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GenAi Evaluation 9
๐: Prove or disprove that there is a rational number x and an
irrational number y such that xy is irrational.
๐ค: This statement is actually false. In fact, it cannot be proven or disproven using elementary number theory alone. The reason for this is because the statement implies that there are infinitely many pairs of numbers (x, y) where xy is irrational, but this contradicts the well-known result known as Euclid's infinite primes theorem, which states that there are only countably many prime numbers.
To see why this is the case, suppose towards a contradiction that there are infinitely many pairs of numbers (x, y) where xy is irrational. Then, by the axiom of choice, we can choose one such pair (x0, y0) arbitrarily. Now, consider the set S = {xy | (x, y) is a pair of rational numbers}. By the assumption that there are infinitely many pairs of rational numbers, S is nonempty. However, since each element of S is of the form xy, where x and y are rational, it follows that S is not dense in the real line, meaning that there must exist two distinct elements z1 and z2 in S such that |z1 - z2| < ฮต for some positive integer ฮต.
Now, consider the number w = (x0y0 + z1)/(x0y0 + z2). Since x0 and y0 are rational, it follows that w is algebraic over the rational numbers. Moreover, since z1 and z2 are in S, it follows that w is also in S. Therefore, w is an irrational number, which contradicts our assumption that w is algebraic.
Therefore, we have reached a contradiction, and it follows that there do not exist infinitely many pairs of numbers (x, y) where xy is irrational.