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amilapsn

  • one year ago

This is about real analysis continuity: (jpeg below in comments)

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  1. amilapsn
    • one year ago
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  2. amilapsn
    • one year ago
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    @Luigi0210 @UnkleRhaukus

  3. UnkleRhaukus
    • one year ago
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    the question is incomplete .... Show that if {A}, and {B}. ?

  4. amilapsn
    • one year ago
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    No it's complete

  5. amilapsn
    • one year ago
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    could you please, check the link again

  6. UnkleRhaukus
    • one year ago
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    i think you've cropper out some words after ".... and discontinues at each point of [ something missing] Q n (0,1)"

  7. amilapsn
    • one year ago
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    No I checked it again with the paper...

  8. amilapsn
    • one year ago
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    I can understand the proposition and I feel it's correct... But I don't know how to convert those feelings into mathematical language..

  9. amilapsn
    • one year ago
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    We just have to prove \(\Large\sf{\lim_{x\rightarrow a}{f(x)}=f(a) \ \forall\ a\in\mathbb{I}\cap(0,1) }\) and \(\Large\sf{\lim_{x\rightarrow a}f(x) }\) doesn't exist \(\Large\sf{\forall a\in\mathbb{Q}\cap(0,1)}\) The problem is I don't know how to put that mathematically....

  10. amilapsn
    • one year ago
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    @Loser66

  11. amilapsn
    • one year ago
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    @ikram002p

  12. ikram002p
    • one year ago
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    ur try is correct, but without finding limts can u write the definition of continuous in ur book here ? i think its the one with " f is continuous at x_0 if for every open interval (a,b) in f(x) containing f(x_0) there is an open interval (c,d) in x(as whole) contains x_0 such that f is continuous" also f is discontinuous at x_0 if there is an open interval in(a,b) in f(x) contains f(x_0) such that for every open interval in x , x_0 in (c,d) f((c,d)) is not sub interval of (a,b)

  13. ikram002p
    • one year ago
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    huh i need ur definition or i'll show u how with mine

  14. amilapsn
    • one year ago
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    wait a sec

  15. amilapsn
    • one year ago
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    our definition for continuity is just the delta epsilon definition for the limit... same idea as your definition... But there's no explicit definition for discontinuity for us as for you. We just have to assume there's a limit and get a contradiction...(contradiction method)

  16. ikram002p
    • one year ago
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    can u take snap for example ? idk what do u mean but getting contradiction with assuming there is a limit

  17. amilapsn
    • one year ago
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    hmmm

  18. ikram002p
    • one year ago
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    does I represent imaginary ?

  19. ikram002p
    • one year ago
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    and Q rational number ?

  20. amilapsn
    • one year ago
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    I-irrational Q-rational

  21. ikram002p
    • one year ago
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    hehehe ok

  22. ikram002p
    • one year ago
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    |dw:1433147116393:dw|

  23. ikram002p
    • one year ago
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    for part a:- we took irrational btw (0,1) let x_0 be any point in I and (0,1) and x_0 in (c,d) f(x_0) in {0} , and f((c,d)) in {0} according to the function definition so its continuous through its domain

  24. amilapsn
    • one year ago
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    shouldn't we have to define (c.d)?

  25. ikram002p
    • one year ago
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    yes u should its the open interval which contains x_0 and subset of I and (a,b) u only have to write in in epsilon delta way

  26. amilapsn
    • one year ago
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    but how can we explain about the rationals contained in (c,d)?

  27. ikram002p
    • one year ago
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    we dont need them see (c,d) is subset of I and (0,1) so it does not have rational at all

  28. amilapsn
    • one year ago
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    can we make an interval without rationals?

  29. ikram002p
    • one year ago
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    yes sure this is the main idea of the question :)

  30. amilapsn
    • one year ago
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    Do we have to prove that we can make an interval without rationals?

  31. ikram002p
    • one year ago
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    its given in the question :O I and (0,1) means all real btw 0 and 1 without rational

  32. amilapsn
    • one year ago
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    no no I'm talking about the interval (c,d)

  33. amilapsn
    • one year ago
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    although we took (c,d) as a subset of I and (0,1) don't we have to show that we "can" REALLY take an interval (c,d) out of the set I and (0,1)

  34. amilapsn
    • one year ago
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    @ikram002p

  35. amilapsn
    • one year ago
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    @ganeshie8

  36. ikram002p
    • one year ago
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    well if u wanna take c,d as subset of real then more work u need to disjoint rational :\ why u wanna do that when u already can define it as subset of (0,1) without rational ?

  37. ikram002p
    • one year ago
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    i have to go but for second part show a contradiction by search of continuity about cretin point for example take ,10/24 ,11/24 and 12/24

  38. ganeshie8
    • one year ago
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    For showing \(f\) is discontinuous at each point of \(\mathbb{Q}\cap (0,1) \) : Consider a sequence \(\{x_n\}\) of irrational numbers in the interval \((0, 1)\) that converge to a rational number \(\frac{a}{b}\) in the interval \((0,1)\). Then, as \(x_n\to \frac{a}{b}\), we have \(f(x_n)\to 0\) because \(f(\text{irrational}) \) is \(0\). But \(f(\frac{a}{b}) = \frac{1}{b^2} \ne 0\) proving the discontinuity.

  39. ikram002p
    • one year ago
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    oh that really works

  40. ganeshie8
    • one year ago
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    yeah that works because rational numbers are dense or something

  41. ikram002p
    • one year ago
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    hehehe yeah woah :O

  42. amilapsn
    • one year ago
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    @ganeshie8 thanks... Would telling that be sufficient in proving our second proposition rigorously?

  43. ganeshie8
    • one year ago
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    Yes, I think that is a rigorous proof for second part.

  44. ganeshie8
    • one year ago
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    may be add this also : Such a sequence \(\{x_n\}\) exists because rational numbers are dense in reals.

  45. amilapsn
    • one year ago
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    I don't understand it... I think I have to study this section more.... After studying I'll medal one of you... Thanks @ikram002p and @ganeshie8 for your time and effort.. :)

  46. ikram002p
    • one year ago
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    plz medal ganesh for second idea he deserves it :)

  47. amilapsn
    • one year ago
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    You know better than me, I'll do it. :)

  48. amilapsn
    • 23 days ago
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    \[\href{ https://en.wikipedia.org/wiki/Radix }{hey}\]

  49. amilapsn
    • 23 days ago
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    \[\href{ hey }{ https://en.wikipedia.org/wiki/Radix }\]

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