Q2:

Prove that $3 + 2\sqrt{5}$ is irrational.

Given: A real number $3 + 2\sqrt{5}$.

To Prove: $3 + 2\sqrt{5}$ is an irrational number.

Step 1: Assumption for Contradiction
Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is a rational number. By the definition of rational numbers, if a number is rational, it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., they have no common factors other than 1).

Step 2: Setting up the Equation
Based on our assumption, we can write:
$3 + 2\sqrt{5} = \frac{p}{q}$

Step 3: Isolating the Irrational Part
We perform algebraic manipulations to isolate the term containing the square root.
Subtract 3 from both sides:
$2\sqrt{5} = \frac{p}{q} - 3$
$2\sqrt{5} = \frac{p - 3q}{q}$

Now, divide both sides by 2:
$\sqrt{5} = \frac{p - 3q}{2q}$

Step 4: Analyzing the Rationality of the Expression
Since $p$ and $q$ are integers, the expression $\frac{p - 3q}{2q}$ must also be a rational number because:
1. The difference of two integers ($p - 3q$) is an integer.
2. The product of two integers ($2q$) is an integer.
3. The quotient of two integers (where the denominator is non-zero) is a rational number.

Step 5: Identifying the Contradiction
From our equation in Step 3, we have:
$\sqrt{5} = \text{a rational number}$

However, we know from the fundamental properties of real numbers that $\sqrt{5}$ is an irrational number. This creates a contradiction because a number cannot be both rational and irrational simultaneously.

Step 6: Conclusion
The contradiction has arisen because of our initial incorrect assumption that $3 + 2\sqrt{5}$ is a rational number. Therefore, we conclude that our assumption is false and $3 + 2\sqrt{5}$ must be irrational.

Final Answer: Since the assumption that $3 + 2\sqrt{5}$ is rational leads to a contradiction, it is proven that $3 + 2\sqrt{5}$ is an irrational number.