# Why is max{n,k}= Ө(n+k) [duplicate]

• Use the definitions. Be mindful of what $n$, $k$ mean here, and how the definition of $\Theta$ extends to two parameters. (Hint: it doesn't, really.) – Raphael Oct 10 '18 at 6:15
Assuming n and k to be non-negative, $$n\leq n+k$$ and $$k\leq n+k$$. Hence,$$\max(n, k) \in \mathcal{O}(n+k)$$. Next, $$n+k≤2\max(n,k)$$. Hence, $$\max(n,k)\in\mathcal{\Omega}(n+k)$$.
Hence, we get that $$\max(n,k)\in \mathcal{\Theta}(n+k)$$.