Question

Prove that given a nonnegative integer n, there is a unique nonnegative integer m such that m^2 ≤ n < (m + 1)^2.

Answer 1

To prove the given statement, we can use a proof by contradiction and assume that there are two values of m that satisfy the inequality. However, this leads to a contradiction, proving that there is a unique value of m.

Step-by-step explanation:

To prove that given a nonnegative integer n, there is a unique nonnegative integer m such that m^2 ≤ n < (m + 1)^2, we can use a proof by contradiction.

Assume that there are two nonnegative integers m and m+1 such that m^2 ≤ n < (m + 1)^2. This means that both m^2 ≤ n and n < (m + 1)^2 are true.

If m^2 ≤ n, then (m + 1)^2 > n. However, this contradicts the assumption that n < (m + 1)^2. Therefore, there can only be one nonnegative integer m such that m^2 ≤ n < (m + 1)^2.