Prove or disprove square root of 2 divided by 6 is rational?

Answer 1

let x= square root 6 + square root 30, I will write S(6) for square root 6, or S(n) for square root n x-S(6)=S(30) now square both sides x^2-2S(6)+6=30 or x^2-24=2S(6) Now square both sides again to get rid of the radical sign x^4-48x+24^2=24 or after we simplify x^4-48x-552=0 Now we know x is a solution of this polynomial because we created the polynomial to make it so! But the rational roots theorem tells us that all the solutions can be written as p/q where p is an integer factor of 552 and q is an integer factor of 1 ( it is 1or -1) So the solution must be a factor of 552 if the solution is rational. ( there are irrational and complex solutions) 2^3x3x23=552 so the only possible rational roots are combinations of these factors, but we know square root of 5 +square root of 30 is <8, we could say <10 and also by approximation. So that means we only have to deal with the factors 2,3,4 and 6 (all the other combinations are >or= 8) But we also know sqrt6+sqrt(30)>2,3 or 4 because sqrt 4+sqrt 4=4 and our sum is greater. So we are left with 6 Well we know our sum can be written as Sqrt(6)[(Sqrt5)+ 1)] which tells us it is not 6. So it is none of the possible rational roots and hence not rational. Lots of ways to do this last part, just show that it can't be any of the rational factors. following the above notation, If S(6) + S(30) were rational, then it's square would be rational as well so we will show that 6 + 2S(180) + 30 is not rational it suffices to show that S(180) is irrational becasue addition is closed in the rational numbers. 180 = 36 times 5 so S(180) = 6S(5). Now we simply need to show that S(5) is irrational. Assume that (a/b)^2 = 5 where a/b is a fraction in Lowest Terms. then a^2 = 5b^2 so a^2 is divisible by 5 which means that a is divisible by 5 (see footnote). for some number k, a=5k. now we have 25k^2=5b^2 dividing through by 5 we see that b is also divisible by five, contradicting the fact that a/b was in lowest terms. Q.E.D. *every integer has a unique prime facotization. If the prime factorization of a is P1 * P2 * P3 * ... * Pn and 5 is not included, then squaring a simply squares all the prime factors, and 5, wich is prime will still not be part of the prime factorization.

Answer 2

Suppose the sqrt(2)/6 is rational. This means that sqrt(2)/6 can be expressed in the form p/q where p and q are co-prime integers.

sqrt(2)/6 = p/q which implies that 2/36 = p^2/q^2 This can be simplified to 18*p^2 = q^2
Now 2 divides the left hand side (LHS) so it must divide the right hand side (RHS).
That is, 2 must divide q^2 and since 2 is a prime, 2 must divide q. That is q = 2*r for some integer r. Then substituting for q gives, 18*p^2 = (2*r)^2 = 4*r^2 Dividing both sides by 2 gives 9*p^2 = 2*r^2.
But now 2 divides the RHS so it must divide the LHS.
That is, 2 must divide 9*p^2 and since 2 is a prime, and 2 does not divide 9, then 2 must divide p. But then we have 2 dividing p as well as q which contradicts the requirement that p and q are co-prime. The contradiction implies that sqrt(2)/6 cannot be rational.