Class 10 - Mathematics
Real Numbers - Exercise 1.3
Top Block 1
Exercise 1.3
Question : 1:Prove that √5 is irrational.
Answer :
Let us suppose that √5 is a rational number. So, it can be represented as p/q form
⇒ √5 = p/q
Take square on both side, we get
⇒ 5 = p2/q2
⇒ 5q2 = p2 …………..1
So p2 is divisible by 5.
⇒ p is divisible by 5.
Let p = 5x (x is a positive integer)
Now p2 = 25c2
From equation 1, we get
5q2 = 25c2
⇒ q2 = 5c2
So, q is divisible by 5
Thus p and q has a common factor 5. It is contradiction of our assumption.
So, √5 is an irrational number.
Question : 2: Prove that 3 + 2√5 is irrational.
Answer :
Let take that 3 + 2√5 is a rational number.
So, we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a – 3b)/b
Now divide by 2 we get
√5 = (a – 3b)/2b
Here, a and b are integer. So (a – 3b)/2b is a rational number.
It means √5 should be a rational number.
But √5 is a irrational number. So, it contradicts the fact.
Hence, 3 + 2√5 is a irrational number.
Mddle block 1
Question : 3:Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
Answer :
(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a, we get
b/a = √2
Here, a and b are integer, so b/a is a rational number. It means √2 should be a rational
number. But √2 is an irrational number. Hence, it is contradict the fact.
Hence, 1/√2 is an irrational number
(ii) Let take that 7√5 is a rational number.
So, we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get,
√5 =a/7b
Here, a and b are integer, so a/7b is a rational number.
It means √5 should be a rational number but √5 is an irrational number.
So, it is contradict the fact.
Hence, 7√5 is an irrational number.
(iii) Let take that 6 + √2 is a rational number.
So, we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side, we get
√2 = a/b – 6
√2 = (a – 6b)/b
Here ,a and b are integer, so (a – 6b)/b is a rational number.
It means √2 should be a rational number but √2 is an irrational number.
So, it is contradict the fact.
Hence, 6 + √2 is an irrational number.
