Product of Negative Numbers

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

Solution

Let AA and BB be two positive real numbers. Subsequently, −A -A and −B-B are the respective additive inverses.

−A⋅−B+−A⋅B=−A⋅(−B+B)-A \cdot -B+ -A\cdot B = -A \cdot (-B+B) −A⋅(−B+B)=0-A \cdot (-B+B) = 0

We then add A⋅BA \cdot B to both sides of the equation:

−A⋅−B+−A⋅B+A⋅B=0+A⋅B-A \cdot -B+ -A \cdot B+A \cdot B = 0+A \cdot B −A⋅−B+(−A+A)⋅B=0+A⋅B-A \cdot -B+(-A + A) \cdot B = 0+A \cdot B

which simplifies to −A⋅−B=A⋅B.-A \cdot -B = A \cdot B.

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

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

#Algebra

Note by Steven Zheng
4 years, 11 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 - 4 years, 11 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 - 4 years, 11 months ago

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