Table of contents
“When mathematicians encounter a problem, they define their way out of it.”
- a wise old saying
When asked to explain multiplication, many of us might remember what we were taught. “Multiplication is repeated addition”. Makes sense at the first glance. Multiplying \(3\) by \(4\) is equivalent to adding \(3\) four times after all. But then, how does one explain why \((-1) \cdot (-1) = 1\)? Surely it cannot be that we add \((-1)\) negative one times. What does that even mean?? So what is multiplication by a negative number? Maybe it’s just defined a certain way, yes?
Nope, absolutely not. To understand how \((-1) \cdot (-1) = 1\), we need to recount an intriguing tale as old as time.
Happily counting
The world was created. Mankind was born. Sets were understood. Not very precisely at first, but intuitively. Sets are after all just well-defined collection of objects: cows, rocks, grains, numbers (those tiny symbols that are just so much easier to keep in mind and can be used for all collections…), you name it.
The beauty of a set lies in what we do with its elements. We can perform operations with the elements of a set. The very first (abstract) set humans were comfortable with was the set of natural numbers. Formally, the set of natural numbers, \(\mathbb{N}\), is defined as the set of numbers generated by adding one to each element, starting with the basis element \(1\), i.e., \(\{1,2,3,...\}\).
\(^1\)This is, in fact, the basis for mathematical induction.
They feel, for lack of a better word, natural! And with this set, comes the operation of counting, just as intuitively. When you can understand counting, it’s not hard to make the leap to addition\(^1\). After all, instead of adding \(1\) to a number, you add two numbers together.
Operations
Addition is an operation on the elements of a set. More specifically, it is a binary operation– it takes in two elements and produces a third element. Mathematically, we can define a binary operation as a function \(f: X \times Y \mapsto Z\) where \(X\), \(Y\) and \(Z\) are sets.
But the interesting thing about natural numbers and the addition operation is that \(X\), \(Y\) and \(Z\) are all natural numbers. You add two natural numbers, you get back a natural number. This is special – it deserves a name. Let’s call it the property of closure. Thus, the set \(\mathbb{N}\) is closed under addition.
A baby step towards algebraic structures
Let move into the abstract world for a bit. A set matched with a certain binary operation that follows the closure property constitutes a magma. Mathematically, a magma \(M\), with a binary operation \(\otimes\) is defined as a set such that
\[\forall \; \; \; a, b \in M \implies a \otimes b \in M\]This forms one of the most basic algebraic structures. What makes a set an algebraic structure, you ask? Well, an algebraic structure is a non-empty set, that has an associated collection of finite arity (binary, ternary, etc) operations and a finite number of identities that it must satisfy.
Figure 1: A visual summary of various algebraic structures
It’s now clear that our set of natural numbers is atleast a magma; it has a binary operation (addition) which follows the closure property. But natural numbers are cooler than that. Turns out, we can define another operation (name it multiplication) which is also closed! And turns out, these two operations don’t just follow the closure property but a bunch of others too!
The addition operation (\(+\)) on \(\mathbb{N}\):
- \(\mathbb{N}\) is closed under addition:
- \(\mathbb{N}\) is assosciative under addition:
Multiplication can be thought as ‘repeated addition’ for this set. Adding a number several times to itself is the same as multiplying it. The multiplication operation (\(\times\)) on \(\mathbb{N}\):
- \(\mathbb{N}\) is closed under multiplication:
- \(\mathbb{N}\) is assosciative under multiplication:
- \(\mathbb{N}\) has an identity for multiplication – yup, the multiplicative identity for natural numbers is 1:
In fact, the coolest part about these two operations lies in the way they interact with each other: the Distributive Law of Multiplication over Addition
\[\forall \; \; \; a, b, c \in \mathbb{N} \implies a \times (b + c) = (a \times b) + (a \times c)\]Enter the Zero
Now, the fact that multiplication on the set of natural numbers had an identity element, yet addition did not, caused great distress to many folks\(^2\). There just had to be an additive identity, right? Symmetry and all that? People hem’ed and haw’ed, trying to figure out what natural number when added to any other just gave the number back. Finally, the concept of \(0\) was invented\(^3\). Caveat being that \(0\) wasn’t a natural number!
\(^2\) Not meant to be historically accurate.
\(^3\) The invention of 0 is a fascinating story by itself!
Instead, a new set was created with natural numbers and \(0\), called whole numbers, \(\mathbb{W} = \{0,1,2,3,...\}\). Addition and multiplication continued to enjoy their initial definitions as well as their properties, with slight extensions – addition now had an identity element. Adding \(0\) to a number resulted in itself (by definition), essentially doing nothing.
Don’t subtract. Be negative instead!
But this new set also brought the question that was brewing in a lot of minds to the fore, “What numbers can I add to get a result of \(0\)?”. Well, there was no answer in the set of whole numbers, apart from the trivial solution of adding \(0\) with itself. You add nothing to nothing and get nothing. That’s unsurprising. But getting nothing from adding something to something? You simply cannot end up with \(0\) from two whole numbers and the addition operation.
What if you change the operation? It seems quite obvious that when you have 5 stones and you want to end up with 0, the key is remove the 5 stones instead of adding anything. Subtraction. But subtraction didn’t quite have any of the attractive properties that addition did. It wasn’t even closed. You can quickly end up in nowhere land if you rely on subtraction. No. It had to be addition.
Mathematicians did the only thing that was decent. They defined a set of numbers called negative integers (gentle reader, note the emphasis on define), \(\{-1,-2,-3,...\}\), with the property that addition of a negative integer with the corresponding positive integer results in \(0\).
Wait, whaat?!!!
Putting it all together
This is huge. Think about this for a moment. We defined \((-5)\) as the quantity which when added to \(5\), results in \(0\). In fact, \((-5)\) is just a symbol with the above property. In other words, \(5\) and \((-5)\) are additive inverses of each other. Combining these newly defined numbers with our set of whole numbers, we get the set of integers, \(\mathbb{Z} = \{...,-3,-2,-1,0,1,2,3,...\}\).
We established \(0\) as the additive identity (adding \(0\) to a number gives the number back) and we saw the reason for the existence of additive inverses. But how do we extend multiplication’s properties with respect to \(0\)? Turns out we can prove that multiplication with \(0\) gives back \(0\)!
We know that,
\(1 \times 0 = 1 \times (0+0) \;\) , since \(0 = 0 + 0\)
\(\implies 1 \times 0 = (1\times 0) + (1\times 0) \;\) , from the distributive law of multiplication over addition!
\(\implies 0 = 1 \times 0 \;\) , on cancelling out \((1 \times 0)\) from both sides!
Since we could have picked any integer instead of \(1\) in the above argument, but resulting in the same way, we can conclude that multiplication with \(0\) gives \(0\) back.
And now, we are equipped to answer the original question we started out with.
Why is \((-1) \times (-1) = 1\)?
We know that,
\((-1) \times 0 = 0 \;\) , since multiplication with 0 gives 0
\((-1) \times (-1 + 1) = 0 \;\) , since -1 and 1 are additive inverses
\(((-1) \times (-1)) + ((-1) \times (1)) = 0 \;\) , from the distributive law of multiplication over addition
\(((-1) \times (-1)) + (-1) = 0 \;\) , since 1 is the multiplicative identity
\(\implies (-1) \times (-1) = 1 \;\) , since -1 and 1 are additive inverses