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Let $R$ be a commutative ring, $a\in R$ and $b\in R-\{0\}$ we say that $b$ divides $a$ (notation: $b\mid a$) if $\exists t\in R$ such that $a=bt$.

In my book I often see the symbol $\dfrac{a}{b}$ and it confuses since it looks like usual division. But I guess that $\dfrac{a}{b}$ is the same as $t$. However, I think that the notation $\dfrac{a}{b}$ is informal. Am I right?

RFZ
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  • Out of curiosity, which book are you referring to? – quasi May 25 '18 at 16:12
  • a/b is usual division. – quasi May 25 '18 at 16:13
  • @quasi, Herstein's book "Topics in algebra" – RFZ May 25 '18 at 16:13
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    Herstein does not use $a/b$ to mean $a|b$. the notation $a/b$ refers to fraction. The notation $a|b$ is a statement asserting that $b$ is a multiple of $a$. See rschwieb's answer. – quasi May 25 '18 at 16:14
  • @quasi, I am just aksing what does $\frac{a}{b}$ mean? For example, what is the following symbol $c=\frac{a}{b}$? – RFZ May 25 '18 at 16:15
  • It means a fraction (the result of division). $c=a/b$ means $a=bc$, whereas $b|a$ means $a=bc$, for some $c$. Note the expression $a/b$ represents a calculation, where as $b|a$ is a statement which is either true or false. – quasi May 25 '18 at 16:16
  • @RFZ Just read existing solutions and ask questions if you have to. I’m not going to repeat content like that. Start here https://math.stackexchange.com/q/349858/29335 – rschwieb May 27 '18 at 01:51

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The symbols $\frac ab$ and $a|b$ are completely different things.

The first one usually denotes the equivalence class of pairs equivalent to the pair $(a,b)$ via the relation $(a,b)\sim (c,d) \iff ad=cb$. This is the same thing as $ab^{-1}$ and also $a/b$. In a nutshell, we are talking about "the result of $a$ divided by $b$" here.

The notation $a|b$ denotes that $a$ and $b$ are related by divisibility, i.e. that $ar=b$ for some suitable $r$. This doesn't produce any "result" like the last one: it just is a statement about how $a$ and $b$ are related.

rschwieb
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  • For example, I need to show that $\text{gcd}(a,b)=\dfrac{ab}{\text{lcm}(a,b)}$. Is it the same thing as $ab=\text{lcm}(a,b)\text{gcd}(a,b)$? – RFZ May 25 '18 at 16:25
  • @RFZ Yes, those two equalities mean the same thing. – rschwieb May 25 '18 at 16:29
  • How does the equivalence relation here refer? We do not use it in the symbol $(a,b)=\dfrac{ab}{[a,b]}$ – RFZ May 25 '18 at 16:30
  • Not really. The first claims more. It usually claims that lcm(a,b) is invertible in the ring. Even if not, It surely claims that gcd(a,b) is unique such that the second equality holds. – C Monsour May 25 '18 at 16:32
  • @RFZ I don't understand what you are asking in the last comment. Maybe this will help: if $a,x,b$ are nonzero elements in a field, and $a=bx$, you can say, simultaneously, that $x=\frac ab$, $b=\frac ax$ and $b|a$ and $x|a$. – rschwieb May 25 '18 at 16:33
  • @rschwieb, Regarding your last comment, I know these facts are true for equation in the field. For example, I need to prove that $(a,b)=\frac{ab}{[a,b]}$ for $a,b\in R-{0}$, where $R$ is euclidean ring. How to prove it? As you said the RHS, namely the fraction $\frac{ab}{[a,b]}$ is the equivalence class, which consists of pairs $(c,d)$ such that $\frac{ab}{[a,b]}\sim \frac{c}{d}$ (in other words, $(ab)d=c[a,b]$). However, the LHS is just $(a,b)$ which is an element of euclidean ring $R$. So we need to prove that element is equal to the equivalence class? – RFZ May 26 '18 at 09:49
  • But it sounds very weird and wrong. Could you clarify my thoughts, please. – RFZ May 26 '18 at 09:49