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anonymous

  • 5 years ago

A piecewise function is defined as follows: for all rational numbers, f(x) = 1, for all irrational numbers, f(x) = 0. will the resulting graph be a line on f(x) = 0 with discontinuities? a line on f(x) = 1 with discontinuities, or a scattering of individual points?

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  1. dumbcow
    • 5 years ago
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    i tend to agree with line on y=0 with discontinuities, there are far more irrational numbers than rational when looking at all real numbers

  2. anonymous
    • 5 years ago
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    is that because they are uncountable?

  3. dumbcow
    • 5 years ago
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    kind of , rational numbers are uncountable too but for every gap between 2 rational numbers there could be an infinite number of irrational numbers

  4. anonymous
    • 5 years ago
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    well you will "see" two lines , y = 1 and y = 0

  5. anonymous
    • 5 years ago
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    it doesnt matter there are more irrationals than rational, so what

  6. anonymous
    • 5 years ago
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    Are there more irrational numbers than rational is really the question

  7. anonymous
    • 5 years ago
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    as far as you are concerned, between any two irrationals is a rational, between any two rationals is an irrational, etc

  8. anonymous
    • 5 years ago
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    or more generally, pick two points, between them is an irrational and a rational

  9. anonymous
    • 5 years ago
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    pick 2 distinct points

  10. anonymous
    • 5 years ago
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    so it is a scattering of individual points, neither line exists

  11. dumbcow
    • 5 years ago
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    very true, imagine we had a microscope ;)

  12. anonymous
    • 5 years ago
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    right, its not a line ,. but at the macro level it "looks" like a line

  13. anonymous
    • 5 years ago
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    So the argument I have heard that because irrational numbers are uncountable, it is a "larger" infinity than the countable set of rational numbers, is false - or that there is only one more

  14. anonymous
    • 5 years ago
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    that is a true statement, but that doesnt mean that you can order them , rational, irrational, rational, irrational, etc

  15. anonymous
    • 5 years ago
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    on the number line. so ...

  16. anonymous
    • 5 years ago
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    it gets even weirder. between any 2 rationals is an irrational. between any two irrationals is a rational. so where exactly are all the "extra" irrationals ?

  17. anonymous
    • 5 years ago
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    between any 2 rational numbers there exists an irrational number. between any 2 irrational numbers there exists a rational number. so you might think, the number of rational and irrational should be about the same, since you can keep going with this statement

  18. anonymous
    • 5 years ago
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    okay, you said, "that is a true statement, but that doesnt mean that you can order them , rational, irrational, rational, irrational, etc" does that mean that there can and cannot be two irrational numbers without a rational between them?

  19. anonymous
    • 5 years ago
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    there cannot be two distinct irrationals without a rational between them, correct

  20. anonymous
    • 5 years ago
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    so its kind of paradoxical

  21. anonymous
    • 5 years ago
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    Do you have a good source where I could read more? Wiki gets too techie too quickly

  22. anonymous
    • 5 years ago
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    yeah, thats true. well you can google cantor's infinity

  23. anonymous
    • 5 years ago
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    but this question is not addressed specifically, its something that my math professor told me

  24. anonymous
    • 5 years ago
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    there are many layman easy websites

  25. anonymous
    • 5 years ago
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    Yeah, I want something between the two levels

  26. anonymous
    • 5 years ago
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    yeah i know the feeling

  27. anonymous
    • 5 years ago
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    Also, I think it's neat to see hebrew letters come into math for this question!

  28. anonymous
    • 5 years ago
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    well college libraries have some gradual immersion books

  29. anonymous
    • 5 years ago
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    yeah, thats cool

  30. anonymous
    • 5 years ago
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    the big question in advanced math, does there exist a cardinality between aleph null ( natural numbers) and the cardinality of the real numbers

  31. anonymous
    • 5 years ago
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    it can be shown that aleph null is the "least" infinity , as weird as that sounds

  32. anonymous
    • 5 years ago
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    and we can make a hierarchy of infinite cardinalities. use the natural number cardinality, and take the cardinality of the power set of the natural numbers

  33. anonymous
    • 5 years ago
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    so we start with | N | , where N = { 1,2,3,...}

  34. anonymous
    • 5 years ago
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    ok

  35. anonymous
    • 5 years ago
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    where |N | means cardinality , that means the size of the set

  36. anonymous
    • 5 years ago
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    right

  37. anonymous
    • 5 years ago
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    now we know for any set , |P (S)| > |S|

  38. anonymous
    • 5 years ago
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    P(S) is the power set of S

  39. anonymous
    • 5 years ago
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    only for finite sets?

  40. anonymous
    • 5 years ago
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    for any set

  41. anonymous
    • 5 years ago
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    ok

  42. anonymous
    • 5 years ago
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    it comes in handy for infinite sets

  43. anonymous
    • 5 years ago
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    what a power set ?

  44. anonymous
    • 5 years ago
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    2^S, right?

  45. anonymous
    • 5 years ago
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    the number of permutations within the set

  46. anonymous
    • 5 years ago
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    right, thats the shorthand for it

  47. anonymous
    • 5 years ago
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    well, its the set of all subsets of the set

  48. anonymous
    • 5 years ago
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    i dont know about permutations, hmmm

  49. anonymous
    • 5 years ago
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    I thought infinity was defined as the set that has the same cardinality as any of its proper subsets

  50. anonymous
    • 5 years ago
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    yes thats a definition of an infinite set

  51. anonymous
    • 5 years ago
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    but im talking about power set, what is the power set operation

  52. anonymous
    • 5 years ago
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    ok say you have { 1, 2 } , the power set is { {}, {1}, {2} , {1,2} }

  53. anonymous
    • 5 years ago
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    S= { 1, 2 } , then P(S)= { {}, {1}, {2} , {1,2} }

  54. anonymous
    • 5 years ago
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    right

  55. anonymous
    • 5 years ago
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    now cantor proved a delightful theorem , that | P ( S) | > | S| for any set S

  56. anonymous
    • 5 years ago
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    so the powerset of integers is larger I mean has a higher cardinality than the set of integers

  57. anonymous
    • 5 years ago
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    exactly

  58. anonymous
    • 5 years ago
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    ok so

  59. anonymous
    • 5 years ago
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    | P ( N ) | > | N | , correct ?

  60. anonymous
    • 5 years ago
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    I sense a trick coming up...

  61. anonymous
    • 5 years ago
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    but, yes

  62. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Cardinality_of_the_continuum

  63. anonymous
    • 5 years ago
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    ok it turns out that |R| which we use shorthand c , it turns out |R| = | P ( N ) | ,

  64. anonymous
    • 5 years ago
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    now the question is, is there an infinity between |N| and | P ( N ) |

  65. anonymous
    • 5 years ago
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    is the real numbers the smallest infinity after the natural numbers

  66. anonymous
    • 5 years ago
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    Wouldn't rational and irrational be smaller than real?

  67. anonymous
    • 5 years ago
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    we have a hierarchy of infinities, |N| < |P(N)| < |P (P(N))| < ...

  68. anonymous
    • 5 years ago
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    oh boy, rational is the same cardinality as natural numbers

  69. anonymous
    • 5 years ago
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    the irrationals are uncountable

  70. anonymous
    • 5 years ago
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    So cantor's work doesn't cover uncountable sets?

  71. anonymous
    • 5 years ago
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    I find it hard to assign the same cardinality to rationals as naturals, which means I only have a working understanding of cardinality and not a real grasp on it.

  72. anonymous
    • 5 years ago
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    help

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