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 Subject: "distributive law" Previous Topic | Next Topic
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JJe (Guest)
guest
Jul-20-01, 09:35 PM (EST)

"distributive law"

 Hi!I was reading this page on this site: . At the bottom, there is a proof for the distributivity of multiplication over addition for natural numbers. However, I think there is a step that seems suspect, that is:x*(y+z) + x = (x*y + x*z) + xThat step looks like it uses the very thing which is being proven here. Is it'so, or am I overlooking something rather trivial?Currently, I'm a third-year student of theoretical physics in university, so I think I should know enough to understand this.JJe---https://mp3.com/jje - Ambient music/miscellanelous experimentshttps://jje.cjb.net - Useless little programs and other stuff

alexb
Charter Member
672 posts
Jul-20-01, 09:47 PM (EST)

1. "RE: distributive law"
In response to message #0

 >However, I think there is >a step that seems >suspect, that is: >>x*(y+z) + x = (x*y + x*z) + x >>That step looks like it uses >the very thing which is >being proven here. Is it >so, or am I overlooking >something rather trivial? Yes. The proof of the distributive law is by induction in z.Just before that derivation I say, "Let M be the set of all z for which the Law holds. We just saw that, by Definition 2, 1 is in M. Let z be in M. Then ..."At this point we do not know yet whether the law holds universaly, but do know that it holds for 1 and have assumed that it also holds for z - a specific but arbitrary number, which means thatx*(y+z) = x*y + x*zby our assumption. From here it follows, as the derivation you pointed to shows, that x*(y+z') = x*y + x*z'and, by induction, it holds for all numbers.>Currently, I'm a third-year student of >theoretical physics in university, so >I think I should know >enough to understand this. That's OK. I think I should have mentioned that the proof is by induction. Sorry.