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Prove that 2 + root6 is irrational or prove that 3 + 8 is irrational
Subtract 3 from 3+√8.your answer will be √8.we can also write √8 as 2√2.now divide it by 2.it will be √2 which is irrational.so whole term is irrational.
I will answer 3+√6 firstLet 3+√6=p/q ;q≠0;p and q are co-prime integers3+√6=p/q√6=p/q -3√6=p-3q /qHera p-3q /q is rationalLet√6=a/b (a rational);b≠0;a and b are co-prime integers √6=a/bSquaring both sides (√6)^2=(a/b)^26=a^2/b^2b^2=a^2/6 this implies 6 divides a^2 as well as aLet a=6cSquaring both sides a^2=(6c)^2a^2=36c^2Substitutingb^2=36c^2/6b^2=6c^2c^2=b^2/6 this implies 6 divides b^2 as well as b From above 6 is a common factor of both a and b. This is a contradiction because of our wrong assumption that √6 is rationalHence √6 is irrational√6=p-3q/qIrrational=rational Which is a contradiction. This happen because of our wrong assumption that 3+√6 is rational Hence 3+√6 is irrational. THE OTHER QUESTIONS answer3+8 is rational
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