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.

Calvin Lin Staff - 3 years, 10 months ago

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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.

Steven Zheng - 3 years, 10 months ago

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