# Fix any errors in my proof (beginner)

Fix any errors in correctness, structure, etc.

Theorem.
Suppose that a and b are nonzero real numbers. Prove that if $a < 1/a < b < 1/b$, then $a < ?1$.

Proof. By contrapositive, if  $a \geq-1$, then it should not be the case that $a < 1/a < b < 1/b$. Consider when a is -1 and b is $\frac{1}{2}$. The statement then would be $-1 < \frac{1}{-1} < \frac{1}{2} < 2$. In this case, $\neg (a < \frac{1}{a} < b < \frac{1}{b})$ because $-1 = \frac{1}{-1}$. Therefore, because the contrapositive is true, it must be the case that if $a < 1/a < b < 1/b$, then $a < ?1$.

Answers can be viewed only if
1. The questioner was satisfied and accepted the answer, or
2. The answer was disputed, but the judge evaluated it as 100% correct.