Algorithm Comparison with Theta-notation We consider two algorithms, Algo1 and Algo2, that solve the same problem. For any input of size n, Algo1 takes time $T_1(n)$ and Algo2 takes time $T_2(n)$. 

Prove or disprove each of the following statements. 

To prove a statement, you should provide a formal proof that is based on the definitions of the order notations. To disprove a statement, provide a counter example and explain it.

Suppose that $T_1(n)$ ∈ $\theta(n^2)$ and $T_2(n)$ ∈ $\theta(n^3)$. 

Does it imply that there exists $n_0$ such that for $n$ ≥ $n_0$, Algo1 runs faster than Algo2 on inputs of size $n$?

By definition of $\theta$-notation,$\\$

If $T_1(n)$ ∈ $\theta(n^2)$, then $c_1 n^2 \leq T_1(n) \leq c_2 n^2$ for all $n \geq n_1$for $c_1, c_2, n_1 &gt; 0$. 

If $T_2(n)$ ∈ $\theta(n^3)$, then $d_1 n^3 \leq T_2(n) \leq d_2 n^3$ for all $n \geq n_2$ for $d_1, d_2, n_2 &gt; 0 $. 

We choose $n_0 = max(n_1, n_2)$

Then, $\\$
$\frac{c_1 n^2}{d_1 n^3} \leq \frac{T_1(n)}{T_2(n)} \leq 
\frac{c_2 n^2}{d_2 n^3} \text{ for all } n \geq n_0$ 

$\lim_{n \rightarrow \infty} \frac{c_1 n^2}{d_1 n^3} = 0$ and $\lim_{n \rightarrow \infty} \frac{c_2 n^2}{d_2 n^3} = 0$

Hence, by Squeeze Theorem, $\lim_{n \rightarrow \infty} \frac{T_1(n)}{T_2(n)} = 0$.
Therefore, Algo1 runs faster than Algo2 for $n &gt;= n_0$

This is my solution, and I am wondering if this is right way to solve this question. 

Any help is appreciated, thank you,
If $T_1(n) = \Theta(n^2)$ and $T_2 = \Theta(n^3)$, then since $n^2 = o(n^3)$, it follows that $T_1(n) = o(T_2(n))$. In particular, $T_1(n) &lt; T_2(n)$ for large enough $n$. In fact, for every $C&gt;1$, $T_1(n) &lt; CT_2(n)$ for large enough $n$ (how large depends on $C$).