Question

Prove the limit assertions in Exercise 20.2.

Since b−a>0, the Archimedean property shows there exists an n∈N such that

n(b−a)>1, and hence bn−an>1.

From this, it is fairly evident that there is an integer m between an and bn, so that (1) holds. However, the proof that such an m exists is a little delicate. We argue as follows. By the Archimedean property again, there exists an integer k>max{∣an∣,∣bn∣}, so that

−k
Then the sets K={j∈Z:−k≤j≤k} and {j∈K:an−k, we have m−1 in K, so the inequality an

Answer 1

To prove assertion (1), we use the Archimedean property and define a set K to find an integer m between an and bn.

Step-by-step explanation:

The provided question is related to proving the limit assertions in Exercise 20.2. To prove assertion (1), we need to show that there exists an integer m between an and bn. To do this, we use the Archimedean property to first find an integer k such that -k > b - a. We then define the set K as {j ∈ Z: -k ≤ j ≤ k}. Since an ≤ a ≤ b ≤ bn, there exists an integer m in K such that an ≤ m ≤ bn.