On page 16 of this algorithms book, it states:
For example, suppose we are choosing between two algorithms for a particular computational task. One takes $f_1(n) = n^2$ steps, while the other takes $f_2(n) = 2n + 20$ steps (Figure 0.2).
He then goes on to say:
This superiority ... (of $f_2$ over $f_1$) ... is captured by the big-O notation: $f_2 = O(f_1)$, because ...
Now my problem is that in the original quote, he said that $f_1(n) = n^2$ steps and $f_2(n) = 2n+20$ steps, so thus $f_1 = O(n^2)$ and $f_2 = O(n)$ (big-O is defined in Section 0.3). But the second quote above states $f_2 = O(f_1)$, which means $f_2 = O(n^2)$ and contradicts his definition of big-O notation. What have I missed?