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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?
letsf:RZ,supposethatxRandyRwithxysuchf(x)f(y)canf(x)beconrinuous?
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(x)=aZf(y)=bZIff()iscontinousthenf(x)musttakesallvaluesbetweenaandb.contradictsf(x)Z.

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