Question

If x is an arbitrary real number, prove that there exists a unique integer n which satisfies the inequalities               n ≤ x < n + 1. 

Answer 1

Let A = {n ∈ Z| n ≤ x}.  i.e A is non-empty. It is also bounded above, so by completeness axiom, s = sup(A) exists. Now, by property of sup, there exists n ∈ A such that s−1 < n ≤ s.

clearly, n ≤ x. Now, s − 1 < n implies that n + 1 does not belong to A. Hence x < n + 1. This show that there is an n which satisfies n ≤ x < n + 1. Now, to show uniqueness, assume on the contrary that there exists m ≠ n such that m ≤ x < m + 1. Then m ∈ A and we must have m ≤ n (otherwise, if m > n, then

m ≥ n+ 1 > x, so m cannot satisfy the said inequality). If m < n then (since both are integers) (n−m) ≥ 1.

Hence, m + 1 ≤ n ≤ x, which contradicts the assumption that m satisfies the given inequality.

Let A = {n ∈ Z| n ≤ x}. A is non-empty. It is also bounded above, so by completeness axiom, s = sup(A) exists. Now, by property of sup, there exists n ∈ A such that s−1 < n ≤ s. clearly, n ≤ x. Now, s − 1 < n implies that n + 1 does not belong to A. Hence x < n + 1. This show that there is an n which satisfies n ≤ x < n + 1. Now, to show uniqueness, assume on the contrary that there exists m ≠ n such that m ≤ x < m + 1. Then m ∈ A and we must have m ≤ n (otherwise, if m > n, then m ≥ n+ 1 > x, so m cannot satisfy the said inequality). If m < n then (since both are integers) (n−m) ≥ 1.

Hence, m + 1 ≤ n ≤ x, which contradicts the assumption that m satisfies the given inequality