Question

Prove the following statements with contrapositive proof. (In each case, think about how a direct proof would work. In most cases contrapositive is easier.) Suppose \(x \in \mathbb{R}\). If \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 \geq 0\), then \(x \geq 0\).

Short answer

The original statement 'If \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 \geq 0\), then \(x \geq 0\)' is proved to be true by proving its contrapositive, 'If \(x < 0\), then \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 < 0\)', to be true.

Step-by-step solution

Understand the contrapositive statement

The contrapositive of the statement 'If P, then Q' is 'If not Q, then not P'. So, the contrapositive of the given statement is, 'If \(x < 0\), then \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 < 0\)'.

Contrapositive Proof

To prove the obtained contrapositive, consider an arbitrary negative real number \(x\). For \(x < 0\), all the terms in the equation become: \(x^{5} < 0, -4 x^{4} > 0, 3 x^{3} < 0, -x^{2} > 0, 3 x < 0\) and \( -4\) is a constant. Adding all these terms, we get a number less than 0, which means \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 < 0\).

Conclusion

We have proved the contrapositive statement to be true and therefore, by the law of contraposition, the original statement, 'If \(x^{5}-4 x^{4}+3 x^{3}-x^{2}+3 x-4 \geq 0\), then \(x \geq 0\)' is confirmed to be true.