# $\Theta(n^2 ) =\Theta(n^2 + 1)$

I'm reading The Algorithm Design Manual and this is one of the excersizes:

Prove or disprove the following statement:

$$\Theta(n^2 ) = \Theta(n^2 + 1)$$

I think this is untrue because the right side of the equation claims to have a multiple $$c$$ such that it is smaller than all the multiples of $$n^2$$, which is untrue. But since the book doesn't have answers and I'm not really sure about this question I had to ask.

Thanks a lot :)

As we have on both sides of equality sets, then it should be prove correspondingly i.e we need to show, that $$\Theta(n^2) \subset \Theta(n^2+1)$$ and $$\Theta(n^2+1)\subset \Theta(n^2)$$. Let's start with first:
Assume $$f \in \Theta(n^2)$$. This means, that for some constants $$A_1, A_2 \gt 0$$ holds
$$A_1 n^2 \leqslant f(n) \leqslant A_2 n^2$$. To obtain inequalities for $$\Theta(n^2+1)$$ we need $$B_1 (n^2+1) \leqslant f(n) \leqslant B_2 (n^2+1)$$ It's easy to see, that it's enough to take $$B_2 = A_2$$ for right hand and $$0 for left hand, as $$A_2>0$$.