(related to Proposition: Addition Of Rational Numbers)
Let \(x\) and \(y\) be rational numbers. By definition, it means that they are represented by some integers \(x=\frac ab\), \(y=\frac cd\), \(a,c\in \mathbb Z\), \(b,d\in \mathbb Z\setminus\{0\}\).
Note that \(ad\), \(cb\) and \(bd\) are all integers, as they are the products of the respective integers \(a,d\), \(c,d\) and \(b,d\). Also note that \(ad + cb\) is an integer, as it is the sum of the integers \(ad\) and \(cb\). Moreover, we have \(bd\neq 0\), since both \(b\neq 0\) and \(d\neq 0\) and their product cannot equal \(0\), since integers form an integral domain. Therefore, the sum \[\begin{array}{rcl} x+y=\frac {ad + cb}{bd}. \end{array}\] exists, because it denotes some new rational number, as it is represented by the integers \(ad + cb\) and \(bd\neq 0\).
It remains to be shown that the addition of rational numbers does not depend on the specific representatives of the numbers \(x\) and \(y\). Suppose, we have different representatives \[\begin{array}{rcl} x=\frac{a_1}{b_1}=\frac{a_2}{b_2},~y=\frac{c_1}{d_1}=\frac{c_2}{d_2}.&&(*) \end{array}\] It follows from the definition of rational numbers that \(a_1=\frac{a_2b_1}{b_2}\) and \(c_1=\frac{c_2d_1}{d_2}\). We have to show that \[x+y=\frac {a_1d_1 + c_1b_1}{b_1d_1}=\frac {a_2d_2 + c_2b_2}{b_2d_2}.\] In the following, we will use the following mathematical definitions and concepts: * definition of rational numbers, * definition of adding rational numbers (hypothesis), * commutativity law of multiplying integers , * integer one is neutral with respect to the multiplication of integers, * distributivity law for integers, and * multiplication of rational numbers:
\[\begin{array}{rcll} x+y&=&\frac{a_1}{b_1}+\frac{c_1}{d_1}&\text{by definition of rational numbers}\\ &=&\frac{a_1d_1+c_1b_1}{b_1d_1}&\text{by hypothesis}\\ &=&\frac{a_1d_1\cdot 1+c_1b_1\cdot 1}{1\cdot b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{(a_1d_1+c_1b_1)\cdot 1}{1\cdot b_1d_1}&\text{by distributivity law for integers}\\ &=&\frac{a_1d_1+c_1b_1}{1}\cdot \frac 1{b_1d_1}&\text{by definition of multiplying rational numbers}\\ &=&(a_1d_1+c_1b_1)\cdot \frac 1{b_1d_1}&\text{by definition of rational numbers}\\ &=&\left(\frac{a_2b_1}{b_2}d_1+\frac{c_2d_1}{d_2}b_1\right)\cdot \frac 1{b_1d_1}&\text{according to }(*)\\ &=&\left(\frac{a_2b_1\cdot d_1}{b_2\cdot 1}+\frac{c_2d_1\cdot b_1}{d_2\cdot 1}\right)\cdot \frac 1{b_1d_1}&\text{by definition rational numbers and of multiplying them}\\ &=&\left(\frac{a_2b_1d_1}{b_2}+\frac{c_2d_1b_1}{d_2}\right)\cdot \frac 1{b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\left(\frac{a_2b_1d_1d_2+c_2d_1b_1b_2}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by hypothesis}\\ &=&\left(\frac{a_2d_2b_1d_1+c_2b_2b_1d_1}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by commutativity law for multiplying integers}\\ &=&\left(\frac{(a_2d_2+c_2b_2)b_1d_1}{b_2d_2}\right)\cdot \frac 1{b_1d_1}&\text{by distributivity law for integers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {b_1d_1\cdot 1}{b_1d_1}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {b_1d_1}{b_1d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}\cdot \frac {1}{1}&\text{by definition of rational numbers}\\ &=&\frac{(a_2d_2+c_2b_2)\cdot 1}{b_2d_2\cdot 1}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2d_2+c_2b_2}{b_2d_2}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{a_2}{b_2}+\frac{c_2}{d_2}&\text{by hypothesis}\\ \end{array}\]
