**Proof that √2 is irrational**

To prove that 2 + 3√2 is an irrational number, we first need to establish that √2 is irrational.

**Proof by contradiction**

We will use a proof by contradiction to demonstrate that √2 is irrational.

Assume √2 is rational, which implies it can be expressed as a fraction in the form a/b, where a and b are integers with no common factors.

√2 = a/b

Squaring both sides, we get:

2 = (a^2)/(b^2)

Rearranging the equation, we have:

a^2 = 2b^2

This implies that a^2 is an even number since it is divisible by 2.

**Lemma: If a^2 is even, then a is also even**

Proof of the lemma:

Assume a is an odd integer. Then, a can be expressed as 2k + 1, where k is an integer.

Substituting this into the equation a^2 = (2k + 1)^2, we get:

a^2 = 4k^2 + 4k + 1

Rearranging, we have:

a^2 = 2(2k^2 + 2k) + 1

Since 2k^2 + 2k is an integer, we can rewrite the equation as:

a^2 = 2m + 1, where m is an integer

This shows that a^2 is of the form 2m + 1, which means it is odd.

However, we assumed a^2 is even, which contradicts our assumption that a is odd.

Therefore, our assumption that a is odd must be false, and a must be even.

**Back to the main proof**

Returning to the equation a^2 = 2b^2, we have established that a is even.

Let a = 2k, where k is an integer.

Substituting this into the equation, we get:

(2k)^2 = 2b^2

This simplifies to:

4k^2 = 2b^2

Dividing both sides by 2, we have:

2k^2 = b^2

This implies that b^2 is also even, and by the same reasoning as before, b must be even.

However, if both a and b are even, they have a common factor of 2, which contradicts our initial assumption that a/b has no common factors.

Therefore, our assumption that √2 is rational must be false.

**Conclusion: √2 is irrational**

Since we have proven that √2 is irrational, we can conclude that 2 + 3√2 is also irrational, as it involves the irrational number √2 in its expression.