A uncountable set of +reals.Show for any real R,there finitly distinct a,b,c,.. in A s.t a+b+c+..>=R?
410641
For the sake of future attempters, the QUESTION is:\nA is an infinite set of positive real numbers. Show that, for any real R, there exist finite and distinct real numbers a,b,c,... in A such that a + b + c + ... >= R\n\nAnswer:\nBy the condition of the problem,\nR lies in A. \n\nSay r is the positive integer closest to R, such that r>R. \n\nNow, since A is the set of positive reals, one subset of the \nset is the set of positive integers. \n\nNow, by the principle of finite induction \nif S is the set of positive integers {1,2,3,...,k} then whenever k lies in S, k+1 also lies in S. \n\nHence, we state r+1 lies in the set A. \nNow since R<r, from the above statement, it follows that \nR+1 lies in A. ... (I)\n \nLet us now assume that, for a real R, and for finite and distinct a,b,c,.....such that a,b,c,.... includes the entire set A except R, \n\na + b + c + ... <  R\nThis implies \na<R\nb<R\nc<R      .......... (II)\n..\n..\nBut as stated in (I), the real number R+1 also lies in set A\nStatements (II), however, inevitably yield\n\nR+1 < R , a contradiction.\nHence, our initial assumption \na+b+c +... < R is incorrect. \n\nThus, there exists atleast one set of finite and distinct reals\na,b,c, such that\na+b+c+... >=R\n------------------------------------------------------\nCorrection: \nYes, I do agree with Ron, my proof is valid only for the special case wherein :\n1. R lies in A.\n2. A is the entire set of real numbers.\nKindly excuse me for the errors, as I attempted the proof in a state of stupor and torpor... :)\n\nThe complete proof requires additional consideration as follows:\n\nFirstly consider that the set A bound by its largest and smallest element, occupies a band on the number line. \n\n1. If R lies outside of A and lies to the left of the A-band, then the proof is trivial. (this accounts for R<0 too) \n2. If R>0 and lies to the right of the A-band,  \nthen, it is possible to find \nn such that n = R/X where X is the smallest element of the set A. let N be the positive integer closest to and larger than n.  \nNow, since R is a finite real number, n and thus N are also finite.\nHence the addition of the first N numbers of the set A arranged in ascending order should give u the required result. \n\n3. If R lies within the A-band (either within or without A), the proof easily follows from the reasoning used in 2 above. \n\nHope this goes towards clarifying.
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
991775
Cantor used a diagonalization argument. We show that the open interval (0,1) is uncountable.  Suppose that f:N->(0,1) is any function. We how it cannot be onto by constructing an element of (0,1) that is not in the image of f. Write the decimal expansions\n\nf(1)=.a_11 a_12 a_13 a_14 .....\nf(2)=.a_21 a_22 a_23 a_24 ....\nf(3)=.a_31 a_32 a_33 a_34 ...\nf(4)=.a_41 a_42 a_43 a_44 ...\n.........\n.........\n\nSo each a_ij is an integer between 0 and 9. There is a slight ambiguity because of the non-uniqueness of decimal expansions. For such cases, we use the expansion with infinitely many 0's, so\n.25000000...\nrather than\n.2499999.....\n\nNow construct a new real number between 0 and 1 as follows:\nLet b_1 be different from a_11 and either 3 or 7.\nLet b_2 be different from a_22 and either 3 or 7.\nLet b_3 be different from a_33 and either 3 or 7.\n.....\nand let\nx=.b_1 b_2 b_3 b_4 .....\nThen x is in (0,1). But by construction, b is not equal to f(1) (since the first decimal digits are different). It is different from f(2) since the second decimal digit is different. Continue and see that x is not f(n) for *any* natural number n. In other words, x is not in the image of the function f!\n\nSince countability requires a function which is both one-to-one and onto, and we have shown that no function from N to (0,1) can be onto, we see that (0,1) is uncountable. Since R is even larger, R is uncountable.\n\nNote: I used the choice of 3 or 7 to make sure to avoid the ambiguity of decmal expansions.
947782
If A is a subset of C and A has the same cardinality as C, then the sets A and C are the same.\n\nI guess as a proof, you could assume that there is an element in C that is not in A...in which case the cardinality of C would be greater than the cardinality of A, which is a contradiction.\n\nThen B gets trapped in the middle.  A = B = C.
271102
Consider a discrete infinite set, like the natural numbers (also known as the counting numbers), or the set of rational numbers. These sets are called countable by mathematicians because they can be associated with the counting numbers in a manner such that each element is uniquely labelled with a single natural number. \nThe set of real numbers and the set of all rotations in the plane cannot be associated with the natural numbers in this manner, and we call such sets uncountable. The first proof of this fact was given by Cantor.\nIn normal conversation, countless simply refers to any infinite set, or even just a very large set of unknown size, ie the number of fishes in the sea or stars in the sky.
785952
So you need to prove that each element (x,y) E br(xo, yo) has a neighbourhood contained in br(xo, yo),\n\nTake a point (xa,ya) E br(xo,yo) and let d = r - ||(xa,ya) - (xo,yo)||.  d is positive since ||(xa,ya) - (xo,yo)|| < r.\nThen the neighbourhood bd(xa,ya) satisfies the condition because for any point (x,y) in bd(xa,ya) we have\n||(x,y) - (xo,yo)||\n<= ||(x,y) - (xa,ya)|| + ||(xa,ya) - (xo,yo)|| (triangle inequality)\n< d + ||(xa,ya) - (xo,yo)|| (because (x,y) E bd(xa,ya))\n= r.\n\nIn plain English: If the point (xa,ya) has distance d from the boundary of the set, then a neighbouroud of (xa,ya) whose elements are all less than d away from (xa,ya) will be contained in the set as well.
561865
A permutation. P(n,r)=(n!)/((n-r)!). C(n,r)=(n!)/(r!(n-r)!)\nor (P(n,r))/(r!). x/something is smaller than x itself,(unless there's negatives or decimals, but we don't have to worry about that because r is positive) so the permutation is biggger than the combination.
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
795252
Let a and b be any two real numbers. Consider the number x defined by\n\nx = ab + (-a)(b) + (-a)(-b).\n\nWe can write\nx = ab + (-a)[ (b) + (-b) ]       (factor out -a)\n  = ab + (-a)(0)\n  = ab + 0\n  = ab.\n\nAlso,\nx = [ a + (-a) ]b + (-a)(-b)      (factor out b)\n  = 0 * b + (-a)(-b)\n  = 0 + (-a)(-b)\n  = (-a)(-b).\n\nSo we have\n      x = ab\nand\n      x = (-a)(-b)\n\nHence, by the transitivity of equality, we have\n\n      ab = (-a)(-b).
771358
3x + 4 < x + 8\n2x < 4\nx < 2\n\n{x | x < 2}
153034
No, nobody really can prove it, with assuming several axioms\nfirst. There are certain things you have to assume are true in your universe, without necessary being able to check them. From those axioms, you can derive conclusions (theorems) that are always true, but only in that particular\nuniverse. The discussed operation of addition is one such theorem. If you assume that the axioms of natural numbers are true, then this addition becomes quite trivival. If you assume those axioms of natural numbers to be false, then the above operation of addition might become false with a different set of supposition. In other words, only with the facts given, one cannot prove, nor disprove, the fact of addition. It all depends on the set of assumptions you would have made before adding the two ones.
203409
k(k+2)! = k.1.2...k.(k+1)(k+2)\n(k+3)! = (k+3)(k+2)! \n=>\nk/(k+3) i think that's enough. that's true if k is a Natural number, there might be some theoremes that say that's true when k is Real but I'm not sure
728108
A plus B equals C. That's the only algebraic phrase I remember.
576001
By the transitive property, a=c
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2