# State true or false --> x = Θ(n⁴), y = Θ(n²) therefore x/y = Θ(n²)

State true or false $$x=\Theta(n^4)$$, $$y=\Theta(n^2)$$ therefore $$x/y=\Theta(n^2)$$

I think the ans is true,but in the book it says false. What is the explanation?

My reason for thinking it to be true is that say for x we have a polynomial $$n^4$$+...... and for y we have a polynomial like $$n^2$$+... so if we divide x/y the ans should have the highest power of $$n^2$$ and the resultant polynomial will be like $$n^2$$+.... So it should be $$\Theta(n^2)$$

It's true. There are positive constants $$c_1$$, $$c_2$$, $$c_3$$ and $$c_4$$ such that, for all large enough $$n$$, $$c_1n^4 \leq x\leq c_2 n^4 \quad\text{and}\quad c_3n^2\leq y\leq c_4n^2\,.$$ Therefore, $$\frac{c_1}{c_4}n^2 \leq \frac{x}{y}\leq \frac{c_2}{c_3} n^2\,,$$ i.e., $$x/y=\Theta(n^2)$$.
By the way, in your argument, you say that $$x$$ and $$y$$ are polynomials. That's not necessarily true – for example, we could have $$x=n^4+\log n$$.
• There is ambiguity in the fact that $x$ and $y$ are functions of $n$ but that isn't shown in the question or answer. It should be $x(n)/y(n)$ I believe. The input to the function should be specified to get a result. This could be why the book says it is false. – ryan Aug 4 at 1:20
• The only problem I would potentially see is, without input, it's not 100% clear that $x$ and $y$ are functions on the same input $n$. For instance $x(n) / y(n^2) \not\in \Theta(n^2)$. – ryan Aug 4 at 23:40
• @ryan If that was the intention, it is completely unreasonable. The only reasonable interpretation that I can see is that $x$ stands for some expression such that the function $f$ that maps $n$ to $x$ is $\Theta(n^4)$ and $y$ similarly defines some some function $g\in \Theta(n^2)$. We're being asked about the expression $x/y$, which corresponds to the function mapping $n$ to $f(n)/g(n)$. Anything else seems like answering the question "Do domestic cats eat elephants?" with "Well, that's kind of ambiguous. There might be a brand of cat food called 'elephants'." – David Richerby Aug 5 at 9:12