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Question Number 159994 by mathocean1 last updated on 23/Nov/21

montrer que le quotient d′un nombe rationnel et dun nombre irrationnel est irrationnel

$$ \\ $$$$\mathrm{montrer}\:\mathrm{que}\:\mathrm{le}\:\mathrm{quotient}\:\mathrm{d}'\mathrm{un} \\ $$$$\mathrm{nombe}\:\mathrm{rationnel}\:\mathrm{et}\:\mathrm{dun}\:\mathrm{nombre}\: \\ $$$$\mathrm{irr}{a}\mathrm{tionnel}\:\mathrm{est}\:\mathrm{irrationnel} \\ $$

Answered by Tokugami last updated on 23/Nov/21

show that the quotient of a rational number and an irrational number is irrational. proof by contradiction: (r_1 /i_1 )=r_2 determinant (((r_1 ,r_2 ∈Q),(i_1 ∈R−Q))) (r_1 /r_2 )=i_1 ((a/b)/(c/d))=i_1 a,b,c,d∈Z ((ac)/(bd))=i_1 this would mean that a ratio of integers is irrational the opposite must be true.

$$ \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{quotient}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{rational}\:\mathrm{number}\:\mathrm{and}\:\mathrm{an}\:\mathrm{irrational} \\ $$$$\mathrm{number}\:\mathrm{is}\:\mathrm{irrational}. \\ $$$$ \\ $$$$\mathrm{proof}\:\mathrm{by}\:\mathrm{contradiction}: \\ $$$$\frac{{r}_{\mathrm{1}} }{{i}_{\mathrm{1}} }={r}_{\mathrm{2}} \begin{matrix}{{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} \in\mathbb{Q}}&{{i}_{\mathrm{1}} \in\mathbb{R}−\mathbb{Q}}\end{matrix}\underbrace{ } \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }={i}_{\mathrm{1}} \\ $$$$\frac{{a}/{b}}{{c}/{d}}={i}_{\mathrm{1}} \:\:{a},{b},{c},{d}\in\mathbb{Z} \\ $$$$\frac{{ac}}{{bd}}={i}_{\mathrm{1}} \\ $$$$\mathrm{this}\:\mathrm{would}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{a}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{is}\:\mathrm{irrational} \\ $$$$\mathrm{the}\:\mathrm{opposite}\:\mathrm{must}\:\mathrm{be}\:\mathrm{true}. \\ $$$$ \\ $$

Commented by mathocean1 last updated on 23/Nov/21

please can you detail sir

$${please}\:{can}\:{you}\:{detail}\:{sir} \\ $$

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