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Why multiplying fractions is equal to multiply the tops, multiply the bottoms? $$\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},$$ And why $$\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},$$ Also why $$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$ I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference.

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    $\begingroup$ We define on $\Bbb R^2$ the operations $$(a,b)+(c,d)=(ad+bc,bd)$$ and $$(a,b)\times(c,d)=(ac,bd)$$ and we verify that $(\Bbb R^2,+,\times)$ is a field. $\endgroup$ – user63181 Oct 25 '14 at 22:25
  • $\begingroup$ As Sami hints, we define fractions in terms of equivalence classes of ordered pairs. You may find this lecture on properties of $\mathbb{Q}$ helpful; I certainly did at one point: youtube.com/watch?v=gTgkrVATzmk $\endgroup$ – Mathemanic Oct 25 '14 at 22:29
  • $\begingroup$ @SamiBenRomdhane Just what I wanted. $\endgroup$ – Dante Oct 25 '14 at 22:30
  • $\begingroup$ See, not a bad question at all! $\endgroup$ – David K Oct 26 '14 at 0:00
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I'll give an abstract look at why these identities hold in arbitrary fields.

In some sense, this is the definition of addition and multiplication of fractions. Specifically, we can define division to be the multiplication of the numerator with the inverse of the denominator.

For example, we can write $\frac{a}{b}=ab^{-1}$, identifying division as taking the multiplicative inverse. Then $$ \frac{a}{b}\cdot \frac{c}{d} =(a\cdot b^{-1})\cdot (c\cdot d^{-1})$$
Now, if we want multiplication to be associative and commutative, then we would find that $$(a\cdot b^{-1})\cdot (c\cdot d^{-1})=(a\cdot c)\cdot (d^{-1}\cdot b^{-1})$$ It is a general fact that $(xy)^{-1}=y^{-1}x^{-1}$, which can be verified directly by multiplying $(xy)$ by both $y^{-1}x^{-1}$ and $(xy)^{-1}$. Then we find that $$ \frac{a}{b}\cdot \frac{c}{d}=(a\cdot c)\cdot (d^{-1}\cdot b^{-1})=(a\cdot c)\cdot (b\cdot d)^{-1}=\frac{a\cdot c}{b\cdot d}$$

Similar justifications can be given for the remaining identities. For example, $\frac{c}{c}=1$ can be verified by $\frac{c}{c}=c\cdot c^{-1}=1$. Again, this is rather definitional.

There's another way in which we can view these identities for fractions as well. There is also another approach, mirroring the construction of the integers. If we're given an integral domain (i.e. a commutative ring in which $ab=0$ implies that one of $a$ and $b$ are equal to $0$) $(R,+,\cdot,0,1)$, where $+$ is some notion of "addition", $\cdot$ some notion of "multiplication", $0$ the identity for addition, and $1$ the identity for multiplication, then we can form a field $\operatorname{Quot}(R)$ called the quotient field or fraction field of $R$.

Specifically, we define the underlying set of $\operatorname{Quot}(R)$ by the quotient $[R\times (R\setminus\{0\})]/\sim$, where $\sim$ is the equivalence relation defined by $(a,b)\sim(c,d)$ if and only if $a\cdot d=b\cdot c$. The idea is that the ordered pairs $(a,b)\in R\times (R\setminus \{0\})$ represent the fractions of elements in $R$, but we also want to identify "equivalent" fractions, and thus we introduce the equivalence relation.

We'll represent the equivalence class of an element $(a,b)$ in $\operatorname{Quot}(R)$ by $\frac{a}{b}$.

Then the definition of addition and multiplication are exactly the commonly given identities for the addition and multiplication of fractions: $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} \quad \text{and} \quad \frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$$

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    $\begingroup$ You meant to write $$(a \cdot b^{-1}) \cdot (c \cdot d^{-1}) = (a \cdot {\color{red}{c}}) \cdot (d^{-1} \cdot b^{-1})$$ in your calculations of $$\frac{a}{b} \cdot \frac{c}{d}$$ $\endgroup$ – N. F. Taussig Oct 25 '14 at 22:41
  • $\begingroup$ @N.F.Taussig indeed, thanks for spotting that typo! I was copy-pasting it over and over again. $\endgroup$ – Hayden Oct 25 '14 at 22:43

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