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

#Algebra

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Comments

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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 - 6 years, 7 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.

Steven Zheng - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$