1

I'm a computer science graduate and I'm not sure to which level of Algebra this question belongs to, whether it is abstract algebra or some other filed, but I'm curious to know and learn.

First, before the usage of negative numbers in multiplication, multiplication was defined as repeated addition of positive numbers(or integers), so we can have the equation $$x^2 -4x + 3 = 0$$

Where I mean by the multiplication in $x^2$ and $4x$ multiplication as repeated addition, I know nothing about negative numbers multiplication yet.

We can rearrange the terms and factorize so we have:

$$(x-2)(x-2) = 1$$

Now, if we think of multiplication as repeated addition, with no mention of negative numbers multiplication, then $(x-2)$ can only be positive, so we have $(x-2) ^2 = 1$ so we have one solution, namely, $x = 3$.

But for me, here comes the problem: there is another solution, namley $x = 1$, and we can find this solution only in case we allowed $(x-2)$ to be negative number, and defined multiplication of negative numbers by negative numbers so for example

$$-1 * -1 = 1$$

in which case $(x-2) = - 1$ and we have $x = 1$.

So how come that we first dealt with multiplication as repeated addition, but then we needed to define(or extend) it in another way so that we can get all the solution of an equation that contained terms multiplied in the sense of repeated addition only?

Can someone recommend a textbook that explain this and related stuff from the beginning, illustrating the motivation and building on this possibly also for complex numbers as well? Again I'm not sure in which branch or level of Algebra these stuff are explained.

Edit:

By the way my question isn't: "why is multiplication of negative numbers defined as it is", which answer is: "to preserve the structure", but rather why must we consider multiplication of negative numbers in the mentioned example equation to arrive at all solutions ! That is it seems that multiplication using negative numbers in the mentioned equation is so inherent and fundamental. I'm not sure if I phrased my question correctly but here it goes...

Loai Ghoraba
  • 259
  • 1
    See https://math.stackexchange.com/questions/156264/building-the-integers-from-scratch-and-multiplying-negative-numbers, https://math.stackexchange.com/questions/9933/why-is-negative-times-negative-positive, and many other questions. – Hans Lundmark Mar 13 '23 at 18:24
  • 2
    Note that we can also find the solution $x=1$ by rearranging the equation to have no instances of subtraction, namely as $$x^2+3=4x.$$ Now verifying that $x=1$ is a solution is entirely within the "positive realm." You should think of the extension of multiplication to negative numbers as being governed by the desire to stay true to algebraic facts we already know. – Noah Schweber Mar 13 '23 at 18:42
  • Thanks all, referring to my last question: "Can someone recommend a textbook that explain this and related stuff from the beginning, illustrating the motivation and building on this possibly also for complex numbers as well? Again I'm not sure in which branch or level of Algebra these stuff are explained." Can someone give an answer? – Loai Ghoraba Mar 13 '23 at 18:59
  • See also Is multiplication not just repeated addition?. The key point is that multiplication (by repeated addition) is linear (satisfies the distributive laws) and we wish this key property to persist [cf. structure preservation] when we adjoin additive inverses ("negatives"), else the new additive inverses will have no relation to the multiplicative structure. Then the Law of Signs follows immediately from distributivity. – Bill Dubuque Mar 13 '23 at 23:23
  • Abstracting from the above leads to the definition of a ring, where the distributive laws are the only axioms connecting the additive and multiplicative structure. – Bill Dubuque Mar 13 '23 at 23:23
  • @BillDubuque thanks a lot, where is this material typically covered? In Abstract algebra or what? Can you recommend a related textbook? – Loai Ghoraba Mar 14 '23 at 05:21
  • Most abstract algebra courses are rather scant on motivational material. If memory serves correct you may find some such material in Numbers, by Ebbinghaus et al.. You may find helpful articles on the historical development of abstract algebra and the axiomatic method, e.g. articles by Leo Corry, e.g. his book Modern Algebra and the Rise of Mathematical Structures. – Bill Dubuque Mar 14 '23 at 06:46
  • Some of these ideas become clearer when one studies category theory, e.g. in terms of universal properties and adjoints, etc, e.g. see Bergman's beautiful book – Bill Dubuque Mar 14 '23 at 06:46
  • @BillDubuque thanks a million, that's was really helpful.

    By the way my question wasn't why multiplication of negative numbers is defined as it is, but rather why we must consider multiplication of negative numbers in the mentioned example equation to arrive at all solutions of the equation, which are themselves positive! That is it seemed to me that multiplication using negative numbers in the mentioned equation is so fundamental.

     – Loai Ghoraba Mar 14 '23 at 08:11
  • @BillDubuque you can see my edit in the question now. – Loai Ghoraba Mar 14 '23 at 08:30
  • 1
    That's simple: completing the square (quadratic formula) transforms to a reduced quadratic $,\bar x^2 = d,$ which has a positive and negative root (when $d>0)$. So even if the original roots $,x,$ are both $> 0,$ the arithmetic takes a detour through negative numbers. Similarly the computation may involve fractions even if the roots are integers, and the cubic formula may take a detour through complex numbers even for real roots. – Bill Dubuque Mar 14 '23 at 09:13
  • 1
    Only by passing to these number systems extending $\Bbb N$ can we obtain a single uniform formula for solving quadratics. In ancient times before negatives, fractions or complexes were invented, they had no such quadratic formula - instead they had to consider numerous cases to ensure they worked only with positive integers. Enabling uniform methods for solving polynomial equations is one of the primary motivations that led to such extended number systems. – Bill Dubuque Mar 14 '23 at 09:13
  • 1
    Well, you could've factorised it as $(2-x)(2-x)=1$ also, at least in this particular example... – Macavity Mar 15 '23 at 18:11

0 Answers0