When multiplying two numbers, there is a set of rules as to negatives and positives. We are taught that two positives give a positive, a negative, and a positive give a negative, and that two negatives multiplied together give a positive value. These rules can be visualized in the following way:
When you multiply a number by a negative number, the result is flipped to the opposite side of the number line. If both numbers are negative, the first flip makes the result negative, and the second flip returns it to positive. While this is a common way to understand multiplying negative numbers, it is not a formal mathematical proof. So why does multiplying two negatives give a positive value?
Before answering that question, let’s define some basic mathematical terms.
In mathematics, an equation, mathematical rule or statment that has not yet been proven true is called a conjecture. In this case, the conjecture might be:
For any real numbers \(a\) and \(b\), if both \(a\) and \(b\) are less than zero, \(a\) times \(b\) results in a value larger than zero.
In more formal writting, this might give:
If \(a,b\ <0\), then \(a\cdot b>0\ \)
Of course, this conjecture has already been proven. Once a conjecture is proven, it is called a theorem.
But how do you prove something as true? What constitutes a proper “proof”? To prove something is true, you can use theorems. But those theorems need proofs of their own, so where did it all start? The answer is axioms. Axioms are the foundation of all theorems and proofs. They are statements that mathematicians take for granted. These statements are often very simple and obviously true. A simple example of an axiom could be: For any real numbers \(a\) and \(b\), \(a+b=b+a\). This would be a statement, a sort of really simple theorem that we can assume is true, allowing us to prove other, more complex theorems.
Axioms can vary depending on context. Choosing different axioms leads to different forms of mathematics. For example, many mathematical problems can be defined in set theory. To solve these problems, you need set theory axioms. The most commonly accepted ones are the nine Zermelo-Fraenkel axioms. You can find more information on these axioms Here.
Here is a little recap of what has been covered:
-> A conjecture is a mathematical fact that has not yet been proven true.
-> A theorem is a conjecture that has been proven true.
-> An axiom is a mathematical fact taken for granted and serves as the starting point for theorems.
-> A proof is a set of logical operations that link axioms and known theorems together to prove some mathematical fact as true.
Now we are ready to prove that two negatives make a positive. We will start by defining the axioms we will use for our proof.
Axiom 1 : \(a+b=b+a\)
Axiom 2 : \(a(b+c)=ab+ac\)
Axiom 3 : \(a+-a=0\)
Axiom 4 : \(a+0=a\)
Axiom 5 : \(a\times0=0\)
Axiom 6 : \(ab=ba\)
Axiom 7 : A negative number is the additive inverse of some positive number. In other words, if \(a > 0\) and \(a+b=0\), then \(b\) is a negative number.
Axiom 8 : If \(a>0\) and \(b>0\), then \(ab>0\) In other words, the product of two postive numbers is positive.
Note that these aren’t axioms mathematicians would typically use in mathematical proofs, but for the purpose of this article, the eight axioms presented above will suffice.
Using these axioms, we need to show that for any two number \(a, b < 0\), the product of those two numbers is greater than zero: \(ab>0\)
We will start by showing: \(ab=-a\cdot-b\)
But to do that, we need to show: \(-ab=(-a)\cdot b\)
Here is how this can be done:
\(-ab=-ab+0\)
\(=-ab+b\times 0\)
\(=-ab+b(a-a)\)
\(=-ab+ba+b\cdot(-a)\)
\(=ba+-ab+b\cdot(-a)\)
\(=ab+-ab+b\cdot(-a)\)
\(=0+b\cdot(-a)\)
\(=b\cdot(-a)+0\)
\(=b\cdot(-a)\)
\(=(-a)\cdot b\)
(Axiom 4, \(a+0=a\) )
(Axiom 5, \(a\times0=0\) )
(Axiom 3, \(a+-a=0\))
(Axiom 2, \(a(b+c)=ab+ac\))
(Axiom 1, \(a+b=b+a\))
(Axiom 6, \(ab=ba\))
(Axiom 3, \(a+-a=0\))
(Axiom 1, \(a+b=b+a\))
(Axiom 4, \(a+0=a\))
(Axiom 6, \(ab=ba\))
This gives us the desired equation: \(-ab=(-a)\cdot b\). Let’s call this equation Theorem 1.
Theorem 1 : \(-ab=(-a)\cdot b\)
From here, we can now show that \(ab=(-a)\cdot(-b)\)
\(ab = ab + 0\)
\(=ab+(-a)\times 0\)
\(=ab+(-a)(b-b)\)
\(=ab+(-a)b+(-a)\cdot(-b)\)
\(=ab+-ab+(-a)\cdot(-b)\)
\(=0+(-a)\cdot(-b)\)
\(=(-a)\cdot(-b)+0\)
\(=(-a)\cdot(-b)\)
(Axiom 4, \(a+0=a\) )
(Axiom 5, \(a\times0=0\) )
(Axiom 3, \(a+-a=0\))
(Axiom 2, \(a(b+c)=ab+ac\))
(Theorem 1, \(-ab=(-a)\cdot b\))
(Axiom 6, \(ab=ba\))
(Axiom 1, \(a+b=b+a\))
(Axiom 4, \(a+0=a\))
This gives us our second desired equation. Let’s call this eqution Theorem 2.
Theorem 2 : \(ab=(-a)\cdot(-b)\)
(Axiom 7) : This implies : \(-a, -b < 0\)
(Axiom 8) : And \(ab>0\)
(Theorem 2): Since \(ab=(-a)(-b)\)
Then we know that \((-a)\cdot(-b)>0\), showing that the product of any two negative numbers must be a positive value.