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lets-f-R-Z-suppose-that-x-R-and-y-R-with-x-y-such-f-x-f-y-can-f-x-be-conrinuous-




Question Number 1196 by 123456 last updated on 13/Jul/15
lets f:R→Z, suppose that ∃x∈R and  ∃y∈R with x≠y such f(x)≠f(y)  can f(x) be conrinuous?
$$\mathrm{lets}\:{f}:\mathbb{R}\rightarrow\mathbb{Z},\:\mathrm{suppose}\:\mathrm{that}\:\exists{x}\in\mathbb{R}\:\mathrm{and} \\ $$$$\exists{y}\in\mathbb{R}\:\mathrm{with}\:{x}\neq{y}\:\mathrm{such}\:{f}\left({x}\right)\neq{f}\left({y}\right) \\ $$$$\mathrm{can}\:{f}\left({x}\right)\:\mathrm{be}\:\mathrm{conrinuous}? \\ $$
Answered by prakash jain last updated on 14/Jul/15
f(x)=a∈Z  f(y)=b∈Z  If f() is continous then f(x) must takes  all values between a and b.  contradicts f(x)∈Z.
$${f}\left({x}\right)={a}\in\mathbb{Z} \\ $$$${f}\left({y}\right)={b}\in\mathbb{Z} \\ $$$$\mathrm{If}\:{f}\left(\right)\:\mathrm{is}\:\mathrm{continous}\:\mathrm{then}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{takes} \\ $$$$\mathrm{all}\:\mathrm{values}\:\mathrm{between}\:{a}\:\mathrm{and}\:{b}. \\ $$$$\mathrm{contradicts}\:{f}\left({x}\right)\in\mathbb{Z}. \\ $$

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