let √2 + √3 = r
for some rational number r.
By squaring both sides, we get:
5 + 2√6 = r^2
so that ,√6 = (r^2 - 5) / 2 is rational.
Now √6 = m/n for some integers m, n, such that m and n are relatively prime.
Squaring both sides, we get
6 = m^2 / n^2
Now, 2 divides the left-hand side, so it divides m^2, but then 2 divides m since 2 is prime. We may write m = 2k for some integer k.
6n^2 = (2k)^2 = 4k^2
3n^2 = 2k^2
Now 2 divides the right-hand side, so 2 divides 3n^2. However, 2 and 3 are relatively prime, so 2 divides n^2, forcing n to be divisible by 2.
However, this is a contradiction to the assumption that m and n are relatively prime.