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We define how to multiply two integers using set theory as follows [(a,b)][(c,d)] = [(ac + bd,ad + bc)] what is the intuition behind defining multiplication like this?

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  • $\begingroup$ What part of set theory are you talking about? The usual definition of integer multiplication as part of Peano arithmetic is nothing like this. $\endgroup$ – DRF Mar 4 '15 at 16:10
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The pair $(a, b)$ (where $a$ and $b$ are both positive integers is supposed to represent the integer $a - b$. (The pair $(a+1, b+1)$ is an equally good representative, as are many other pairs). We know, from our experience with numbers in everyday life, that $$ (a-c)(b-d) = ab + cd -ad - bc = (ab+cd) - (ad + bc) $$ This tells us that $(ab+cd, ad+bc)$ would be a good representation for the product...provided that both entries of the ordered pair are positive integers. Fortunately, they are in fact both positive, so we're good.

(Your book may specify "nonnegative" rather than "positive" ... both can be made to work).

You might want to ask yourself "What's wrong with choosing $(ab-ad, cd - cb)$ as the representation?" as a way to verify that you understand what's going on here.

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  • $\begingroup$ the reason if we specify that (ab - ad,cd - cb) is that (we would have (a + c)(b + d) = ac + ad + cb + cd so we have that (ab + cd) + (ad + bc) this would work only for positive numbers not for negatives aswell right? $\endgroup$ – user111750 Mar 4 '15 at 18:48
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    $\begingroup$ Well...not really. The problem with that second formulation is that if $d > b$, then $ab - ad$ isn't defined (because we're working only with POSITIVE integers, and the result of this subtraction would have to be negative). $\endgroup$ – John Hughes Mar 5 '15 at 20:02

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