One thing that many of my students (at all levels!) seem to have trouble with are those pesky minus signs. For some reason people get very phobic around them and very often assume that when a minus sign comes up then they must have done something wrong. The fact is that minus signs are a very normal thing to arise – but that still seems to be no consolation to many.

One question that a lot of people find bewildering is the fact that when two negative numbers are multiplied together then a positive answer results – yes that’s right, for example $-3\times-4=12$ (NOT $-12$ as I’m sure I’ve said many hundreds of times during my time as a maths tutor). How is this so? It does seem a little counter-intuitive. Well…here’s a proof of it. The proof that follows will show, once and for all, that two negative numbers (real numbers if you must know, but then does it make sense anyway to talk about negative complex numbers or quaternions or anything like that?).

Here we go…

Throughout the proof I will assume the axioms of real numbers found in Mary Hart – Guide 2 Analysis Second Edition (i.e. that the real numbers are an abelian group under addition and an abelian group under multiplication, multiplication distributes over addition, the real numbers satisfy the order axioms and the completeness axiom).

First I want to prove that $0t=0$ for any $t \in \mathbb{R}$

$0t = (0+0)t$ as $0$ is the additive identity and therefore $0+0=0$

$0t = 0t + 0t$ by distributivity

$0t+(-0t)=(0t+0t)+(-0t)$

$0=0t+(0t+(-0t))$ by associativity of addition and $0t$ and $-0t$ are additive inverses

$0=0t+0$

$0=0t$  as required (*)

Next I want to prove that $-(-s)=s$ for all $s \in \mathbb{R}$

$-(-s)+(-s)=0$ since they are additive inverses

$(-(-s)+(-s))+s=s$

$-(-s)+((-s)+s)=s$ by associativity of addition

$-(-s)+0=s$ since $s$ and $-s$ are additive inverses

$-(-s)=s$ since $0$ is the additive identity (**)

Thirdly I need to prove that $s(-t)=-(st)=(-s)t$ for any $s,t \in \mathbb{R}$

$st+(-s)t = (s+(-s))t$ by distributivity

$st+(-s)t= 0t$ since $-s$ is the unique inverse of $s$

$st+(-s)t= 0$ by (*)

$-(st)=(-s)t$

similarly $-(st)=s(-t)$

and so $s(-t)=-(st)=(-s)t$  as required  (***)

Now I need to show that $(-s)(-t) = st$ for any $s,t \in \mathbb{R}$

$(-s)(-t)+(-((-s)(-t)))=0$ since they are additive inverses

$(-s)(-t)+(-(-s))(-t) = 0$ by (***)

$(-s)(-t)+s(-t)=0$ by (**)

$(-s)(-t) = -(s(-t))$ by uniqueness of inverses

$(-s)(-t) = s(-(-t))$ by (***)

$(-s)(-t) = st$ by (**)

It may appear that I am done here but in fact I have not said that (despite appearances) $-s$ and $-t$ are indeed negative; however, if we now assume that $s>0$ and $t>0$ then by the order axioms we have that $st>0$. But is it true that $-s$ and $-t$ are negative? Let’s find out…

since $s>0$ then by the order axioms we have that $s+(-s)>-s$. But $s+(-s)=0$ since they are inverses so $0>-s$. In other words $-s<0$ and thus $-s$ is negative. Similarly it can be shown that $-t$ is also negative hence $(-s)(-t)$ is indeed the product of two negative real numbers and thus since $(-s)(-t)=st$ and $st>0$ then $(-s)(-t)>0$ and the product $(-s)(-t)$ is positive. The proof is complete.

This proves that the fact that two negatives multiplied together give a positive is not just some rule that someone made up for a laugh to make maths difficult for GCSE and A-Level students – it is entirely consistent with the axioms of the real number system. Edmund Landau gives a much more thorough treatment of this in his book Foundations of Analysis, which I mentioned in my previous post on $2 \times 2 = 4$ in which the author builds gradually up to the real number system from the natural numbers and doesn’t assume the axioms that I have stated above. However this way of doing things is much longer to carry out taking Landau almost 90 pages to do; conciseness is desirable but must give way to thoroughness from time-to-time.

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