# Product of two negative numbers is positive [duplicate]

What is the practical proof for $-1(-1)=+1$. Actually multiplication is repetitive addition. I am struggling how can I provide an activity to prove practically $-1(-1)=+1$

• It's not a matter of proof -- but a convention that happens to make the laws of arithmetic hold with greater generality than other possible conventions for the product of negative numbers would. – Henning Makholm Jul 26 '14 at 16:35
• I'm not sure why people keep claiming that multiplication is repeated addiction. Or exponentiation is repeated multiplication. All of these operation are only defined that way on natural number, but there is no reasons to expect that they are still defined the same way once it is extended, nor even the same property. See also: exponentiation with exponent being matrices; fractional derivative. – Gina Jul 26 '14 at 16:41
• Again, it is hard to find an "intuitive" insight. We can start with multiplication as repeted addition. Then, move on to multiplication of a negative number for a positive one: thus, $5 \times (-2)=-10$ becuse we add five time the quantity $-2$, and this is "obviously" a negative quantity (we move to the left with reference to $0$). Finally, we must think to mult for $-1$ as reversing the direction. Thus $(-5) \times (-2) = 10$ becuse it is : $(-1) \times 5 \times (-2) = (-1) \times (-10)$. – Mauro ALLEGRANZA Jul 26 '14 at 19:06

Law of Signs proof: $\rm\,\ (-x)(-y) = (-x)(-y) + x(\overbrace{-y + y}^{\large =\,0}) = (\overbrace{-x+x}^{\large =\,0})(-y) + xy = xy$

Equivalently, evaluate $\rm\:\overline{(-x)(-y) +} \overline{ \underline {x(-y)}} \underline{ +\,xy}\,$ in 2 ways, noting each over/under term $= 0$

Said more conceptually, $\rm\:(-x)(-y)\$ and $\rm\:xy\:$ are both inverses of $\rm\ x(-y)\$ so they are equal by uniqueness of inverses: if $\,a\,$ has two additive inverses $\,\color{#c00}{-a}\,$ and $\,\color{#0a0}{-a},\,$ then

$$\color{#c00}{-a}\, =\, \color{#c00}{-a}+\overbrace{(a+\color{#0a0}{-a})}^{\large =\,0}\, =\, \overbrace{(\color{#c00}{-a}+a)}^{\large =\,0}+\color{#0a0}{-a}\, =\, \color{#0a0}{-a}$$

This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.

These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.

• ❤️ last paragraph – garageàtrois Oct 15 at 19:50

$$1 + (-1) = 0$$ $$-1(1 + (-1)) = -1(0)$$ $$-1(1) + -1(-1) = 0$$ $$-1 + -1(-1) = 0$$ $$1 + (-1) + -1(-1) = 0 + 1$$ $$0 + -1(-1) = 1$$ $$-1(-1) = 1$$

• Note that the proof implicitly assumes that the distributive law holds for all integers, not just positive integers. – Bill Dubuque Jul 26 '14 at 17:28

If you assume some basic properties of addition and multiplication:

$$(-1) \times 0 = 0$$ $$(-1) \times ((-1) + 1) = 0$$ $$(-1) \times (-1) + (-1) \times 1 = 0$$ $$(-1) \times (-1) + (-1) = 0$$ $$(-1) \times (-1) + (-1) +1 = 0+1$$ $$(-1) \times (-1) + 0 = 1$$ $$(-1) \times (-1) = 1$$

• Note that the proof implicitly assumes that the distributive law holds for all integers, not just positive integers. – Bill Dubuque Jul 26 '14 at 17:28