Why is O(2n) treated the same as O(n)?

Write the claim in full: $T(n)=O(g(n))$ means there are a constant $c>0$ and a size $n_0$ such that $T(n)\le c\,g(n)$ for every $n\ge n_0$. The $2$ in $2n$ is not sitting outside that inequality. It is one more number that can be folded into $c$. To show $2n$ is $O(n)$, take $g(n)=n$. Choose $c=2$ and $n_0=1$. Then for every $n\ge1$, $2n\le2n$, which is true by equality. The claim is satisfied, and nothing about the factor $2$ has been denied.

Put numbers in and the scale is plain. At $n=1{,}000{,}000$, a cost of $2n$ is $2{,}000{,}000$ operations and a cost of $n$ is $1{,}000{,}000$. The ratio is $2$. A quadratic cost at the same size is $1{,}000{,}000{,}000{,}000$ operations, which is $500{,}000$ times the linear one: $10^{12}/2{,}000{,}000=500{,}000$. The factor $2$ is a difference of two; the shape is a difference of five hundred thousand. At the scale where growth classes are meant to be read, the second difference is the one that decides.

The natural guess is that $O(2n)$ records the $2$ and $O(n)$ does not, so the first must be a larger class. That treats the expression inside $O$ as if it were the running time itself. It is not. $O(2n)$ is not twice $O(n)$; the notation does not multiply that way. Both expressions name the same set of functions, the ones bounded by a straight line for large enough $n$. If the $2$ were meant to survive, the question would be about the exact cost, or about which of two linear procedures is faster, and $O$ is deliberately blind to both.

The equality is not just one absorption; it is mutual. If $T(n)=O(2n)$, then some $c$ and $n_0$ give $T(n)\le c(2n)$ for $n\ge n_0$. Set $c'=2c$. Then $T(n)\le c'n$, so $T(n)=O(n)$. Conversely, if $T(n)=O(n)$ with constant $c$, then $T(n)\le c\,n=(c/2)(2n)$, so $T(n)=O(2n)$. A factor of two can be moved from the function into the constant or from the constant into the function. That is why the two claims are the same claim.

None of this says the factor $2$ is imaginary. It says the factor belongs to the implementation and the machine, and to the small inputs, rather than to the growth class. A procedure that does $2n$ work is twice as slow as one that does $n$ work at the same $n$, if the constant is the same. On a list of ten items that is the difference between twenty operations and ten. If the input never gets large, that difference is the whole story, and asking for the growth class is asking the wrong question. The convention is a statement about the limit, not a promise about today.

A larger constant can even reverse the early winner. Compare $100n$ with $n^2$. At $n=10$, $100n=1{,}000$ and $n^2=100$, so the linear procedure is ten times more work. At $n=100$, both are $10{,}000$. At $n=10{,}000$, $100n=1{,}000{,}000$ and $n^2=100{,}000{,}000$, so the quadratic procedure is one hundred times more work. The constant set the crossing point at $100$ and the shape controlled everything after it. Dropping constants is safe for the eventual comparison and unsafe for the early one.