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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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