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 ⇒ 𝑎2 = 5𝑏2
Therefore, a2 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 = 5𝑏2 ⇒ 5𝑘2 = 𝑏2
This means that b2 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.
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