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Real Numbers Ex-1.3 Ch-1 [Solutions] NCERT Maths Class 10th

Exercise 1.3

Q1. Prove that √5 is irrational.

Answer: Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 … (i)

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get
b2/5 = c2
Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.

Q2. 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 so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.

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 so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a 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 so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a 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 so √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.



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