Question

Prove that there exists an integer m such that \({m^2} > {10^{1000}}\). Is your proof constructive or nonconstructive?

Short answer

It is proved by constructive proof that there exists a number \(m\) such that \({m^2} > {10^{1000}}\).

Step-by-step solution

Describe the approach

This problem van be solved by constructive approach, which means that try to create own method to prove the statement.

Prove that \({m^2} > {10^{1000}}\)

Let \(m = {10^{500}} + 1\).

So, the inequality will be as follows.

\(\left( {{{10}^{500}} + 1} \right) > {10^{500}}\)

Square each side of inequality.

\(\begin{array}{c}{\left( {{{10}^{500}} + 1} \right)^2} > {\left( {{{10}^{500}}} \right)^2}\\{\left( {{{10}^{500}} + 1} \right)^2} > {10^{1000}}\\{m^2} > {10^{1000}}\end{array}\)

Therefore, it is proved by constructive proof that there exists a number \(m\) such that \({m^2} > {10^{1000}}\).