Prove why the product of two negative real numbers is a positive real number.

**Solution**

Let \(A\) and \(B\) be two positive real numbers. Subsequently, \( –A\) and \(–B\) are the respective additive inverses.

\[-A \cdot -B+ -A\cdot B = -A \cdot (-B+B)\] \[-A \cdot (-B+B) = 0\]

We then add \(A \cdot B\) to both sides of the equation:

\[-A \cdot -B+ -A \cdot B+A \cdot B = 0+A \cdot B\] \[-A \cdot -B+(-A + A) \cdot B = 0+A \cdot B\]

which simplifies to \[-A \cdot -B = A \cdot B.\]

To prove why the quotient of two negative real numbers is a positive real number, just treat either \(A\) or \(B\) as the multiplicative inverse **(division is the multiplication of reciprocals)**.

Check out my other notes at Proof, Disproof, and Derivation

## Comments

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TopNewestYou should include the definition of a positive number, and explain why the product of 2 positive numbers is also positive. – Calvin Lin Staff · 2 years, 11 months ago

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– Steven Zheng · 2 years, 11 months ago

This problem assumes that you only know that the product of two positive reals is a positive real, and that there exists an additive inverse element for every real number. Maybe I should include that, but I assume the reader knows algebra. They just have to prove why this is so.Log in to reply