Solve it in English language and step by step

1. Prove that $\sqrt{5}$ is irrational.

Step 1: Assume, for the sake of contradiction, that $\sqrt{5}$ is a rational number.
If $\sqrt{5}$ is rational, it can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (their greatest common divisor is 1).

$$\sqrt{5} = \frac{a}{b}$$

Step 2: Square both sides of the equation.

$$(\sqrt{5})^2 = \left(\frac{a}{b}\right)^2$$ $$5 = \frac{a^2}{b^2}$$

Rearrange the equation:

$$5b^2 = a^2$$

This implies that $a^2$ is a multiple of 5. If a prime number $p$ divides $a^2$, then $p$ divides $a$. Thus, $a$ is a multiple of 5.

Step 3: Since $a$ is a multiple of 5, we can write $a = 5c$ for some integer $c$.
Substitute $a = 5c$ into the equation $5b^2 = a^2$:

$$5b^2 = (5c)^2$$ $$5b^2 = 25c^2$$

Step 4: Divide both sides by 5.

$$b^2 = 5c^2$$

This implies that $b^2$ is a multiple of 5. Therefore, $b$ is also a multiple of 5.

Step 5: From Step 2, we found that $a$ is a multiple of 5. From Step 4, we found that $b$ is a multiple of 5.
This means that $a$ and $b$ have a common factor of 5.

Step 6: This contradicts our initial assumption in Step 1 that $a$ and $b$ are coprime.
Therefore, our initial assumption that $\sqrt{5}$ is rational must be false.
Hence, $\sqrt{5}$ is an irrational number.
$\boxed{\text{Proven}}$

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

Step 1: Assume, for the sake of contradiction, that $3 + 2\sqrt{5}$ is a rational number.
If $3 + 2\sqrt{5}$ is rational, it can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime.

$$3 + 2\sqrt{5} = \frac{a}{b}$$

Step 2: Isolate the $\sqrt{5}$ term.
Subtract 3 from both sides:

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

Step 3: Divide by 2 to completely isolate $\sqrt{5}$.

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

Step 4: Analyze the right-hand side of the equation.
Since $a$ and $b$ are integers, $a - 3b$ is an integer, and $2b$ is a non-zero integer.
Therefore, $\frac{a - 3b}{2b}$ is a rational number.

Step 5: This implies that $\sqrt{5}$ is a rational number.
However, we know that $\sqrt{5}$ is irrational.

Step 6: This is a contradiction. Our initial assumption that $3 + 2\sqrt{5}$ is rational must be false.
Therefore, $3 + 2\sqrt{5}$ is an irrational number.
$\boxed{\text{Proven}}$

3. Prove that the following are irrationals: