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