10th Maths Exercise 1.3

1.  Prove that √5 is irrational.

Solutions: Let us assume, that √5 is rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x 

Squaring on both sides, we get

  (y√5)² = x²

⇒5y² = x²................................ (1)

Thus, x² is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get, 5y² = (5k)²

⇒y² = 5k²

𝒚^𝟐 is divisible by 5 it means y is divisible by 5.

                      Therefore, x and y are co-primes. Since, our assumption about √5 is rational is incorrect. Hence, √5 is irrational number.

Q2. Prove that 3 + 2√5 + is irrational.

 

Sol: Let us assume 3 + 2√5 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y Rearranging, we get,

Since, x and y are integers, thus,

                                    

Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational. So, we conclude that 3 + 2√5 is irrational.

Q3.  Prove that the following are irrationals:

(i) 1/√2          (ii)    7√5              (iii)    6 + √2

Solutions:

(i) 1/√2

 

Sol: Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational. Hence, we can conclude that 1/√2 is irrational.

(ii)    7√5

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational. Hence, we can conclude that 7√5 is irrational.

(iii)    6 +√2

                         Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅ Rearranging, we get,

√2 = (x/y) - 6

Since, x and y are integers, thus (x/y) - 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.