Well, in this question it is said that 100√√3+√2+100√√3−√2, and the owner asks for "alternative proofs" which do not use rational root theorem. I wrote an answer, but I just proved 100√√3+√2∉Q and 100√√3−√2∉Q, not the sum of them. I got (fairly) downvoted, because I didn't notice that the sum of two irrational can be either rational or irrational, and I deleted my (incorrect) answer. So, I want help in proving things like √5+√7∉Q, and (1+π)−π∈Q, if there is any "trick" or rule to these cases of summing two (or more) known irrational numbers (without rational root theorem).
Thanks.
Answer
To prove that √5+√7 is irrational:
√5+√7=ab
a2b2=12+√35
a2−12b2b2=√35
35=(a2−12b2)2(b2)2
35|a2−12b2
352|(a2−12b2)2
352|(b2)2
Both the numerator and denominator are multiples of an even power of 2. Contradiction.
The method can be extended to many other sums of nth roots.
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