📘 Important Concepts & Key Theorems

• Rational Number: A number in the form a/b, where a and b are integers, co-prime (gcd(a, b) = 1), and b ≠ 0.
• Theorem 1.2: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
• Method of Contradiction: To prove a statement is true, we assume it is false (e.g., assuming a number is rational), show that this leads to a logical contradiction, and thereby conclude the original statement is true.

Solution:

Let us prove this by contradiction. Assume to the contrary that √5 is rational.

Therefore, we can find co-prime integers a and b (where b ≠ 0) such that:

√5 = a / b

Squaring both sides, we get:

5 = a² / b²

⇒ 5b² = a²   --- (Equation 1)

This means 5 divides a². By Theorem 1.2, 5 also divides a.

So, we can write a = 5c for some integer c.

Substituting a = 5c in Equation 1:

5b² = (5c)²

5b² = 25c²

⇒ b² = 5c²

This means 5 divides b². By Theorem 1.2, 5 also divides b.

Therefore, a and b have at least 5 as a common factor. But this contradicts our assumption that a and streamb are co-prime (have no common factor other than 1).

This contradiction has arisen because of our incorrect assumption that √5 is rational.

Hence, √5 is irrational. (Proved)

Solution:

Let us assume, to the contrary, that 3 + 2√5 is rational.

Then, there exist co-prime integers a and b (b ≠ 0) such that:

3 + 2√5 = a / b

Rearranging the equation to isolate √5:

2√5 = (a / b) - 3

2√5 = (a - 3b) / b

√5 = (a - 3b) / 2b

Since a and b are integers, a - 3b and 2b are also integers, making (a - 3b) / 2b rational.

This implies that √5 is rational. However, this contradicts the known fact that √5 is irrational.

Therefore, our assumption is false.

Hence, 3 + 2√5 is irrational. (Proved)

(i) 1 / √2

Solution:

Let us assume to the contrary that 1 / √2 is rational.

Then, there exist co-prime integers a and b (where a, b ≠ 0) such that:

1 / √2 = a / b

Taking the reciprocal on both sides:

√2 = b / a

Since a and b are integers, b / a is a rational number. This implies that √2 is rational.

But this contradicts the fact that √2 is irrational.

Hence, 1 / √2 is irrational. (Proved)

(ii) 7√5

Solution:

Let us assume to the contrary that 7√5 is rational.

Then, there exist co-prime integers a and b (b ≠ 0) such that:

7√5 = a / b

Rearranging the terms:

√5 = a / (7b)

Since a, b, and 7 are integers, a / (7b) is rational.

This implies that √5 is rational, which contradicts the fact that √5 is irrational.

Hence, 7√5 is irrational. (Proved)

(iii) 6 + √2

Solution:

Let us assume to the contrary that 6 + √2 is rational.

Then, there exist co-prime integers a and b (b ≠ 0) such that:

6 + √2 = a / b

Rearranging the equation:

√2 = (a / b) - 6

√2 = (a - 6b) / b

Since a and b are integers, (a - 6b) / b is rational.

This implies that √2 is rational, which contradicts the fact that √2 is irrational.

Hence, 6 + √2 is irrational. (Proved)

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