Discuss the reflexive, symmetric, and transitive properties for asymptotic comparisons \((O, \Omega, \theta, o)\)

In the context of asymptotic notation, 'reflexivity' implies that a function will always be bounded by itself. \n For the given notations: \n - \(O\) notation is reflexive, because for any function \(f(n)\), \(f(n) = O(f(n))\). The function is certainly an upper bound for itself. \n - \(\Omega\) notation is reflexive, because for any function \(f(n)\), \(f(n) = \(\Omega\)(f(n))\). The function is an acceptable lower bound for itself. \n - \(\theta\) notation is reflexive as well, because for any function \(f(n)\), \(f(n) = \(\theta\)(f(n))\). The function is both an upper and lower bound for itself. \n - \(o\) notation is not reflexive, because for any \(f(n)\), \(f(n) \neq o(f(n))\). This is because \(o(f(n))\) implies that \(f(n)\) grows strictly slower than \(o(f(n))\). Therefore, a function cannot grow strictly slower than itself.

Syntactically, symmetry refers to the property of a binary relation where, if \(f(n)\) is related to \(g(n)\), then \(g(n)\) is related to \(f(n)\). \n For the given notations:\n - \(O\) notation is not symmetric since if \(f(n) = O(g(n))\), it does not imply \(g(n) = O(f(n))\).\n - \(\Omega\) notation is also not symmetric.\n - \(\theta\) notation is symmetric. If \(f(n) = \(\theta\)(g(n))\), then \(g(n) = \(\theta\)(f(n))\), by definition.\n - \(o\) notation is also not symmetric.

In terms of asymptotic notation, transitivity means that if \(f(n)\) is related to \(g(n)\) and \(g(n)\) is related to \(h(n)\), then \(f(n)\) is related to \(h(n)\). For the given notations:\n - \(O\) notation is transitive. If \(f(n) = O(g(n))\) and \(g(n) = O(h(n))\), then \(f(n) = O(h(n))\).\n - \(\Omega\) notation is transitive as well.\n - \(\theta\) notation is also transitive.\n - \(o\) notation, however, is not transitive. If \(f(n) = o(g(n))\) and \(g(n) = o(h(n))\), it does not necessarily imply that \(f(n) = o(h(n))\).