# Product of Negative Numbers

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

Note by Steven Zheng
3 years, 10 months ago

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You should include the definition of a positive number, and explain why the product of 2 positive numbers is also positive.

Staff - 3 years, 10 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.

- 3 years, 10 months ago