Question

prove the following are irrational

(1) 1/√2
(2) √3+√5
(3) 6 + √2
(4) √5
(5) 3 + 2√5

Answer 1

1) 1/√2=p/q (assume that it is a rational )

 √2=Q/p  ... So it is irrational......

2) √3+√5=p/q (assume that it is a rational )

squaring on both sides 

(√3+√5)*2=(p/q)*2

3+5+2√3*√5=p*2/q*2

2√15=p*2/q*2-8

√15=p2-8q2/2q*2  hence it is irrational

3) 6 + √2
 6 + √2=p/q
√2=p/q-6

hence it is irrational   

5)3 + 2√5
3 + 2√5=p/q


 2√5=p/q-3

√5=(p/q-3)/2

hence it is irrational...........

Answer 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 is contradict

Hence result is 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 is contradict

Hence result is 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 is contradict

Hence result is 6 + √2 is a irrational number