Final answer:
To show that 8n is Ω(5n) using Omega notation, we demonstrated that for constants c = 1 and n_{0} = 1, the inequality 8n ≥ 5n holds for all n ≥ 1. This confirms that 8n grows at least as quickly as 5n, thus proving 8n is Ω(5n).
Step-by-step explanation:
To prove using the definition of Omega notation that 8n is Ω(5n) or not, we must show that 8n grows at least as quickly as some constant multiple of 5n for sufficiently large values of n. By definition, f(n) is Ω(g(n)) if there exist positive constants c and n0 such that for all n ≥ n0, f(n) ≥ c · g(n). Let's take f(n) = 8n and g(n) = 5n.
To prove that 8n is Ω(5n), we need to find a constant c such that 8n ≥ c · 5n holds. Let's choose c = 1. It's clear that for all n ≥ 1, 8n ≥ 1 · 5n holds true, as 8n will always be larger than or equal to 5n. Therefore, we have proven that 8n is Ω(5n), satisfying the condition for Omega notation with c = 1 and n0 = 1.