Belongs to the unit Real Numbers

**Prove that**√**5 is irrational.**

**Solution:**

Let √5 is a rational number.

Therefore, we can find two integers a, b (b ≠ 0) such that √5 = 𝑎/𝑏

Let a and b have a common factor other than 1.

Then we can divide them by the common factor, and assume that a and b are co-prime.

𝑎 = √5𝑏 ⇒ 𝑎^{2} = 5𝑏^{2}

Therefore, a^{2} is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

(5𝑘)^{2} = 5𝑏^{2} ⇒ 5𝑘^{2} = 𝑏^{2}

This means that b^{2} is divisible by 5 and hence, b is divisible by 5.

This implies that a and b have 5 as a common factor. And this is a contradiction to the fact that a and b are co-prime. Hence, √5 cannot be expressed as 𝑝/𝑞 or it can be said that √5 is irrational.

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

**Solution:**

Let 3 + 2√5 is rational.

Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 3 + 2√5 = 𝑎/𝑏

⇒ 2√5 = (𝑎/𝑏) − 3

⇒ √5 = ½ x ( (𝑎/𝑏) − 3)

Since a and b are integers, ½ x ( (𝑎/𝑏) − 3) will also be rational and therefore, √5 is rational.

This contradicts the fact that √5 is irrational.

Hence, our assumption that 3 + 2√5 is rational is false.

Therefore, 3 + 2√5 is irrational.

**3. Prove that the following are irrationals : (i) 1/**√**2 (ii) 7** √**5 (iii) 6+** √2

**Solution:**

**(i) 1/**√**2**

Let 1/√2 is rational.

Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 1/√2 = 𝑎/𝑏 Or √2 = 𝑏/𝑎; 𝑏/𝑎 is rational as *a *and *b* are integers.

Therefore, √2 is rational which contradicts to the fact that √2 is irrational.

Hence, our assumption is false and 1/√2 is irrational.

(ii) 7√5

Let 7√5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 7√5 = 𝑎/𝑏

⇒ √5 = 𝑎/7𝑏; 𝑎/7𝑏 is rational as *a and b *are integers.

Therefore, √5 should be rational.

This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.

(iii) 6 + √2

Let 6 + √2 be rational.

Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 6 + √2 = 𝑎/𝑏

⇒ √2 = [(𝑎/𝑏) – 6]

Since *a and b* are integers, [(𝑎/𝑏) – 6] is also rational and hence, √2 should be rational.

This contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence, 6 + √2 is irrational.

## 1 thought on “Real Numbers 1.3”

Comments are closed.