Chapter 1 — REAL NUMBERS

Proof (by contradiction):

Assume that \(\sqrt{5}\) is rational. Then it can be written as:

\[ \sqrt{5} = \frac{a}{b}, \quad b \ne 0,\; a,b \in \mathbb{Z},\; \gcd(a,b)=1 \]

Step 1: Square both sides

\[ (\sqrt{5})^2 = \left(\frac{a}{b}\right)^2 \] \[ 5 = \frac{a^2}{b^2} \] \[ a^2 = 5b^2 \quad \text{(1)} \]

Step 2: Analyze equation (1)

Since \(a^2 = 5b^2\), \(a^2\) is divisible by 5. Therefore, \(a\) must also be divisible by 5.

Let \(a = 5c\), where \(c\) is an integer.

Step 3: Substitute into (1)

\[ a^2 = (5c)^2 = 25c^2 \] Substitute into (1): \[ 25c^2 = 5b^2 \] \[ b^2 = 5c^2 \quad \text{(2)} \]

Step 4: Analyze equation (2)

From \(b^2 = 5c^2\), \(b^2\) is divisible by 5. Hence, \(b\) is also divisible by 5.

Step 5: Contradiction

Thus, both \(a\) and \(b\) are divisible by 5. This means they have a common factor 5, which contradicts the assumption that \(\gcd(a,b)=1\).

Conclusion:

\[ \therefore \; \sqrt{5} \text{ is irrational.} \]

Method 1: Using properties

• \(3\) and \(2\) are rational numbers.
• \(\sqrt{5}\) is irrational.
• \(2\sqrt{5}\) is irrational (product of non-zero rational and irrational).
• \(3 + 2\sqrt{5}\) is irrational (sum of rational and irrational).

\[\therefore \; 3 + 2\sqrt{5} \text{ is irrational}\]

Method 2: Proof by contradiction

Assume that \(3 + 2\sqrt{5}\) is rational.

\[ 3 + 2\sqrt{5} = \frac{a}{b}, \quad b \ne 0,\; \gcd(a,b)=1 \]

Rearranging:

\[ 2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \] \[ \sqrt{5} = \frac{a - 3b}{2b} \]

Since \(a\) and \(b\) are integers, the RHS is rational. This implies \(\sqrt{5}\) is rational, which is a contradiction.

\[\therefore \; 3 + 2\sqrt{5} \text{ is irrational}\]

Proof (by contradiction):

Assume that \(\frac{1}{\sqrt{2}}\) is rational.

\[ \frac{1}{\sqrt{2}} = \frac{a}{b}, \quad b \ne 0,\; \gcd(a,b)=1 \]

Step 1: Square both sides

\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \left(\frac{a}{b}\right)^2 \] \[ \frac{1}{2} = \frac{a^2}{b^2} \] \[ b^2 = 2a^2 \]

Step 2: Analyze

\(b^2\) is divisible by 2 ⇒ \(b\) is divisible by 2.

Let \(b = 2c\)

Step 3: Substitute

\[ b^2 = (2c)^2 = 4c^2 \] \[ 4c^2 = 2a^2 \Rightarrow a^2 = 2c^2 \]

Thus, \(a\) is also divisible by 2.

Step 4: Contradiction

Both \(a\) and \(b\) are divisible by 2, contradicting \(\gcd(a,b)=1\).

Conclusion:

\[ \therefore \; \frac{1}{\sqrt{2}} \text{ is irrational} \]

Proof (by contradiction):

Assume that \(7\sqrt{5}\) is rational.

\[ 7\sqrt{5} = \frac{a}{b}, \quad b \ne 0,\; a,b \in \mathbb{Z},\; \gcd(a,b)=1 \]

Step 1: Divide both sides by 7

\[ \sqrt{5} = \frac{a}{7b} \]

Since \(a\) and \(7b\) are integers, \(\frac{a}{7b}\) is rational.

Step 2: Contradiction

This implies that \(\sqrt{5}\) is rational, which is false.

Conclusion:

\[ \therefore \; 7\sqrt{5} \text{ is irrational} \]

Proof (by contradiction):

Assume that \(6 + \sqrt{2}\) is rational.

\[ 6 + \sqrt{2} = \frac{a}{b}, \quad b \ne 0,\; a,b \in \mathbb{Z},\; \gcd(a,b)=1 \]

Step 1: Rearranging

\[ \sqrt{2} = \frac{a}{b} - 6 \] \[ = \frac{a - 6b}{b} \]

Since \(a\) and \(b\) are integers, \(\frac{a - 6b}{b}\) is rational.

Step 2: Contradiction

This implies \(\sqrt{2}\) is rational, which is false.

Conclusion:

\[ \therefore \; 6 + \sqrt{2} \text{ is irrational} \]