Minus minus and negative times negative

mathematics
Simple algebraic proofs demonstrating why subtracting a negative yields a positive and why multiplying two negatives results in a positive
Published

February 25, 2019

Conceptual artwork showing hands holding a central blue sphere with a white plus sign, surrounded by floating white spheres bearing minus signs.

KJ Runia, BSc (Hons) MInstP AMIMA

Holds a Bachelor of Science (Honours) degree in Mathematics and Physics from the School of Mathematics and Statistics and the School of Physical Sciences at The Open University, Walton Hall, Milton Keynes in the United Kingdom. Studies currently for an MPhys (Master of Physics). Is a Member of the Institute of Physics (IOP) and an Associate Member of the Institute of Mathematics and its Applications (IMA).

Minus minus is plus. And negative times negative is positive. Two negatives make a positive. You may have heard or uttered these expressions many times. Even though you will know this already, here you will find an algebraic proof, just for your reference. Requirements: simple algebra from secondary school.

Minus minus is plus

We all learnt in school that subtracting a negative number is equivalent to adding the positive counterpart of that number. For example:

\[1 - (-2) = 1 + 2 = 3.\]

Using variables instead of specific numbers, we can state this generally as:

\[a - (-b) = a + b,\]

where \(a\) and \(b\) are arbitrary real numbers.

Let us prove this by contradiction. Suppose, instead, that the opposite were true:

\[a - (-b) = a - b.\]

Subtracting \(a\) from both sides:

\[-(-b) = -b.\]

Placing brackets around \(-b\) on the right-hand side makes the structure clearer:

\[-(-b) = (-b).\]

Let \(c = (-b)\). Substituting \(c\) into the equation gives:

\[-c = c,\]

which is absurd for any non-zero real number (for instance, \(-1 = 1\)). Because the assumption leads directly to an impossible contradiction, the original statement must hold:

\[a - (-b) = a + b.\]

Negative times negative is positive

We also learnt that multiplying two negative numbers yields a positive product. We will prove that:

\[(-a)(-b) = ab,\]

for any real numbers \(a\) and \(b\).

We begin with the basic identity:

\[(-a)(-b) = (-a)(-b).\]

Adding zero does not alter the value of an expression:

\[(-a)(-b) = (-a)(-b) + 0.\]

Since any real number multiplied by zero equals zero (\(a \times 0 = 0\)), we rewrite zero as \(a(0)\):

\[(-a)(-b) = (-a)(-b) + a(0).\]

Now, we express \(0\) as the difference \((b - b)\):

\[(-a)(-b) = (-a)(-b) + a(b - b).\]

Expanding the brackets using the distributive property (\(a(b - b) = ab + a(-b)\)):

\[(-a)(-b) = (-a)(-b) + ab + a(-b).\]

Commuting the second and third terms:

\[(-a)(-b) = (-a)(-b) + a(-b) + ab.\]

Now look at the first two terms on the right-hand side:

\[(-a)(-b) = \Big[ (-a)(-b) + a(-b) \Big] + ab.\]

Factoring out the common factor \((-b)\) from the bracketed terms:

\[(-a)(-b) = (-b)\Big[(-a) + a\Big] + ab.\]

Because \((-a) + a = 0\), the first term vanishes completely:

\[(-a)(-b) = (-b)(0) + ab = 0 + ab = ab.\]

Therefore:

\[(-a)(-b) = ab. \quad \blacksquare\]

Image credits and references

  • Featured image: Conceptual 3D spheres illustration via Pixabay (CC0 Public Domain).
  • Algebraic proofs and derivations by KJ Runia.