3 ms·
You just have to take out square factors from each, and combine terms with matching sqrt(n), and that will be unique
by dooglius 5y ago
You just have to take out square factors from each, and combine terms with matching sqrt(n), and that will be unique
- edflsafoiewq 5y agoHow do you know its unique?
- aliceryhl 5y agoI don't think there's an elementary proof, but you will see a proof if you study algebraic number theory.
- deleted 5y ago[deleted]
- jepler 5y agoThat is to say, rewrite each sqrt(N_i) as P_i * sqrt(Q_i) where P and Q are integers and Q contains no squares. So 2 becomes 1 * sqrt(2), 4 becomes 2 * sqrt(1) and 8 becomes 2 * sqrt(2). Now, you can subtract equal terms from each side of the equation and if you can reach 0=0 then the numbers are equal. If you're left with something like sqrt(3) = 5 * sqrt(2) the the numbers are unequal. This stems from the fact, that I give without proof, that for integers X, Y and Z that contain no squares that sqrt(x) + sqrt(y) is never equal to sqrt(z). So there's no way to (say) add a bunch of square roots of 2 and have it become equal to a square root of 3 or 5. A number contains no squares if its prime factorization contains no repeated factors. Since this seems to involve factorization, plus a step of matching up numbers from both sides, the computational complexity would seem to be at least the complexity of factorization. The term-matching step is presumablty the easier step of the two.