Okay, I understand that aleph_0 is supposed to represent the amount of geometric curves that exist, but how is that proven?
132555
First of all, aleph_0 is the cardinality of the natural numbers (also, the cardinality of the integers, the rationals, and the algebraics). The cardinality of the real numbers is usually denoted by C (for the continuum). The continuum hypothesis is the conjecture that C=aleph_1. It turns out that this is independent of the usual axioms of set theory. In other words, whether C=aleph_1 or not is undecidable in the sense of Godel.\n\nEven if C=aleph_1, it is not provable that the collection of subsets of R (which is the same as the collection of geometric curves, suitably defined) is the same as aleph_2. So what you ask for is not only not proven, it is not provable. There is another hypothesis called the Generalized Continuum Hypothesis that states that the collection of subsets of a set with cardinality aleph_n has cardinality aleph_{n+1}. This result turns out to be independent of set theory also.
17241
Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)
406201
Actually, kaksi_guy seems to be onto something. Unfortunately, not only does his proof have the flaw that you noticed, but even if that were rectified, he would still have only proved that f is zero at infinitely many points on the interval, not that f is zero on every point in the interval. Still, his post is the inspiration for what follows:\n\nFirst, we need to assume that g(x) is defined everywhere on the interval between the two points. We don't need to assume that g(x) is continuous, but if we allow a function that is undefined at even one point, such as g(x)=(x-3/x)/2 (which is undefined at zero), we may easily generate exceptions to the theorem (in this case, f(x)=x²-1).\n\nWith that out of the way, we can begin our proof. Suppose f(x) = 0 at two points a and b. By the extreme value theorem, f(x) has both a maximum and a minimum on the closed interval [a, b]. Let us denote the minimum value f(x) attains on [a, b] as min(f(x)) (and analogously for max(f(x))). Now, since f(a)=f(b)=0, min(f(x))≤0 and max(f(x))≥0. Suppose that min(f(x))<0. Then the minimum does not occur on the endpoints, so it must occur on a point c in the open interval (a, b). It is readily shown that this implies f'(c)=0. Therefore, from the original differential equation, f''(c) - f(c) = 0 and thus f''(c) = f(c). However, f(c)<0, so this implies the function is concave down at c, contradicting the fact that c is a minimum. Thus min(f(x))=0. Similarly, if max(f(x))>0, then there would be a point c on (a, b) which is a maximum, but is concave up, which is also a contradiction. Thus, min(f(x))=max(f(x))=0, and a function whose minimum and maximum are both the same is a constant function, in this case f(x)=0 on the interval [a, b]. Q.E.D.
269879
You **did** say that the Pythagorean Theorem was a given, right? I think that your INTENT is to prove the converse of the Pythagorean Theorem, right?\n\nLet x, y, and z be positive integers such that\nx^2 + y^2 = z^2.\nClearly z > x and z > y (otherwise either x^2 or y^2 must be less than zero; prove this for yourself if you must).\n\nLet x and y be the sides of a right triangle that form the right angle.\nThen, by the Pythagorean Theorem the length of the hypotenuse h is\nh = √(x^2 + y^2)\nh = √(z^2) . . . . . . . . (given equation above)\nh = z . . . . . . . . . . . (given z is positive) \n        \nTherefore, by side-side-side congruence, any triangle whose sides are of lengths, x, y, and z, where x^2 + y^2 = z^2, must be a right triangle with hypotenuse z.\n\n\nASIDE TO HOOD...: You are using the Pythagorean Theorem to prove the Pythagorean Theorem. Although the questioner's intent is not perfectly clear, I don't think the intent was to prove "if x, then x". If so there are easier ways to do so.
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I think you mean that: Given a figure (function in the form of a curve), and the value of ordinate (y), does a value of abscissa (x) exist corresponding to that y?  \n\nIf the figure is an open curve, like y=x or y=ax^3, the answer is yes.\n\nBut there may be many other cases. For the curve y=x^2, which is a parabola, x may assume any real value; but y is always non-negative.\n\nAgain, x^2+y^2=a^2, is a circle and a closed figure; x can take up values between -a and +a. y also takes values in the same range.\n\nFor this curve, given a value of y, within the admissible range of -a to +a, there are two values of x. \nFor y=+a or y=-a, there will be only one value of x; and for y>a or y<-a, there will be no value of x.\n\nFinally, for periodic curves like y=sin x, for admissible values of y (between -1 and +1) there exist infinitely many values of x.\n\nSo, you see there is not one answer but many depending on the nature of the figure!
383443
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
465957
I agree.\n\nHere is one proof not involving geometrical bodies.  Too long to paste so just go there:\n\nhttp://fermatslasttheorem.blogspot.com/2006/02/eulers-formula.html
235482
yeah, looks like the circle is just a 0, which would mean x^0 + 7 = 1 + 7 = 8.  It does look weird, though.
59168
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
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Graph them both.  Notice that, after graphing the one, you only need to turn your "page" 90 deg counterclockwise and you'll get a mirror image of f^-1(x).  \n\nYou'll also notice that each has exactly one rectangular area that the other does not.  The first, f(x), is a "piano"-shape, formed by a "pseudo triangle" and a rectangle, the latter of which measures 8x5, because 8=12-4.  Now take that rectangle away, and look from the point of view of f^-1(x).  This "triangle" lacks a rectangle 4x4 in dimensions, where the second 4 is the height, from f^-1 (9) = 4, and the first 4 is from 4=9-5.\n\nThus, we subtract 8x5=40, and add 4x4=16, to get our answer (which I'm actually too lazy to finish giving; it's 24 less than the first area).
375703
The first thing to do is to extract from this problem a list of the \n"givens" (the things you are told are true). \nSo here's what we are given:\n\nGIVEN - Segment AB is the diameter of a circle with center O\n          Line AC is tangent to circle O at A\n          Line BD is tangent to O at B\n\nTO PROVE - AC and BD are parallel\n\nSTATEMENTS                                     REASONS\n\n1.  Segment AB is a diamter of circle O        Given\n2.  Line AC is tangent to O at A                     Given\n3.  Line BD is tangent to O at B                    Given\n\n4.  AC is perpendicular to AB                  Tangent line is perpen-\n                                                                 dicular to the radius\n\n5.  BD is perpendicular TO BA                  Theorem as stated in \n                                                                  #4 (Theorem 9-1 - If a\n                                                                   line is tangent to a\n                                                                  circle, then the line \n                                                                   is perpendicular to \n                                                                   the radius drawn to\n                                                                  the point of tangency)\n\n6.  AC is perpendicular to AB                  Two lines perpendicu-\n                                                                 lar to same line are\n                                                                 parallel (Theorem 3-7\n                                                                  In a plane two lines\n                                                                  perpendicular to the \n                                                                   same line are parallel\n\n\n\n\n\nLet me know if you need any more help, I can e-mail you more in-depth info.....\n\n\nHope this helps!!!
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A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
640439
no, a circle has infinite lines of symetry. For an equilateral triangle, i believe there r only 3
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it can also be defined more geometrically this way:\n\nconsider the graph of\n\ny = 1/x\n\ne is the only point on the x-axis > 1 such that\nthe integral of y=1/x from x=1 to x=e is 1.\n\nin other words, for any number b >= 1, let A_b\nbe the area inside the region between x=1, x=b, y=0\n(the x-axis) and y=1/x. So for b=1, A_b = 0, and as b gets\nbigger, A_b gets bigger. In fact one can prove that\nA_b becomes as large as you want by making b big enough. So there's some point b=e where A_e = 1. That's e.\n\nThis is related to the logarithm, as the function\nf(b) = A_b\nis actually the natural log function, i.e.\nlog(b) = A_b.\n\nand, it's not a ratio of integers (not a rational number), i.e. there's no integers m, n such that e = m/n. this is also the case with pi. i guess when you say pi is a ratio, you mean the ratios to do with the circle. e is a ratio in that sense, for instance e = e/1, or e = (pi*e)/pi. but i'm not aware of any useful geometric ratio.\n\ne and pi also have the property (along with most real numbers) that there's no polynomial with integer coefficients p(x) such that p(e)=0 or p(pi)=0. contrast this with say sqrt(11), which isn't a rational number, but does give 0 when plugged in the polynomial\n\nP(x) = x^2 - 11.\n\nthis is the definition of transcendental versus algebraic. e and pi are trans, whereas sqrt(11) is algebraic. there are "more" transcendental numbers than algebraic numbers, in a quite reasonable sense of "more".
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What everyone has said is correct, but look at it this way: Consider a square. We know from the theorem of pythagoras that the ratio of the hypotenuese to the length of one side = sqrt(2) or, if you like, the ratio of one side to the hypotenuese = 1/sqrt(2). These constant ratios are independent of the size of the square. We have the ratio of the perimeter of the square to the hypotenuese = 4/sqrt(2) = 2*sqrt(2), which is independent of the size of the square.  Similarly the ratio of the circumference of a given circle to the longest chord(the diameter) has an exact value, and similarly this value is independent of the size of the circle. As has been mentioned, for any regular n-gon(square, pentagon, hexagon etc) the ratio of the perimeter to the length of the longest line segment that can be drawn within the n-gon will have a specific value depending on n, and this value can be determined. As we let n approach infinity, the n-gon approaches a circle and the ratio of the perimeter of the n-gon to the longest "chord" approaches a specific limit and this limit is pi, the ratio of the circumference to the diameter.  And, as mentioned above, analytic techniques can (and have) determined the value of the constant pi to a trillion or so digits.\n\nFootnote: SK, determining the limit of a series or even that the limit exists does not necessarily require calculus.\nFor example, the sum of a geometric series such as Sum[1/(2^n)] = 1/2 + 1/4 + 1/8+....+1/2^n is proved without calculus to be 1- 1/2^n and the limit of it (without calculus) as n---> inf. = 1. Check out the below link regarding pi.\n\n http://mathworld.wolfram.com/ArchimedesAlgorithm.html
211006
Consider K_3,3, which is the graph obtained when you draw all lines between 3 points {a,b,c} and three other points {x,y,z}. So there is a line between a and x, but not one between a and b.\n\nAn automorphism, f, will take a to one of the six points. There are two possibilities: either f(a) is in {a,b,c} or f(a) is in {x,y,z}. In the first case, {f(a), f(b), f(c)} is a permutation of {a,b,c} and {f(x),f(y),f(z)} is a permutation of {x,y,z}. In the second case, it goes the other way around. Furthermore, any two permutations in both sets will lead to an automorphism of your graph.\n\nThus, the order of the automorphism group will be 6*6*2=72.\nThere is a subgroup isomorphic to S_3 cross S_3 of index 2. Just checking commutativity tells us that the reflection between {a,b,c} and {x,y,z} is not in the center, and in fact, conjugation by the reflection reverses the order of elements in S_3xS_3. Thus, the automorphism group is the semi-direct product of S_3 cross S_3 with Z_2 where Z_2 reverses the order of the cross product.\n\nI hope this helps!
218797
the area under the curve f(x)!