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

Step 1: Assume $\sqrt{5}$ is rational

To prove that $\sqrt{5}$ is irrational, we use the method of contradiction. We start by assuming the opposite, that $\sqrt{5}$ is a rational number. A rational number can always be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ have no common factors other than 1 (i.e., they are coprime).

Step 2: Square both sides and rearrange

If $\sqrt{5} = \frac{a}{b}$, then squaring both sides gives $5 = \frac{a^2}{b^2}$. Multiplying both sides by $b^2$ gives us the equation $5b^2 = a^2$. This equation implies that $a^2$ is a multiple of 5, which means that $a$ itself must be a multiple of 5.

Step 3: Substitute $a=5c$ and find a contradiction

Since $a$ is a multiple of 5, we can write $a = 5c$ for some integer $c$. Substituting this into the equation $5b^2 = a^2$, we get $5b^2 = (5c)^2$, which simplifies to $5b^2 = 25c^2$. Dividing both sides by 5 gives $b^2 = 5c^2$. This implies that $b^2$ is a multiple of 5, and therefore $b$ must also be a multiple of 5.

Step 4: Conclude $\sqrt{5}$ is irrational

We have shown that both $a$ and $b$ are multiples of 5. This means that $a$ and $b$ have a common factor of 5. However, our initial assumption was that $a$ and $b$ are coprime (have no common factors other than 1). This is a contradiction. Therefore, our initial assumption that $\sqrt{5}$ is rational must be false. Hence, $\sqrt{5}$ is irrational.

Step 5: Assume $3 + 2\sqrt{5}$ is rational

Now we need to prove that $3 + 2\sqrt{5}$ is irrational. Again, we use the method of contradiction. Let's assume that $3 + 2\sqrt{5}$ is a rational number. This means it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime.

Step 6: Isolate $\sqrt{5}$

From the equation $3 + 2\sqrt{5} = \frac{p}{q}$, we can isolate $\sqrt{5}$. First, subtract 3 from both sides: $2\sqrt{5} = \frac{p}{q} - 3$. Combine the terms on the right side: $2\sqrt{5} = \frac{p - 3q}{q}$. Finally, divide by 2: $\sqrt{5} = \frac{p - 3q}{2q}$.

Step 7: Conclude $3 + 2\sqrt{5}$ is irrational

Since $p$, $q$, 2, and 3 are all integers, the expression $\frac{p - 3q}{2q}$ must be a rational number. This means that if $3 + 2\sqrt{5}$ were rational, then $\sqrt{5}$ would also be rational. However, we have already proven that $\sqrt{5}$ is irrational. This is a contradiction. Therefore, our assumption that $3 + 2\sqrt{5}$ is rational must be false. Hence, $3 + 2\sqrt{5}$ is irrational.

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