does any1 know anything bout commutive, associative, identity,zero, or distributive property in math?
595704
frum de way u rite i think u need a lot of help
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Brackets, Orders, Division, Multiplication, Addition, Subtraction.\nIn some countries its BEDMAS, where E is exponents.
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/\n\nhttp://www.math.armstrong.edu/MathTutorial/
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0 is the additive identity.\n1 is the multiplicative identity.\n2 is the only even prime.\n3 is the number of spatial dimensions we live in.\n4 is the smallest number of colors sufficient to color all planar maps.\n5 is the number of Platonic solids.\n6 is the smallest perfect number.\n\nLook here for the continuation:\n\nhttp://www.gotfacts.com/numbers1.php
500151
The distributive property (of multiplication over addition) is an axiom that states that for any numbers a, b, and c, the equality a*(b + c) = a*b + a*c is true. This can be used to distribute or factor (read the equation from right to left. Remember that if a=b, then b=a).\nFor example, 3*1.99 = 3*(2.00 + (-0.01)) = 6.00 + (-0.03) = 6.97.\nLikewise, x^2 + x = x*(x + 1).
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There are two seminar papers by Magnus Hestenes that should give you a good start:\n\n"A  Ternary Algebra with Applications to Matrices and Linear Transformation"\n\n"On a Ternary Algebra"
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Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
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I assume by unity, you mean that 1 is the multiplicative identity and that R has a ring structure (+, *) which is the simplest algebraic structure I know with two binary operations.\n\nSuppose 1+1 = 0. Then 1 = -1, which implies c = -c. But then c + c = 0 = 1 + 1, so 1 + 1 = c.\nSuppose 1 + 1 = 1. Then 1 = 0 which implies 1 + 1 = 0 + 0 = 0. Since 1 = 0, this implies c = 1*c = 0*c = (1 + (-1))*c = c + (-c) = 0, so again 1 + 1 = c. Note this is the trivial ring R = {0}.\nQED
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The top 2 terms have a in common; factor it out by Distributive Property: a(a+4)\n\nThe bottom 2 terms have b in common, do the same: b(a+4)\n\nNow as long as a doesn't equal -4, the (a+4) on top cancels the (a+4) on bottom, leaving a/b
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Product of Powers Property: This property states that to multiply powers having the same base, add the exponents. \n\nFor example, take x² times x³.\n\nx² is x times x.\n\nx³ is x times x times x.\n\nx² times x³ is therefore (x times x) times (x times x times x), or x to the fifth power.\n\nSo when you multiply two numbers with the same base, you add the exponents.
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http://mathworld.wolfram.com/ComplexNumber.html\n\nfind definitions of absolute value (||), power and product.\nIt should be straightforward from there.
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its simple distributive law.\nwhat has happened is they have taken (a-2) common out of both the terms .\neg: cx-x = x(c-1)\nthe same thing has happened here with x=(a-2) and c=a^2\nso, in step 2, we have: a^2(a-2)-(a-2)\ntaking (a-2) common outside; you end up with (a-2)*((a^2)-1)\nwhihc is [3]\nthen as you know, (a^2)-1 is same as (a^2)-(1^2) and then we apply the algebraic identity (a^2)-(b^2)=(a+b)*(a-b) to get (a-2)(a+1)(a-1) which is [4]
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Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
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Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
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This proof is trivial. The number -1 is defined as the number satisfying the equation 1 + (-1) = (-1) + 1 = 0. \nMultiply both sides by -1:\n(-1)*1 + (-1)*(-1) = 0\nby the distributive property.\nBy the fact that 1 is the multiplicative identity and the definition of additive inverse, we get:\n-1 + (-1)*(-1) = 1 + (-1)\nAdd 1 to both sides (additive inverse property):\n(-1)*(-1) = 1.\nQED.
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A ring and a field both have two operations, usually denoted by * and +. For both, associativity of both * and + are assumed. It is assumed that + has an identity, 0, and inverses, x|-> -x and that * is distributive over +.\n\nUsually, it is assumed that a ring has an identity, 1, under *, but books vary in this so be careful. \n\nFor fields, * is assumed to be commutative and that every element other than 0 has an inverse under *. For rings, * is not necessarily assumed to be commutative. if you drop the commutativity assumption for fields but keep the inverses, you get what is called a skew-field.\n\nExamples: The integers are a ring, but not a field since 2 (for example) doesn't have a multiplicative inverse. The rationals form a field. The collection of nxn matrices form a non-commutative ring which is not a skew-field, and the quaternions form a skew-field.
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The algebra of sets develops and describes the basic properties and laws of sets, the set-theoretic operations of union, intersection, and complementation and the relations of set equality (mathematics) and set inclusion. It also provides systematic procedures for evaluating expressions, and performing calculations, involving these operations and relations.