How do you prove an absolute maximum?
817024
its the biggest of all relative maximums
740329
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
711149
Is it lim (x approaches 0)?
796937
The infinity norm is just the maximum absolute row sum - ie take each row, add together the absolute value of each element. Take the maximum over all your sums.\n\nThe reason its important(ish) is because it is one of the 'extreme' norms. Every matrix norm is "between" the 1-norm (which is the same thing, but for columns) and the infinity-norm.\nFor example, the square of the 2-norm is <= the product of the 1 norm and infinity-norm.
834504
You have made an error in the description of the problem, because inf  A is not the lowest upper bound of A because that is Sup A.\n\nGiven: two non empty sets of real numbers;\nA , B   ; For all a in A & for all b in B  : a ≤ b  (####)\n\na#). . .if for all a ε A & a2 ε A :  a ≤  a2 => a2 = Sup A \nb#). . .if for all b ε B & b1 ε B:   b1 ≤  b => b1 = Inf B\nNow:\na#) & b#) =>  a2 ≤ b1 <=> because of (####) <=> Sup A ≤ Inf B\n\nProof of question 1):\nThere exist exactly one number c ε R : b ≤ c ≤ a for all a ε A & b ε B\n(@@@)  b ≤ c ≤ a  & Sup a ≤ Inf B =>\n Inf B  ≤ Sup A & Sup A ≤ Inf B =>\nSup A = Inf B  put this into (@@@) => \nInf B ≤ c ≤  Sup A & Sup A ≤ c ≤ inf B => \nc = Sup A = Inf B (%%%)\n\n2) Is already in proved look (%%%)\n\n3) I suppose you mean b in B instead t in T. \na  ≤  b =>  0  ≤  b-a   this will always be true because Sup A =  Inf B, but I can't say it is true for all z in Z+ because, I don't now anything about Sup B , but if it has now upper limit it will be true for all z in Z+
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
1136716
2r<22\nr<11\nAnswer= (-infinity,11)\n\n* don't have the infinity symbol on computer
520801
Given: f,g: R → R continuous\nSo,|x - c| < &#92;del ==> |f(x) - f(c)| < &#92;epi.             (i)\nand\n|x - c| < &#92;del ==> |g(x) - g(c)| < &#92;epi.                (ii)\n\nNow,\nLet h(x)=max {f(x),g(x)}.\n\nTo Prove: h(x) is continuous.\n\nProof:\nTo show h(x) to be continuous it is enough to show that the following holds...\nfor |x - c| < &#92;del ==> |h(x) - h(c)| < &#92;epi.           (iii)\n\nObserve from (i) and (ii) that \nfor |x - c| < &#92;del ==> |max{f(x),g(x)} - max{f(c),g(c)}| < &#92;epi                                                   (iv)\n\nSince f(x) and g(x) is continuous in (x - &#92;del, x + &#92;del) interval. Max{} is bounded. And hence (iv) is true.\n\nSo by (iii) h(x) is continous.
766935
The easy way is to note that both numbers have to be equal so 56/2=28\n\n\nthe hard calculus way is to maximize xy given that y=56-x\nx+y=56 implies\nxy=x(56-x)=56x-x^2= to find maximum of 56x-x^2\ndifferentiating with respect to x gives 56-2x. The critical point is when  56-2x=0 so x=28.y=28
365052
Suppose a continuous open map f: R -> R is not monotonic.  Then we may assume, without loss of generality, that there are points x < y < z in R such that f(y) is the maximum (or minimum) element of f((x,z)).  This implies that f((x,z)) is not open, contradicting the assumption that f is open.
754882
Maximum value is obtained by subtracting the smallest number in the set from the largest number i.e. 35/6 - (-4) = 59/6\n\nMinimum value is obtained by subtracting the largest number in the set from the smallest number i.e. -4 -35/6 = -59/6
973577
given x < y\n\nnot sure if this would be using mathmatical induction but this is how i would do it\n\ntake\nx < 1/2 (x+y) < y\n2x< x + y < 2y\nx < y < 2y\n\nx<y is stated to be true and 2y is certainly bigger than y\ntherefore x < 1/2 (x+y) < y
247429
there has been alot of work on showing that at least a certain percent of zeros must lie on the line. I think we're up to around 90%. It is unclear to me as to whether or not a proof can be found in this way. My guess is no
872821
You can prove it geometrically:\nhttp://www.mathwords.com/m/median_of_a_trapezoid.htm
365360
Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
928789
numbers are 25, taken twice.\n\nway to prove it is to write x*(50-x), take a derivative and set it to zero.
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.