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I've seen numerous explanation of why two negatives make a positive and I did understand most of them. Now, I found the following explanation :

If now we consider the product arising from the multiplication of the two quantities $a-b$, and $c - d$, we know that it is less than that of $(a-b)\times c$, or of $ac-bc$; in short, from this product we must subtract that of $(a - b) \times d$; but the product $(a - b)\times(c - d)$ becomes $ac-bc-ad$, together with the product of $-b\times-d$ annexed; and the question is only what sign we must employ for this purpose, whether $+$ or $-$. Now we have seen that from the product $ac-bc$ we must subtract the product of $(a-b)\times d$, that is, we must subtract a quantity less than $ad$; ◈ we have therefore subtracted already too much by the quantity $bd$; this product must therefore be added; that is, it must have the sign $+$ prefixed; hence we see that $-b\times-d$ gives $+bd$ for a product; or $-$ minus multiplied by $-$ minus gives $+$ plus.

The ◈ mark is the part where I'm not having the same conclusion... Does it seem convincing to you or valid ?

If so, could you rephrase it or explain it for me?? I'm not sure of following why this is THE conclusion!Thank you!

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    $\begingroup$ How old is that text...? $\endgroup$ – Pedro Tamaroff Dec 31 '13 at 2:07
  • $\begingroup$ 19th century I guess... not sure. It comes from Euler's elements of algebra, as a note on the bottom. (This note is not from Euler.) Why the question? $\endgroup$ – user108343 Dec 31 '13 at 2:08
  • $\begingroup$ What is the main question of yours? $\endgroup$ – mathlove Dec 31 '13 at 2:11
  • $\begingroup$ The part after the red line I'm not understanding. I'm not sure what the text is trying to say... A reformulation or another explanation would really help me :/ thank you again! $\endgroup$ – user108343 Dec 31 '13 at 2:13
  • $\begingroup$ You mean you want to know the proof that $-\cdot -=+$? or you want to know what the text is trying to say? $\endgroup$ – mathlove Dec 31 '13 at 2:20
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I'm going to write what the text is trying to say.

1) $(a-b)\times (c-d)$ is less than $(a-b)\times c=ac-bc.$

2) $(a-b)\times (c-d)=(a-b)\times c-(a-b)\times d=(a-b)\times c-ad+(-b)\times (-d)$.

3) $(a-b)\times d$ is less than $ad$.

4) $(-b)\times (-d)$ has to be plus.

Obviously the auther assumes $a,b,c,d\gt0$.

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  • $\begingroup$ Yeah, that's what I understood. it's only the fourth part which I'm not following.... $\endgroup$ – user108343 Dec 31 '13 at 2:43
  • $\begingroup$ OK. You can compare $(a-b)\times c-(a-b)\times d$ with $(a-b)\times c-ad+(-b)\times (-d)$. Since $(a-b)\times d\lt ad$, $-ad$ means 'substract too much'. $\endgroup$ – mathlove Dec 31 '13 at 2:45
  • $\begingroup$ In other words, $(a-b)\times c-(a-b)\times d\gt (a-b)\times c-ad.$ So, $(-b)\times (-d)$ has to be plus. $\endgroup$ – mathlove Dec 31 '13 at 2:48
  • $\begingroup$ i think the inequality sign is a mistake... $\endgroup$ – user108343 Dec 31 '13 at 2:53
  • $\begingroup$ No. This is fine. Since $(a-b)\times d\lt ad$, we have $-(a-b)\times d\gt -ad.$ Hence, $(a-b)\times c-(a-b)\times d\gt (a-b)\times c-ad.$ $\endgroup$ – mathlove Dec 31 '13 at 2:55
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The author makes a poor argument here, because he assumes that we somehow already know that $(-b)\times(-d)$ equals $bd$ or $-bd$ and we just have to figure out which alternative is true.

Anyway, he says that (assuming $a,b,c,d>0$) we have $(a-b)d<ad$ and hence $ac-bc-ad>(a-b)(c-d)$, from which it follows that $(-b)(-d)>0$.

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Cool article. Maybe if you put it in an inequality:

$$(a - b)*c > (a-b)(c-d)$$

then:

$$ac - bc < ac - da - bc \;?\; bd$$

Reorder and group:

$$(ac - bc) < (ac - bc) - da \;?\; bd$$

$$da<\;?\;bd$$

The right side can't be less than the left side so if all a,b,c,d are positive integers (although it wasn't stated explicitly in the text above so I'm assuming) $bd$ must added to keep the right side greater. This is what I gathered.

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