Question

Show that the product of two of the numbers \(6{5^{1000}} - {8^{2001}} + {3^{177}}\),\(7{9^{1212}} - {9^{2399}} + {2^{2001}}\), and \(2{4^{4493}} - {5^{8192}} + {7^{1777}}\)is nonnegative. Is your proof constructive or non-constructive? (Hint: Do not try to evaluate these numbers!)

Short answer

The multiplication of two of the numbers is non-negative.

And, as the exact explanation is not given, it is non-constructive proof.

Step-by-step solution

Introduction

The multiplication of two negative numbers is non-negative and the multiplication of two non-negative numbers is non-negative.

\(\begin{aligned}{}{\bf{a = 6}}{{\bf{5}}^{{\bf{1000}}}}{\bf{ - }}{{\bf{8}}^{{\bf{2001}}}}{\bf{ + }}{{\bf{3}}^{{\bf{177}}}}\\{\bf{b = 7}}{{\bf{9}}^{{\bf{1212}}}}{\bf{ - }}{{\bf{9}}^{{\bf{2399}}}}{\bf{ + }}{{\bf{2}}^{{\bf{2001}}}}\\{\bf{c = 2}}{{\bf{4}}^{{\bf{4493}}}}{\bf{ - }}{{\bf{5}}^{{\bf{8192}}}}{\bf{ + }}{{\bf{7}}^{{\bf{1777}}}}\end{aligned}\)

As a, b, c are three numbers then each number can have two signs either negative or non-negative.

It may be possible that at least two of these numbers are negative or non-negative.

Cases for the given values

There are eight cases for these numbers.

\(\begin{aligned}{}{\bf{a > 0,b > 0,c > 0}}\\{\bf{a > 0,b > 0,c < 0}}\\{\bf{a > 0,b < 0,c > 0}}\\{\bf{a > 0,b < 0,c < 0}}\end{aligned}\)

\(\begin{aligned}{}{\bf{a < 0,b > 0,c > 0}}\\{\bf{a < 0,b > 0,c < 0}}\\{\bf{a < 0,b < 0,c > 0}}\\{\bf{a < 0,b < 0,c < 0}}\end{aligned}\)

From all the above cases, either of the two facts if true.

Therefore, the multiplication of two of the numbers is non-negative.

As the exact explanation is not given, it is non-constructive proof.