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Multiplication of:

  1. $3\times 4=12$, obvious

  2. $-3 \times 4=-12$, obvious

  3. $-3 \times -4=12$, not so obvious. We are just told to memorises it.

I would like to demonstrate that the multiplication of two negatives numbers is positive.

Here is my steps:

$5-3=2$

let $a-b=1$

$5-2(a-b)=3$

$5-2a-2(-b)=3$

$-2(-b)=3-5+2a$

$-2(-b)=2(a-1)$ where $a-1=b$

$-2(-b)=2b$

Are my steps correct? If not valid, I would like to see your method of showing the multiplication of two negative numbers is positive.

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    $\begingroup$ Usually, I think multiplication with negative number e.g; $-2$ as multiplying the number by $2$ and rotating $180$ deg. in the complex plane, keeping the length same. As, $i^2=-1$ and multiplying by $i$ can be interpreted as rotating by $90$ deg. keeping the magnitude unchanged. $\endgroup$ Apr 16, 2019 at 10:55
  • $\begingroup$ That it is a very interesting way of looking at it $\endgroup$
    – Endgame
    Apr 16, 2019 at 10:57
  • $\begingroup$ By associativity, $-a\cdot -b=(a)(-1)(b)(-1)=ab\cdot(-1)(-1)$. So the only real question is, why is it that $(-1)(-1)=1$? $\endgroup$ Apr 16, 2019 at 13:51
  • $\begingroup$ is repeated addition giving the same result as multiplication obvious ? is subtraction of a negative being addition of a positive obvious ? $\endgroup$
    – user645636
    Apr 16, 2019 at 14:15
  • $\begingroup$ Why is 2 in your opinion more obvious than 3? $\endgroup$
    – user
    Apr 16, 2019 at 21:55

2 Answers 2

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$\overbrace{\bf\ Law\ of\ Signs}^{\rm\Large {(-x)(-y)}\ =\ xy} $ proof: $\rm\,\ (-x)(-y) = (-x)(-y) + \color{#c00}x(\overbrace{\color{#c00}{-y} + y}^{\Large =\,0}) = (\overbrace{-x+\color{#c00}x}^{\Large =\,0})(\color{#c00}{-y}) + xy = xy.$

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  • $\begingroup$ He is aware of that law, just want a convincing explanation ig. $\endgroup$ Apr 16, 2019 at 10:45
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    $\begingroup$ His approach lacks rigour and proper notation. It is handwavey at best. If his argument used my approach he would not need convincing. $\endgroup$ Apr 16, 2019 at 10:50
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I would like to see your method

Multiplication of $p$(suppose in $\mathbb{Z}$) by $i=\sqrt{-1}$ can be interpreted geometrically as rotating the line joining $(p,0)$ and $(0,0)$ by $90$ deg. without scaling the line. Suppose we are multiplying $3$ and $-1$. As, multiplying by $-1=i^2$ is same as multiplying by $i$ twice, hence we can interpret multiplication by $-1$ as rotating the line made by $(0,0)$ and $(3,0)$ by $180$ deg. without scaling it and looking at the point where the line cuts the imaginary axis.

Now, if you multiply $3$ with $-2$, think it like $3\times 2\times -1$, i.e; multiply the multiplicand by the magnitude of the multiplier and then rotate the line $(6,0)$ by $180$ deg.

Same for $-3\times -2$. We can rewrite like $3\times 2\times -1\times -1$. Multiplying $3$ and $2$ gives $6$, and multiplying by $-1$ twice means rotating the line joining $(0,0)$ and $(6,0)$ by $180$ deg twice, or a total $360$ rotation and you will land on $(6,0)$ again!

I think this way will convince you.

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