Show that there does not exist a rational number such that x^2=6. Is sqrt2 + sqrt3 a rational number ?
My answer : Let x = (a/b)in lowest term. (a/b)^2 = 6 a^2=6b^2 --> a is multiple of 3 so let a=3n (3n)^2 = 6b^2 9n^2 = 6b^2 3n^2 = 2b^2 -->2b^2 is a multiple of 3 , so b also multiple of 3 Conclusion : Since both a and b share same factor so x could not be rational.
Assume sqrt2 + sqrt3 is rational r^2 = (sqrt2 + sqrt3 )^2 (2 sqrt6) + 5 = r^2 sqrt 6 =( r^2 -5)/2 Conclusion: Since sqrt6 is irrational,so (r^2-5)/2 also irrational which bring a cotradiction . Hence , sqrt2 +sqrt3 is irrational .
OR Assume sqrt2 is irrational, sqrt2 = a/b in lowest term square both side , a^2 = 2b^2 --> a is multiple of 2 so let a =2n (2n)^2 = 2b^2 4n^2=2b^2 b^2 = 2n^2 -->so b is also multiple of 2 Since a and b has same factor so sqrt2 cannot be rational
sqrt 3 = a/b a^2 = 3b^2 -->a is a multiple of 3, so let a =3n (3n)^2 = 3b^2 9n^2 = 3b^2 b^2= 3n^2 -->b is also multiple of 3 CONCLUSION: since a and b share the same factor hence sqrt 3 not rational
Conclusion: Since sqrt2 and sqrt3 is not rational hence sqrt2 + sqrt3 is also not rational
For the second part i don't know which answer is correct , so i write both
The first part is correct. That is, you proved correctly that $\sqrt6$ is irrational and that $\sqrt2+\sqrt3$ is irrational too.
The second part (your other proof of the fact that $\sqrt2+\sqrt3$ is irrational) is wrong. The sum of two irrational numbers may well be rational. Take $\sqrt2$ and $-\sqrt2$, for instance.