NCERT Solutions For Class 10 Maths Chapter – 1 Exercise – 1.3

NCERT Solutions For Class 10 Math Chapter – 1 Exercise – 1.3

 

Q1. Prove that 5 is irrational.

Solution:-

 

Let take √5 as irrational number.

If a and b are two co – prime and b is not equal to 0.

We can write√5 = a/b

b√5 = a

To remove root , squaring on both sides , we get

5b2 = a2

 

Therefore , 5 divides a2 and according to theorem of rational number , for any prime number p which is divides a2  then it will divides a also.

 

That means 5 will divides 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 divisible 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 +25 is irrational.

Solution:-

 

Let take that 3 + 2√5 is a irrational 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 integers 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.

 

Q3. Prove that the following are irrationals:

  • 1/2

Solution:-

 

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.

 

  • Let take that 7√5 is 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 numbers so √5 should be a rational number but √5 is irrational number so it contradicts.

Hence , 7√5 is a irrational number.

 

 

  • 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 sides 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 rational number.

But √2 is a irrational number so it contradicts.

Hence 6 + √2 is a irrational number.