Addition and Multiplication of Reals

Samia CHALLAL

2 Addition and Multiplication of Reals

$\bullet$ Two operations, addition and multiplication, are defined on $\mathbb{R}$ and satisfy the laws of

$\begin{eqnarray*} &&\textbf{closure} \qquad \forall u,v \in \mathbb{R} \qquad \Longrightarrow \qquad u+v \in \mathbb{R} \quad \hbox{ and } \quad u v \in \mathbb{R} \\ \\ &&\textbf{commutativity} \qquad \forall u,v \in \mathbb{R} \qquad \Longrightarrow \qquad u+v= v+ u \quad \hbox{ and } \quad u v= v u \\ \\ &&\textbf{associativity} \qquad \forall u,v, w \in \mathbb{R} \qquad \Longrightarrow \qquad (u+v)+w= u + (v+w) \quad \hbox{ and } \quad (u v) w = u (v w) \\ \\ &&\textbf{distributivity} \qquad \forall u,v, w \in \mathbb{R} \qquad \Longrightarrow \qquad u (v+ w)= u v + u w \quad \hbox{ and } \quad (u+ v) w = u w + v w \\ \\ &&\textbf{identity} \qquad \forall u\in \mathbb{R} \quad \Longrightarrow \quad u+0 = 0+ u=u \quad \hbox{ and } \quad 1. u = u. 1 = u \\ \\ &&\textbf{inverse} \qquad * \quad \forall u\in \mathbb{R} \quad\exists\,\, -u\in \mathbb{R} \quad \hbox{ such that }\quad u+(-u) = (-u)+ u=0 \\ \\ &&\qquad \qquad \qquad *\quad \forall u\in \mathbb{R}\setminus \{0\} \quad\exists\,\, u^{-1}\in \mathbb{R} \quad \hbox{such that }\quad u^{-1}. u = u. u^{-1} = 1\\ \\ && - u \quad \hbox{is the additive inverse of } u , \qquad \qquad u^{-1} \quad \hbox{ is the multiplicative inverse of } u.\end{eqnarray*}$

$\bullet$ The following properties are true for any $u,\, v,\, w\in \mathbb{R}$

$\begin{eqnarray*} && \quad v+ u = v+w \qquad\Longrightarrow\qquad u= w \qquad (cancellation) \\ \\ && \quad \hbox{The equation } \quad x+ v= u \quad \hbox{ has the only solution } \quad x= u-v \\ \\ && \quad 0 . v = v. 0 = 0 \qquad \qquad \qquad \qquad \\ \\ && \quad (-1) \,v = - v \\ \\ && \quad u. v =0 \qquad\Longrightarrow\qquad u= 0 \qquad \hbox{ or } \qquad v=0 \end{eqnarray*}$

$\bullet$ The operation of subtraction is defined as : $\qquad \qquad u-v = u+ (-v)$

$\bullet$ The operation of division is defined as : $\qquad \qquad\displaystyle{\frac{u}{v}= u \div v = u\times v^{-1} }$ and satisfies

$-\frac{u}{v} = \frac{-u}{v}=\frac{u}{-v} = - \frac{-u}{-v}\qquad \qquad \qquad \qquad \qquad \frac{-u}{-v}= \frac{u}{v}$

$\frac{u}{v}= \frac{s}{t} \qquad \Longleftrightarrow\qquad u t = v s \qquad\qquad \qquad\qquad \qquad \frac{u}{v}= \frac{ku}{kv}\qquad \forall k \in \mathbb{R}\setminus\{0\}$

Watch a Video

View on YouTube

Show/Hide Examples

Exercise 1

Show/Hide Solution.

i) $\quad$ We have

$\displaystyle{\Big( {3} + \frac{{1}}{{5}} \Big) \Big( {2} - \frac{{5}}{{3}}\Big) = ({3})({2}) - \frac{ ({3}). ({5}) }{{3}} +\frac{({1})({2}) }{{5}} -\frac{({1})({5}) }{({5})({{3}})} }$

$\displaystyle{= \frac{1}{ ({5}) ({3}) } \Big[ ({3})({2}) ({5})({3}) - ({3}). ({5}).({5}) + ({1})({2}) ({3}) - ({1})({5}) \Big] =\frac{{16}}{{15}} }.$

ii) $\quad$ We have

$\displaystyle{ \Big( \frac{{2}}{{3}} - \frac{{1}}{{6}}\Big) \Big( \frac{{2}}{{3}} + \frac{{1}}{{6}} \Big) = (\frac{ {2} }{ {3} } )^2 - (\frac{ {1} }{ {6} } )^2 = \frac{({2})^2}{({3})^2} - \frac{({1})^2}{({6})^2} = \frac{({2})^2({6})^2 - ({3})^2 ({1})^2 }{({3})^2 ({6})^2} = \frac{{135}}{{324}} }.$

iii) $\quad$ We have

$\displaystyle{ \frac{{4}}{{5}} \Big(\frac{{3}}{{4}} - \frac{{1}}{{3}} \Big) = \frac{{4}}{{5}} .\frac{({3})({3}) - ({4}) ({1}) }{({4}) ({3})} =\frac{ {4}(({3})({3}) - ({4}) ({1}) )}{({5})({4}) ({3})}=\frac{{20} }{{60}} }.$

Exercise 2

Show/Hide Solution.

i) $\quad$ Counter-example:

$2 \in \mathbb{N}$ and $5 \in \mathbb{N}$ but $2- 5= - 3 \not\in \mathbb{N}$ .

$2 \in \mathbb{N}$ but $2- 2= 0\not\in \mathbb{N}$ .

ii) $\quad$ Counter-example:

$\sqrt{2} \in \mathbb{R}\setminus \mathbb{Q}$ but $\sqrt{2} . \sqrt{2} = 2 \not\in \mathbb{R}\setminus \mathbb{Q}$ .

iii) $\quad$ Let $a\in \mathbb{Q}$ and $b\in \mathbb{Q}$ . We have

$a= \frac{m}{n} , \quad \quad b=\frac{k}{l}$ with $m, n, k, l \in \mathbb{Z}$ and $n, \, l \not= 0$ . Then

$a. b = \frac{m}{n} . \frac{k}{l} = \frac{m k}{n l} \in \mathbb{Q}$

since $m k \in\mathbb{Z}$ and $n l \in\mathbb{Z}$ with $n. l \not= 0$ .

Exercise 3

Show/Hide Solution.

i) $\quad$ We have

$\displaystyle{ \frac{ \frac{1}{{7}} }{ \frac{1}{{2}} + \frac{1}{{3}} } = \frac{ \frac{1}{{7}} }{ \frac{{2}+{3}}{({2}) ({3})} }=\frac{1}{{7}}. \frac{({2}) ({3})}{{2}+{3}}= \frac{({2}) ({3})}{({7})( {2}+{3})} =\frac{{6}}{{35}} }$ .

ii) $\quad$ We have

$\displaystyle{ \frac{ \frac{1}{{6}} + \frac{1}{{3}} }{ \frac{1}{{4}} + \frac{2}{5} } = \frac{5({4}) ({3}+ {6}) }{ ({6}) ({3}) (5 + 2 ({4}) ) } = \frac{{180}}{{234}} }$ .

iii) $\quad$ We have

$\displaystyle{ \frac{ 2}{\frac{1}{{2}} } - \frac{\frac{2}{{8}}}{2} } = 2 ( {2}) - \frac{1}{{8}} = \frac{2 ({2}) ({8}) -1 }{ {8}} = \frac{{31}}{{8}} } .$

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

License

Share This Book