k+2)+2n + 8 = 5k + 12 = 5(k + 2) + 2, not divisible by 5.
  • n+12=5k+16=5(k+3)+1n + 12 = 5k + 16 = 5(k + 3) + 1, not divisible by 5.
  • n+16=5k+20n + 16 = 5k + 20, divisible by 5.

Thus, only n+16n + 16 is divisible by 5.

Q31: Show that there is no positive integer ‘p’ for which Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1is rational.

Sol: 

If possible let there be a positive integer p for which Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1 = 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 - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1
Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1

Now

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

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

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

Q32: Prove that Class 10 Maths Chapter 1 Question Answers - Real Numbers - 1is irrational, where p and q are primes.

Sol: 

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

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

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

1. What are real numbers?
Ans. Real numbers are all the numbers on the number line, including rational numbers (such as fractions and integers) and irrational numbers (such as square roots and pi).
2. How do you classify real numbers?
Ans. Real numbers can be classified into rational numbers (which can be expressed as a fraction) and irrational numbers (which cannot be expressed as a fraction).
3. Can you give examples of real numbers?
Ans. Examples of real numbers include integers (-3, 0, 5), fractions (1/2, 3/4), decimals (0.25, 3.14), and irrational numbers (sqrt(2), pi).
4. How are real numbers used in everyday life?
Ans. Real numbers are used in everyday life for measurements (such as length, weight, and temperature), calculations (such as budgeting and shopping), and understanding quantities (such as money and time).
5. What is the importance of real numbers in mathematics?
Ans. Real numbers are fundamental in mathematics as they form the basis for arithmetic operations, algebraic equations, and geometry. They are essential for solving real-world problems and understanding the principles of mathematics.
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