How to find binary relation which is reflexive and symmetric, but not transitive?
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Case 0 -- NOT TRANSITIVE\n"is no more than 1 away from"\nSymmetric:  If A is no more than 1 away from B, then B is... A.\nReflexive:  A is no more than 1 away from A\nNot Transitive:  If A is no more than 1 away from B, and B is no more than 1 away from C... you can't conclude that A is no more than 1 away from C.  (Take 1,2,3 as an example)\n\nCase 1 -- NOT SYMMETRIC\n"greater than or equal to"\nReflexive: A >= A\nTransitive: if A >= B and B >= C then A >= C\nNot Symmetric:  A >= B does not imply B >= A (e.g. 5,3)\n\nCase 2 -- NOT REFLEXIVE\n"is a sister to"\nSymmetric:  A is a sister to B, so B is a sister to A\nTransitive: A is a sister to B and B is a sister to C, so A is a sister to C\nNot Reflexive:  A is *not* a sister to A.\n\n----- \nEdit:  Other examples of Case 0 (not transitive):\n"knows" as in two people know each other.\nReflexive:\nA knows A\nSymmetric:\nA knows B, implies B knows A.\nNot transitive:\nA knows B and B knows C, doesn't imply that A knows C.\n\n"Has the same grandparent"\nReflexive:\nA has the same grandparent as A\nSymmetric:\nA has the same grandparent as B --> B ... A\nNot transitive:\nA ... B and B ... C, but A may not have the same grandparent as C.\n\n"classmates"\nReflexive:\nA was in the same class as A\nSymmetric:\nA took the same class as B, so B took the same class as A\nNot Reflexive:\nFor example, A took French with B, and B took Biology with C, but A and C never had any classes in common.
296681
either A, B, or AB is possible
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
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This is an easy derivation. \n\nMultiplicative Principle: \nOne fundamental principle of counting is that if you have n of one type of object (N) and m of another type of object (M), then the number of ways to pair objects of type one with objects of type two with regard to order is n*m. This is argued by saying that one can first pair one of the N-type objects with all of the M type objects for m pairs, and then do the same for the remaining N type objects, giving m + m + ... + m n times, or n*m pairings. \n\nPermutations:\nSuppose you have k objects and you want to know how many ways you can order them in a line. You have k ways to choose the first object, k - 1 ways of choosing the second object, and so forth until you run out of objects. By the fundamental principles of counting, this makes the number of possible orderings k*(k - 1)*...*1 = k! which we will call P(k).\n\nCombinations:\nSuppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted (by the fundamental principles of counting). Let P(k) be the number of ways of ordering k objects (this is just k! = k*(k-1)*...*1 ). Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!].
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emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged
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Are you looking for "equal" and "opposite?"
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Most of you I think misinterpretd the question(or maybe it was me), but I understood this:\n"is that function odd or even (referring to symmetry)?"\nWell if that is the question, do this:\nF(x) is even (symmetyric with respect to the y-axis) if F(-x)=F(x)\nF(x) is odd (symmetric with respect to the origin) if F(-x)= -F(x)\nThe function does have symmetry if it doesn't fill the above situations.\n\nIn this case by simple substitution the function is not symmetric at all.\n\nNOTE: There is a third type of symmetry for relations (not functions!) which is with respect to the x-axis.
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Additive equivalence is a symmetric relation over the reals. See the source link.
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in decimal its 2\nin binary its  0\nin other like octal hexa etc its 2\nin logical or its 1\nnow wat do u want????
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I've seen one variant that goes like this:\nLet a and b be equal non-zero quantities\n\n          a = b\n\nMultiply through by a\n\n          a2 = ab\n\nSubtract b2\n\n          a2 − b2 = ab − b2\n\nFactorize both sides\n\n          (a − b)(a + b) = b(a − b)\n\nDivide out (a − b)\n\n          a + b = b\n\nObserving that a = b\n\n          b + b = b\n\nCombine like terms on the left\n\n          2b = b\n\nDivide by the non-zero b\n\n          2 = 1\n\nQ.E.D.\n\nThe problem here is that dividing by (a-b) is dividing by zero, which is an undefined operation.
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NO !!\nwhere is the double negatives ?\nu just replaced " Behind" with " in front" !!\n\nin Logic,   if X = Y  then   not not X = y  \nbut u cannt say    ( not X =y )
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nything that doesn't involve either zero or U
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I would call it anterior to posterior/dorsal to ventral bilateral symmmetry - that is, if you drew a line through the animal from the front to back connecting its top and bottom, the resulting halves would be symmetrical.  However, I don't know if there are really specific types of bilateral symmetry - usually things are defined as asymmetrical, bilaterally symmetrical, or radially symmmetrical.
130202
regular, reverse and atomic
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Here you go...\nhttp://www.mathsisfun.com/platonic_solids.html
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None. The two numbers have no common factors, bar -1. You mean exactly (with no remainder) not evenly.
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Symmetric about a: f(x)=f(2a-x)\nSymmetric about b: f(x)=f(2b-x)\nBoth: f(x)=f(2a-x)=f(2b-(2a-x))=f((2a-2b)+x),\nso f has a period of 2a-2b.