Prove that √2 + √3 is irrational.

To show that \( \sqrt{2} + \sqrt{3} \) is irrational, we can use a proof by contradiction. Let's assume the opposite: that \( \sqrt{2} + \sqrt{3} \) is rational. This means it can be expressed as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers with no common factors, and \( b \neq 0 \).

Step 1: Isolate one square root

From our assumption, we can rearrange the equation:

\( \sqrt{2} = \frac{a}{b} - \sqrt{3} \)

Step 2: Square both sides

Next, we square both sides to eliminate the square root:

\( 2 = \left(\frac{a}{b} - \sqrt{3}\right)^2 \)

Expanding the right side gives:

\( 2 = \frac{a^2}{b^2} - 2\frac{a}{b}\sqrt{3} + 3 \)

Step 3: Rearranging the equation

Now, we can rearrange this to isolate the term with \( \sqrt{3} \):

\( 2 - 3 - \frac{a^2}{b^2} = -2\frac{a}{b}\sqrt{3} \)

Which simplifies to:

\( -1 - \frac{a^2}{b^2} = -2\frac{a}{b}\sqrt{3} \)

Step 4: Solving for \( \sqrt{3} \)

Now, we can express \( \sqrt{3} \):

\( \sqrt{3} = \frac{1 + \frac{a^2}{b^2}}{2\frac{a}{b}} \)

This shows that \( \sqrt{3} \) can also be expressed as a fraction, which implies that \( \sqrt{3} \) is rational.

Step 5: Contradiction

However, we know that \( \sqrt{3} \) is irrational. This contradiction means our initial assumption that \( \sqrt{2} + \sqrt{3} \) is rational must be false.

Final Statement

Thus, we conclude that \( \sqrt{2} + \sqrt{3} \) is indeed irrational.