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anonymous

  • 4 years ago

is anybody good at discrete math?

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  1. perl
    • 4 years ago
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    me

  2. anonymous
    • 4 years ago
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    Thanks!!

  3. anonymous
    • 4 years ago
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    alright my question is how do you figure out if anything is antisymmetric

  4. perl
    • 4 years ago
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    thats not very specific

  5. anonymous
    • 4 years ago
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    like okay R1 (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)

  6. perl
    • 4 years ago
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    In mathematics, a binary relation R on a set X is antisymmetric if, for all a and b in X if aRb and bRa then a = b,

  7. anonymous
    • 4 years ago
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    yeah but how does that explain if this is antisymmetric or not

  8. perl
    • 4 years ago
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    well first you have to state what the relation is, equality? less than, greater than? division modulo n?

  9. perl
    • 4 years ago
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    I would say this is not anti-symmetric because you have (3,4) ( 4,1) but 3 does not equal 1

  10. perl
    • 4 years ago
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    if you delete the point (3,4) then it is anti-symmetric

  11. anonymous
    • 4 years ago
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    what about (1,2) and (2,1) and (1,1)

  12. anonymous
    • 4 years ago
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    Wait so basically its anti symmetric if no points equal each other like a is not equal 2 b?

  13. perl
    • 4 years ago
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    1R2 and 2R1 so 1 =1 is true

  14. anonymous
    • 4 years ago
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    ohh so in that situation a = b?

  15. anonymous
    • 4 years ago
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    and if thats the case then it would be antisymmetric

  16. perl
    • 4 years ago
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    well the order relation <= is anti-symmetric

  17. perl
    • 4 years ago
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    so the definition of anti-symmetry generalises what we find in the usual order relation <= . for instance if a<=b and b <= a , then a = b , where a,b are integers and <= means less than or equal (usual ordering). but notice you can have unusual relations, like subset. if A is a subset of B and B is a subset of A, then A = B

  18. anonymous
    • 4 years ago
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    i dnt get that sorry this type of math is really confusing it took me 3 weeks to understand sets and now im trying to understand relations bim pretty much behind in class :/

  19. anonymous
    • 4 years ago
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    That make sense but how does that apply to that problem

  20. perl
    • 4 years ago
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    well we try to find in math the most general or abstract definitions. we find definitions that strip the terms of any specific meaning, so we can generalise to many different types of relations.

  21. perl
    • 4 years ago
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    so we find many 'things' that behave in math just like the ordering <= where 2<= 3 . in advanced math we look for patterns

  22. anonymous
    • 4 years ago
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    Yeah thats why im being confused because im so used to specific examples that this abstract way of thinking leaves me lost in the dark

  23. anonymous
    • 4 years ago
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    ohhh okay

  24. perl
    • 4 years ago
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    so how do you define <= . when you define it, you realise that other mathematical objects behave similiar to how 2 <= 3 . for instance {1,2} is a subset of {1,2,3}

  25. anonymous
    • 4 years ago
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    Yeah i get that

  26. perl
    • 4 years ago
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    so you could say {1,2} < {1,2,3} , or you could say <= to indicate that it is not a proper subset

  27. anonymous
    • 4 years ago
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    alright

  28. perl
    • 4 years ago
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    so using our austere definition we have R is antisymmetric if whenever (a,b) is in R and (b,a) is in R, then it must be the case that a = b

  29. anonymous
    • 4 years ago
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    OHHHH IGET IT !!!!!!!

  30. anonymous
    • 4 years ago
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    lol thanks but hold on lol hold on

  31. perl
    • 4 years ago
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    lol

  32. perl
    • 4 years ago
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    right, aRb is the same thing as saying (a,b) is in R. or equivalent

  33. anonymous
    • 4 years ago
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    wait but if thats the case where (a,b) element of R and (b,a) an element of R and a = b then isnt that kinda like being symmetric

  34. anonymous
    • 4 years ago
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    because (1,2) is symmetric to (2,1)

  35. perl
    • 4 years ago
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    symmetric is different

  36. anonymous
    • 4 years ago
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    how so

  37. perl
    • 4 years ago
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    R is symmetric when if (a,b) is in R, then (b,a) is in R

  38. anonymous
    • 4 years ago
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    thats the same thing i said tho

  39. perl
    • 4 years ago
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    antisymmetry is different

  40. perl
    • 4 years ago
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    right, sorry i didnt see what you wrote

  41. anonymous
    • 4 years ago
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    yeah but using that.. isnt that the same as antisymmetric

  42. perl
    • 4 years ago
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    ok so here are some examples one sec , thinking

  43. anonymous
    • 4 years ago
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    okay

  44. perl
    • 4 years ago
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    R = { (1,1) (2,1) (2,1) } is antisymmetric. R= { (1,1) , (1,2) (2,1) } is symmetric

  45. perl
    • 4 years ago
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    A relation can be both symmetric and antisymmetric

  46. perl
    • 4 years ago
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    woops

  47. perl
    • 4 years ago
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    R= { (1,1) ( 1,2) } is anti symetric

  48. anonymous
    • 4 years ago
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    a null set would be everything but i dnt get why the first one is antisymmetric

  49. anonymous
    • 4 years ago
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    a = 2 right?

  50. anonymous
    • 4 years ago
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    and b = 1?

  51. anonymous
    • 4 years ago
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    lol careful you are speaking to an open mind at the moment

  52. perl
    • 4 years ago
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    R= { (1,1) ( 2,1) } is antisyemmetric

  53. perl
    • 4 years ago
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    The terms 'symmetric' and 'anti-symmetric' apply to a binary relation R: R symmetric means: if aRb then bRa. R anti-symmetric means: if aRb and bRa, then a=b.

  54. anonymous
    • 4 years ago
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    okay i get that now because a in the R = {(1,1), (2,1)} is basically 1 and b is 1 as well so a is 1 in the (1,1) and b is 1 in the (2,1) right?

  55. perl
    • 4 years ago
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    well it satisfies it because it doesnt violate it

  56. perl
    • 4 years ago
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    it satisfies it trivially, since there is no occurence of both (a,b) and (b,a) in R R = {(1,1), (2,1)}

  57. perl
    • 4 years ago
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    the antisymmetry is an 'AND' condition. if you have both (a,b) AND (b,a) , then a=b. the symmetry condition is if-then condition

  58. anonymous
    • 4 years ago
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    hmm i guess sheesh why is this math so hard i swear who thinks of creating stuff like this

  59. perl
    • 4 years ago
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    well they are both if then conditions, but the antisymmetry has a little extra in it

  60. anonymous
    • 4 years ago
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    okay so symmetry mean something must happen first then something else happen.. its like a program

  61. perl
    • 4 years ago
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    R is symmetric means: if (a,b) is in R then (b,a) is in R. R anti-symmetric means: if (a,b) and (b,a) is in R, then a=b.

  62. anonymous
    • 4 years ago
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    okay so what about this one (3,4) what properties you give this in my eyes i see no reflexive no symmetry no antisymmetry and no transitive

  63. perl
    • 4 years ago
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    yes its like rules for a checking program , you could program it to check

  64. anonymous
    • 4 years ago
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    okay

  65. perl
    • 4 years ago
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    so just that one point? R = { ( 3,4) } ?

  66. anonymous
    • 4 years ago
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    yea

  67. perl
    • 4 years ago
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    then it is not symmetric, but it is anti-symmetric (trivially)

  68. anonymous
    • 4 years ago
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    its in an example in my book i hate that they dnt explain stuff

  69. anonymous
    • 4 years ago
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    Yes thats what they said but it makes no sense

  70. perl
    • 4 years ago
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    it is anti symetric why? do you see, because it is an if then condition

  71. anonymous
    • 4 years ago
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    how its only (3,4)

  72. perl
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    if (a,b) and (b,a) are in R, then a=b. but you dont have (a,b) and (b,a) in R

  73. anonymous
    • 4 years ago
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    yea okay you just have (a,b)

  74. perl
    • 4 years ago
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    because 'if-then' is true when the 'if' part is false

  75. perl
    • 4 years ago
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    i think thats called trivial or vacuously true. we just say true though :)

  76. anonymous
    • 4 years ago
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    are you talking about the logic statement if p is false but q is true then the whole statement ends up being true?

  77. perl
    • 4 years ago
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    p-> q is true when p is false.

  78. perl
    • 4 years ago
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    or another way to read the definition... is ,

  79. perl
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    Program: if you can find two points in your relation (a,b) & (b,a) .... then it must be the case that a = b. else true

  80. perl
    • 4 years ago
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    or true otherwise

  81. anonymous
    • 4 years ago
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    and a specific example would be (3,4) so basically any single point except (1,1) or other any points similar to that are antisymmetric

  82. perl
    • 4 years ago
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    Program: if you can find two points in your relation (a,b) & (b,a) .... then it must be the case that a = b. (if a does not equal B then your anti-symmetry is false.) otherwise true (if you cant find any two points (a,b) (b,a) )

  83. perl
    • 4 years ago
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    right , any relation R = { (a,b) } is antisymmetric , trivially , since there is only one point

  84. perl
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    since you need at least two points to even check the condition

  85. anonymous
    • 4 years ago
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    true cool so one answer down for the rest of my life a single point is antisymmetric

  86. perl
    • 4 years ago
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    lol

  87. perl
    • 4 years ago
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    well, you have to be more precise. the relation R= { (a,b)} is anti symmetric

  88. perl
    • 4 years ago
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    and also better say its a binary operation, and 'point' is an element of your relation aRb or (a,b) ordered pair

  89. anonymous
    • 4 years ago
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    yeah lol that .. R = { (a,b) }

  90. perl
    • 4 years ago
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    now, how can we make your original relation antisymmetric. can we delete something

  91. anonymous
    • 4 years ago
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    okay i get you

  92. perl
    • 4 years ago
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    R= { (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4) } now what are the least amount of points we can delete to make this anti-symmetric its no fair if we delete all but one , lol

  93. anonymous
    • 4 years ago
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    lol okay

  94. anonymous
    • 4 years ago
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    all try my best

  95. anonymous
    • 4 years ago
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    ill*

  96. perl
    • 4 years ago
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    ask yourself, why does it fail anti-symmetry (clue)

  97. anonymous
    • 4 years ago
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    (4,40 and (1,1) and (2,2) and (3,4)

  98. anonymous
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    (4,4) i meant

  99. anonymous
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    everything left will make it anti symmetric

  100. perl
    • 4 years ago
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    let's see

  101. anonymous
    • 4 years ago
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    wait no keep (4,4)

  102. anonymous
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    so the elements shall be (1,2), (2,1), (4,1), (4,4)

  103. perl
    • 4 years ago
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    right, i guess there is more than one solution, ok can you write out the new R

  104. perl
    • 4 years ago
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    actually (4,4) is fine

  105. anonymous
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    so R = { (1,2), (2,1) , (4,1) , (4,4)}

  106. anonymous
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    wow thanks your like a teacher right.. you have to be.. if not you should Seriously!!! lol

  107. anonymous
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    huh but wont you need (4,1) to be with (4,4)

  108. perl
    • 4 years ago
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    thanks. we have to be careful though

  109. perl
    • 4 years ago
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    lets look at the definitoin again

  110. anonymous
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    ohh yeah i forgot its a single element once the the other antisymmetric elements are taken out of play.. right?

  111. perl
    • 4 years ago
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    R anti-symmetric means: if (a,b) and (b,a) is in R, then a=b.

  112. anonymous
    • 4 years ago
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    Ohh and even so a = 1 and b= 1 ohh okay!!

  113. anonymous
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    Yeah so its the one!!

  114. perl
    • 4 years ago
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    but you have (1,2) (2,1) so 2 = 1 , that is false

  115. anonymous
    • 4 years ago
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    so (4,4) can be out or in right?

  116. perl
    • 4 years ago
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    or (1,2) and (2,1) are in R , but 1 != 2, i use != for not equal

  117. perl
    • 4 years ago
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    (4,4) can be in

  118. perl
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    ok lets start from left to right

  119. anonymous
    • 4 years ago
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    i think i get it.. it might be misconstrued but i get it better than how this conversation began

  120. anonymous
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    okay

  121. perl
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    (1,1) is fine . now (1,2) and (2,1) is a problem, so we have to delete one or the other

  122. perl
    • 4 years ago
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    remember we only look at pairs of (a,b) , (b,a)

  123. perl
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    we dont care about points that are not in the form (a,b) (b,a)

  124. anonymous
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    ohh okay

  125. perl
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    so thats the only problem i see , (1,2) (2,1)

  126. perl
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    so there are two possible solutions for problem outline above R = { (1,1), (2,1),(2,2),(3,4),(4,1),(4,4) } R= { (1,1),(1,2) ,(2,2),(3,4),(4,1),(4,4) }

  127. perl
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    thats two antisymmetric relations , which i deleted minimal number of terms

  128. perl
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    given the original R

  129. anonymous
    • 4 years ago
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    See this is where i mess up and im just realizing i do this i look at (1,2) and call that (a,b) then i look at (2,1) and call that (b,a) lol like i see it in reverse.. okay am i looking at it wrong or what

  130. anonymous
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    ahh i see so an easy way to look at it is to see where a = b or b = a?

  131. perl
    • 4 years ago
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    yes if (a,b) is (1,2) then (b,a) is (2,1) , correct

  132. perl
    • 4 years ago
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    so there is your only potential 'problem'

  133. perl
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    because clearly 1 is not equal to 2

  134. anonymous
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    yeah true

  135. anonymous
    • 4 years ago
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    okay i understand

  136. anonymous
    • 4 years ago
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    okay ill right out a set

  137. perl
    • 4 years ago
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    ok :)

  138. anonymous
    • 4 years ago
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    R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }

  139. anonymous
    • 4 years ago
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    well set A = {1,2,3,4}

  140. anonymous
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    So that R 5 is a relation of the set

  141. perl
    • 4 years ago
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    why do you call it R5, thats just a label , like R_5

  142. anonymous
    • 4 years ago
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    ehh i just say that because i cant make a sub script

  143. perl
    • 4 years ago
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    no problem, so you want to know if R5 is antisymmetric?

  144. anonymous
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    and okay This is Reflexive, but not symmetric, this is transitive, but not antisymmetric

  145. perl
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    it looks antisymmetric, i dont see any 'inverse pairs', like (a,b) (b,a) .

  146. perl
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    where the (a,b) are switched

  147. anonymous
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    well i thought it has to be for all values of a and b not just for some?

  148. perl
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    whatever is in your relation ,

  149. anonymous
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    so the antisymmetric part would be (3,4) and (4,4)

  150. perl
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    no because (3,4) and (4,3) are not in the relation

  151. anonymous
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    ohh i see your point sorry i mixed that up

  152. perl
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    you have to find a pair like (1,2) (2,1) when it fails antisymmetry ,

  153. anonymous
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    (1,2) and (2,2)

  154. perl
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    dont want to call it 'anti anti-symmetry'. lets just say it is 'not' antisymmetric if we can find a pair (a,b) (b,a)

  155. perl
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    nope, (1,2) and (2,2) are fine. (1,2) and (2,1) are a problem

  156. anonymous
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    yeah we cant find a pair so because of that its antisymmetric?

  157. perl
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    yes !!

  158. perl
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    so it 'trivially' satisfies the antisymmetric condition

  159. anonymous
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    ohh okay i get this abstract thing however where my other deductions correct?

  160. anonymous
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    were*

  161. perl
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    I found an equivalent condition. might be easier to read R is antisymmetric if whenever (a,b) is in R with a ≠ b then (b,a) must not be in R

  162. anonymous
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    okay makes sense.. so what about the transitive property

  163. perl
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    So if you can find at least two points (a,b) (b,a) with a not equal to b, then it is not antisymmetric relation. Otherwise it is antisymmetric

  164. perl
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    transitive, if aRb and bRc , then aRc if (a,b) is in R and (b,c) is in R, then it must be that (a,c) is in R

  165. anonymous
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    SO we have (2,3) (3,4) and (2,4) is that what makes it transitive

  166. perl
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    you have to check all possibilities R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }

  167. perl
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    yes you are right

  168. anonymous
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    ohh and i have a question as long as i find one thing inside the relation does that make the whole relation transitive or symmetric or whatever?

  169. perl
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    (2,3) (3,4) are in R, so it must be that (2,4) in R. (and it is) , similarly we check other candidates

  170. perl
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    if you find one example? no you have to check to make sure that if (a,b) (b,c) are in R then (a,c) is in R. for any a,b,c

  171. perl
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    but you can do shortcuts. if you have (a,b) and (b,b) then transitive says (a,b) must be in it, and clearly that is already there

  172. anonymous
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    ohh okay so (1,2) (2,3) (1,3)

  173. perl
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    yes that works

  174. anonymous
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    (1,3) (3,4) (1,4) that makes 3

  175. anonymous
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    but thats it how will i know when i satisfied the quota to make it transitive or any other property

  176. perl
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    so basically we have to check (1,2) (2,3) --> (1,3) is in R (1,3) (3,4) ---> (1,4) is in R...

  177. perl
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    when there is nothing left to check

  178. anonymous
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    but their is some thing left to check (1,1) (2,2) (3,3) (4,4)

  179. perl
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    and we dont need to check (a,a)

  180. anonymous
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    and those are not transitive they are reflexive

  181. anonymous
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    ohh okay

  182. anonymous
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    make sense

  183. perl
    • 4 years ago
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    let me check reflexive relation, i believe a relation is reflexive when..

  184. anonymous
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    great so i now understand transitive and antisymmetric and oddly enough the symmetric property thanks to you ^_^

  185. perl
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    In mathematics, a reflexive relation is a binary relation on a set for which every element is related to itself

  186. perl
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    so we need (1,1 ) (2,2) (3,3) (4,4)

  187. perl
    • 4 years ago
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    if you are going to use elements 1,2,3,4 in your R

  188. perl
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    if we dont use 4 at all in our R, then we dont need (4,4)

  189. anonymous
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    ohkay

  190. anonymous
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    makes sense

  191. perl
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    for instance, im going to delete terms

  192. anonymous
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    alright

  193. perl
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    R = { (1,1), (1,2) , (1,3) , ( 2,2) , ( 2,3) , (3,3) }

  194. perl
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    this is antisymmetric, transitive, reflexive

  195. anonymous
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    okay shall i prove it

  196. perl
    • 4 years ago
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    sure , you can check exhaustively

  197. perl
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    that just means you exhaust the possibilities

  198. anonymous
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    (claps hands) time to work..

  199. anonymous
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    its antisymmetric because (2,3) and (3,3) and its reflexive because (1,1) (2,2) (3,3) and its transitive because (1,2) (2,3) (1,3)

  200. perl
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    it is antisymmetric because there are no pairs (a,b) and (b,a) in R . so it satisfies antisymmetry trivially

  201. anonymous
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    dammit i always mess that one up .. lol

  202. perl
    • 4 years ago
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    heheh

  203. anonymous
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    ill learn over dam lol

  204. anonymous
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    ill learn over time i meant to say

  205. perl
    • 4 years ago
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    try the new equivalent definition, whenever (a,b) is in R then (b,a) cannot be in R. if it is in R , then it is not antisymmetric relation

  206. perl
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    if both (a,b) and (b,a) are in R, then it is not antisymmetric , (where a and b are distinct)

  207. anonymous
    • 4 years ago
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    ohh okay but if (b,a) is in R then a must equal to b for to be antisymmetric

  208. perl
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    so in other words, you are looking for counterexamples to antisymmetry. if you cannot find any counterexamples (a,b) and (b,a) , then it is true antisymmetry

  209. perl
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    yes, which brings you back to the old definition

  210. anonymous
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    great this makes sense now

  211. perl
    • 4 years ago
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    many paths that lead to Rome

  212. anonymous
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    by the way what is irreflexive?

  213. perl
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    An irreflexive, or anti-reflexive, relation is the opposite of a reflexive relation. It is a binary relation on a set where no element is related to itself. An example is the "greater than" relation (x>y).

  214. perl
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    so R = { (1,1) } is not irreflexive R = { (1,2) } is irreflexive

  215. anonymous
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    ohh that easy!!

  216. anonymous
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    okay i understood that off rip

  217. perl
    • 4 years ago
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    actually, reflexive, irreflexive, and antisymmetry are easy to check. the one that takes a bit of work is transitive

  218. anonymous
    • 4 years ago
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    yeah now that i understand them it does take me a bit longer to figure out the transitive ones

  219. perl
    • 4 years ago
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    now, we define an 'equivalence relation' as a relation which is reflexive, symmetric, and transitive.

  220. anonymous
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    ahh yes i was just reading about that

  221. perl
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    there are also 'order' relations,. partially orders , and total orders

  222. anonymous
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    yeah i have no idea about those havent read up to that part yet

  223. perl
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    so now we see the point of all this work . we can define equivalence relations, and order relations using the definitions we used earlier (transitive, reflexive, symmetric, antisymmetric, etc)

  224. anonymous
    • 4 years ago
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    Yeah just about to bad those i might have to get into tomorrow its like 4 am right nw

  225. perl
    • 4 years ago
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    well now you a grasp of this :)

  226. anonymous
    • 4 years ago
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    Yup!! :)

  227. anonymous
    • 4 years ago
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    thanks

  228. perl
    • 4 years ago
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    yeah the book should be clear about the 'trivial' satisfying . that a relation can trivially satisfy antisymmetry (since there are no pairs (a,b) (b,a) ).

  229. perl
    • 4 years ago
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    this gets into the logic thing, p->q is true in the case that p cannot be satisfied. for instance , the empty set is a subset of any set.

  230. anonymous
    • 4 years ago
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    it does but i understand best when some one is pointing me in the right direction especially when i have questions which the book doesnt ever want to answer

  231. perl
    • 4 years ago
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    its like, the definition says for all a,b in R if (a,b ) & (b,a) is in R then a=b. well, what if there are no (a,b), (b,a) in R. then it is taken to be true vacuously or trivially/

  232. perl
    • 4 years ago
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    but it is easier to just look for (a,b) (b,a) where a not equal to b. If you find it, it is not antisymmetric, if you cannot find it then it is antisymmetric

  233. perl
    • 4 years ago
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    It is often easier to use 'equivalent' definitions. also for programming :)

  234. anonymous
    • 4 years ago
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    Hmm how so?

  235. perl
    • 4 years ago
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    computer programmers have different needs than mathematicians . mathematicians want precise rigor , to be able to prove things. but at a cost , it might be harder to understand what the mathematician is talking about .

  236. perl
    • 4 years ago
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    computer programmers are not as concerned with proving things . computer scientists, maybe. :)

  237. anonymous
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    I am a programmer well tryna be

  238. perl
    • 4 years ago
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    here is the way mathematicians define antisymmetry , http://en.wikipedia.org/wiki/Antisymmetric_relation

  239. anonymous
    • 4 years ago
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    lol well thanks for the help i must go to bed thanks ^_^

  240. perl
    • 4 years ago
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    gnite :)

  241. anonymous
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    yea hthe upside down A means for all right

  242. perl
    • 4 years ago
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    right

  243. perl
    • 4 years ago
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    R(a,b) means (a,b) is in R

  244. perl
    • 4 years ago
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    I like this part R is antisymmetric .. if R(a,b) with a ≠ b, then R(b,a) must not hold.

  245. perl
    • 4 years ago
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    i got that from wikipedia. thats the easiest way to look at it

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