Question

Prove that given a nonnegative integer n, there is a unique nonnegative integer m such that \({m^2} \le n < {\left( {m + 1} \right)^2}\).

Short answer

It is shown that \({m^2} \le n < {\left( {m + 1} \right)^2}\).

Step-by-step solution

Describe the given information

It is given that n is a nonnegative integer.

Prove that \({m^2} \le n < {\left( {m + 1} \right)^2}\)

It is given that \(n\) is a non-negative integer. So, take the square root of \(n\), which is \(\sqrt n \). It is known that \(\sqrt n \) is a real number and any real number lies between two consecutive integers. Therefore, there exists an integer \(m\) such that,

\(m \le \sqrt n < m + 1\)

Since \(\sqrt n \) is non-negative, the integer \(m\) also has to be nonnegative.

Square each side of the inequality \(m \le \sqrt n < m + 1\) then it gives,

\({m^2} \le n < {\left( {m + 1} \right)^2}\)

Therefore, it is shown that \({m^2} \le n < {\left( {m + 1} \right)^2}\).