Question

Prove that there exist a sequence of rational number that converges to real number

Answer 1

Answer:

Without loss of generality we may assume the real number a is >0. (If a<0, we can apply the argument below to |a| and then switch signs.) We sketch a fairly formal proof, based on the fact that the reals are a complete ordered field. In one of the remarks at the end, we give an easy informal but incomplete "proof."

Let n be a natural number. Let m=m(n) be the largest positive integer such that mn<a. Then m+1n≥a, and therefore |a−m/n|<1/n.

Let rn=m/n. It is easy to show from the definition of limit that the sequence (rn) has limit a.

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