Let $ A = \{ 2,3,4, \dots , 100 \} $ and $ B = \{ a^{1/b}| a,b \in A \} $. What is the maximal cardinality of a subset $ S $ of $ B $ such that $ xy \notin \mathbb{Q} $ for any distinct numbers $ x, y \in S $?
I tried to see what are the conditions that distinct $ a,b,c,d \in A $ must satisfy in order for $ a^{1/b} \cdot c^{1/d} $ to be rational. So suppose $ a^{1/b} \cdot c^{1/d} = \frac{x}{y} \in \mathbb{Q} $. Then by rising this to the power $ bd $, we must have that $ \big(\frac{x}{y}\big)^{bd} \in \mathbb{N} $ hence $ y $ divides $ x $. Therefore $ a^{1/b} \cdot c^{1/d}$ is a positive integer. Note that this is similar to the well known result that if $ a,b $ are positive integers, then $ a^{1/b} $ is either a positive integer or irrational.
So for distinct $ a,b,c,d \in A $ such that $ a^{1/b} \cdot c^{1/d} \in \mathbb{Q} $, we must have that $ a^{1/b} \cdot c^{1/d} = n \in \mathbb{N} $. I then showed that if this is the case, then the product of all the primes dividing $ ac $ must divide $ n $. Now since $$ 2^{1/2} \leq a^{1/b} \leq 100^{1/2} \quad \text{for any }a^{1/b} \in B $$ we must have that $$ 2 \leq n \leq 100 $$ Hence the product of all the primes dividing $ ac $ cannot be larger than $ 100 $. This helps us exclude some pairs $ (a,c) $ but still, it doesn't add that much and it is not the cleanest method I would say.
I thought identifying conditions for which products of two elements of $ B $ are rational might help in deducing the maximal cardinality of the subset $ S $ from the hypothesis but maybe this is not the best path.
I would appreciate any help in solving this problem. Thank you!