Question about the proof that if x is a real number then there exists a postive integer n such that -n < x < n.

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One of my homework questions of real analysis went like this:

Prove that if $x$ is a real number then there exists a postive integer $n$ such that $-n < x < n$.

My answer was this:

First notice that $-n < x < n$ is equivalent to $|x|<n$.

Proof: We know that if $x$ is a real number then $|x|$ is also a real number, so by the Archimedean Principle there exits a natural number $n$ such that $|x|<n$, thus $-n<x<n$.

But my teacher told not to use my answer and instead use his without telling me why.

His answer:

$x$ is real, by the Archimedean Principle there exits an $n_1$ such that $x < n_1$. Also, $-x$ is a real, hence there exists a natural $n_2$ such that $n_2>-x$ so $-n_2<x$.

Take $n=max(n_1 , n_2)$, so we get $-n<x<n$.

I understand his answer, but what bothers is the fact that he told me not to use my method without any further explanation.