# Fix any errors in my proof (beginner)

*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 $a < ?1$.

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