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Question Number 211434 by Davidtim last updated on 09/Sep/24

0.9999....=?(a/b)

$$\mathrm{0}.\mathrm{9999}....=?\frac{{a}}{{b}} \\ $$

Answered by Rasheed.Sindhi last updated on 09/Sep/24

n=0.9999....  10n=9.9999  10n−n=9  n=(9/9)=1=(1/1)

$${n}=\mathrm{0}.\mathrm{9999}.... \\ $$$$\mathrm{10}{n}=\mathrm{9}.\mathrm{9999} \\ $$$$\mathrm{10}{n}−{n}=\mathrm{9} \\ $$$${n}=\frac{\mathrm{9}}{\mathrm{9}}=\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}} \\ $$

Commented by Davidtim last updated on 09/Sep/24

why we can not get comletely (a/b) which  may produce 0.9999 again?

$${why}\:{we}\:{can}\:{not}\:{get}\:{comletely}\:\frac{{a}}{{b}}\:{which} \\ $$$${may}\:{produce}\:\mathrm{0}.\mathrm{9999}\:{again}? \\ $$

Commented by Frix last updated on 09/Sep/24

d_j ∈{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}  (d_1 /9)=0.d_1 d_1 d_1 ...=0.d_1 ^(•)   (9/9)=0.999...=0.9^(•)   But we also have (9/9)=1 ⇒ 0.9^• =1

$${d}_{{j}} \in\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8},\:\mathrm{9}\right\} \\ $$$$\frac{{d}_{\mathrm{1}} }{\mathrm{9}}=\mathrm{0}.{d}_{\mathrm{1}} {d}_{\mathrm{1}} {d}_{\mathrm{1}} ...=\mathrm{0}.\overset{\bullet} {{d}_{\mathrm{1}} } \\ $$$$\frac{\mathrm{9}}{\mathrm{9}}=\mathrm{0}.\mathrm{999}...=\mathrm{0}.\overset{\bullet} {\mathrm{9}} \\ $$$$\mathrm{But}\:\mathrm{we}\:\mathrm{also}\:\mathrm{have}\:\frac{\mathrm{9}}{\mathrm{9}}=\mathrm{1}\:\Rightarrow\:\mathrm{0}.\overset{\bullet} {\mathrm{9}}=\mathrm{1} \\ $$

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