Is this a valid definition of the rationals?

Question

$$\mathbb{Q}=\left\{\sum_{n=1}^k f(n)\mid k,n\in\mathbb{N}\land f\text{ is a finite composition of $+$, $-$, $\div$, $\times$}\right\}$$

Reasoning:

Any real number can be described by a (sometimes infinite) sum of rational numbers. If such a sum is taken to be $q=\sum_{n=1}^k f(n)$, then every real number which is not rational can be approximated to arbitrary precision by increasingly large $k$. If $f(n)$ is composed solely of elementary arithmetic operations ($+,-,\div,\times$), then $q$ remains rational for all $k<\infty$.

If the quotient of any two rational numbers is also rational, then it follows that for elementary functions $f$ and $g$, the quotient of the summations $\sum_{n=1}^k f(n)$ and $\sum_{n=1}^k g(n)$ is always rational even as $k$ tends towards infinity.

Intuitively, it would seem that the quotient of any two such summations is always rational even if $k=\infty$. However, this is not the case, as there are many infinite sums satisfying the above conditions which are irrational.

Therefore, the sum is rational iff the upper bound $k$ is finite.

Becuase any real number may be represented as a summation, it follows that any rational number can be represented as a summation.

Thus, every rational number can be represented as a finite sum of elementary functions.

Answer 1

If $k$ is finite, then as noted above there's no need to use the summation notation at all. The right thing to say in my opinion is:

$\mathbb{Q}$ is the smallest set of reals containing $1$ and closed under $+,\times,-,$ and $\div$.

One direction is easy to prove: A positive rational ${a\over b}$ can always be represented as $(1+...+1)\over (1+...+1)$ with $a$-many $1$s in the numerator and $b$-many $1$s in the denominator. $0$ can be gotten as $1-1$, and this lets us take negatives. So $\mathbb{Q}$ is contained in any such set. Conversely, clearly $1\in\mathbb{Q}$ and $\mathbb{Q}$ is closed under $+,\times,-,$ and $\div$, so we're done.

Meanwhile, every real can be represented as an infinite sum of rationals, and so allowing the sum to be infinite does indeed get all of $\mathbb{R}$. However, there are two caveats worth noting. First, not every infinite sum corresponds to a real (an infinite sum can diverge or oscillate). Second, some infinite sums do still correspond to rationals, contra your claim "the sum is rational iff the upper bound $k$ is finite." For example, consider $$\sum_{i=1}^\infty{1\div 2\times ...\times 2\mbox{ ($i$ times)}},$$ more clearly written as $\sum_{i=1}^\infty 2^{-i}$, which is of course just $1$.