Use proof by contradiction. Assume that the sum is rationial, that is
2–√+5–√=ab2+5=ab
where aa and bb are integers with b≠0b≠0. Now rewrite this as
5–√=ab−2–√.5=ab−2.
Squaring both sides of this equation we obtain
5=a2b2−22–√ab+2.5=a2b2−22ab+2.
Now, carefully solve for 2–√2 and obtain
2–√=−3b2a+a2b.2=−3b2a+a2b.
This implies that 2–√2 is a rational number which is a contradiction. Thus
2–√+5–√2+5
is an irrational number.
