One way to remove the feeling of obviousness from the statement is to reformulate it: in every complete ordered field F the integers (which could be defined as the additive subgroup of F generated by the multiplicative identity) form an unbounded set. This reformulation encourages us to think about the question abstractly, rather than relying on our normal, concrete representation of the real numbers.
Here is the usual proof: suppose that the integers were bounded in F. By the completeness axiom, there would have to be a least upper bound, x, say. Since no number smaller than x is an upper bound, there must be an integer n greater than x-1. But then n+1 is greater than x, contradicting the fact that x is an upper bound.
Was it really necessary to use the completeness axiom in the above proof? Yes it was, and the way to demonstrate this is to give an example of an ordered field in which the integers, still defined as the additive subgroup generated by the multiplicative identity, are bounded. Obviously the completeness axiom will have to fail for such a field.
The most familiar example of an ordered field failing the completeness axiom is Q, the field of rational numbers. However, the integers are certainly unbounded in Q.
Here is a field in which the integers are bounded! Let F be the field of all rational functions in an indeterminate x - that is functions of the form p(x)/q(x), where p and q are polynomials (with real coefficients) and q is not identically zero. I do not insist that these functions should be everywhere defined - for now they can be thought of as formal expressions - though in fact they will be defined at all but finitely many points. Strictly speaking I am talking about equivalence classes of these formal expressions, so I regard p/q as the same as r/s if ps is the same polynomial as qr.
Such functions can be added and multiplied in the obvious ways, so it is not hard to check that F really is a field. We would now like to show that F is an ordered field, by defining a suitable ordering on the functions p/q. This we can do by saying that p/q is less than r/s if there exists a d > 0 such that p(x)/q(x) is less than q(x)/r(x) for every x in the open interval (0,d). Loosely speaking, we say that one rational function is smaller than another if it is smaller `just to the right of zero'.
Various facts about this ordering must be checked. One basic one is that this is a total ordering - that is, given two functions p/q and r/s, then either they are equal or one is less than the other. To see this, note that if neither is less than the other, then as x approaches zero from above, the sign of p(x)/q(x) - r(x)/s(x) must change infinitely often, so that p(x)/q(x) - r(x)/s(x) is zero infinitely often. But this means that the polynomial ps-qr has infinitely many roots, and is therefore the zero polynomial, which implies that p/q equals r/s. The other axioms for an ordered field are easy to check.
What are the `integers' in such a field? Well, the multiplicative identity is the constant function 1, and the subgroup it generates consists of all integer-valued constant functions. Thus, in this field, 26 means the function that always takes the value 26, and so on. What would be an upper bound for all of these functions in the ordering just defined? An obvious example is the function 1/x, which, if you are near enough to zero, is at least as big as any fixed integer.
Notice also that in this ordered field there are some very small elements as well. We can identify the real number t with the constant function f(x)=t, and then the function f(x)=x is smaller than all real numbers, but not zero. Thus, for example, the sequence 1/n does not converge to zero in this field (because if we let epsilon be the function f(x)=x, then epsilon > 0 and there is no constant function 1/n which is smaller than epsilon).
Two further remarks: (i) in the field F it is not true that every element has a decimal expansion; (ii) we could have defined the ordering by saying that one function is smaller than another if it is smaller for sufficiently large x instead, and the argument, with obvious changes, would have worked just as well.
This example shows that it is not wholly obvious that the integers are unbounded in R. Any proof must use the completeness axiom in some form, because otherwise it will not distinguish between R and the ordered field F just constructed in which the integers are bounded.