Answer
See solution
Work Step by Step
**Proof by Contradiction:**
Assume, for the sake of contradiction, that
\[
\sqrt{2} + \sqrt{3}
\]
is rational. Then there exists a rational number \( r \) such that
\[
\sqrt{2} + \sqrt{3} = r.
\]
We can isolate one of the square roots. For example, subtract \(\sqrt{2}\) from both sides:
\[
\sqrt{3} = r - \sqrt{2}.
\]
Now, square both sides of the equation:
\[
3 = (r - \sqrt{2})^2 = r^2 - 2r\sqrt{2} + 2.
\]
Rearrange the equation to isolate the term involving \(\sqrt{2}\):
\[
3 - r^2 - 2 = -2r\sqrt{2},
\]
\[
1 - r^2 = -2r\sqrt{2},
\]
\[
2r\sqrt{2} = r^2 - 1.
\]
Now solve for \(\sqrt{2}\):
\[
\sqrt{2} = \frac{r^2 - 1}{2r}.
\]
Since \( r \) is rational, both \( r^2 - 1 \) and \( 2r \) are rational. Therefore, \(\sqrt{2}\) would be rational.
However, it is a well-known fact that \(\sqrt{2}\) is irrational. This contradiction shows that our assumption must be false.
Thus, \(\sqrt{2} + \sqrt{3}\) is irrational.
