Prove the statement for two couples.?
94604
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). "
498942
what proof are you looking for? can you specify more details about the actual question?
691967
The couple themselves.
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
1155743
Mutually exclusive - They both cannot occur at the same time.\nTherefore P(A and B) = 0\nP(A or B) = .5
219570
Prove, patent and present
491741
you arguments are good, however i swear that proving FLT isn't so easy . the best will that you yourself prove FLT in a single page checking out every arguments.\nafter all it took 350 yrs to find a proof !!\nkeep on trying ,good luck.
108416
none if they are false
1048575
Maintain a sense of reasonable doubt. Accept proofs that work beyond a reasonable doubt as real. Accept cooberated circumstantial evidence, with physical evidence.\n\nTest everything except your wife's, and God's love.
57033
the answer is 3 because a-b|a-b & a-b|a so a-b|b
60524
Um, no!  I don't think you could prove something like that one way or the other.  But does it matter?
609321
One is the inverse of the other
29042
milkyway: \nI think your answer is incomplete. Unfortunately, I don't have my own proof, but let me criticize yours. \nSorry.:-)\n\nWhat is definitely proven by induction in your post is that you are able to construct an acceptable set of N points. However, in my view, your reasoning does not prove the statement for a given set.\n\n1. From the very beginning, you have N points. How can one realize the induction scheme in this case? Perhaps, you mean that you can temporarily erase N-3 points (keeping in mind their positions) and start with 3 points forming a triangle, or "3 couples" in your terms. Then you restore the erased points one by one at their positions.  I believe this should be explained in your proof. \n\n2. Note that if you start with 3 aligned points, by "restoring" the 4th point, you can generate an allowed set of N+1 points out of the prohibited set of N points. This is missing in your proof, where you deal only with allowed N-sets.  \n\nBut this is not the most serious flaw there.\n\n3. You write: "now, this way there always are the old two couples, plus two new couples (new point-non aligned point of each of such two couples). So going from N to N+1, the number of acceptable couples is always >=3."  \n\nI disagree with this statement. \n\nIndeed, imagine that the three couples, which work at the stage N, form a triangle. Aligning the newly "restored" point with one side of the triangle, you kill one working couple. Still you have the two remaining old couples and get a possibly working new couple (new point -- non aligned point of the triangle). However, it is not at all guaranteed that there is no other point lying on the line going through this new possibly working couple. If such point exists, the number of working couples in this case gets reduced from 3 to 2 when going from N to N+1. At the next step, this way I will reduce the number of working couples from 2 to 1 and so on.\n\nThe transition from N to N+1 is therefore not justified.\n\n4. A general question. For N=infinity the statement is false for points located at nodes of a regular \n(say, square) lattice. Doesn't it contradict to the very idea of proving this statement by induction?\n\n\n------------------------------\n\n\n\n\nHere is my proof.\n\nGiven a set of N points, the number of all possible couples is M=N(N-1)/2.\n\nFor each couple draw a straight line going through it. We get M lines.\n\nLet us assume that the statement we are going to prove is false: i.e., none of those M lines goes through ONLY TWO points.\n\nThen, of course, each of those M lines coincides with at least the two other lines, so the actual number of distinguishable lines K < M.\n\nWe have N points and K different lines.\n\nEach of the NK pairs (point + line) is characterised by a distance S(n,k) between  the point n and the line k. \n\nThis distance S(n,k) is defined as a length of a segment perpendicular to the line k with one of the ends lying on the line k and the other end coinciding with the point n.\n\nLet us list all the NK distances S(n,k). Since NK is a finite integer number, one can find the shortest value (or several equal shortest values) of S(n,k). \n\nTake a closer look at the point P and the line L with the shortest S(n=P,k=L).\n\nBy assumption, there are at least three points (L1,L2, and L3) lying on the line L. Therefore, at least two of these three points (say, L1 and L2) can be found to the one side (say, to the right) of the point A where the perpendicular to the line L starting at the point P hits the line L. The case when one of the three points (say, L1) coincides with the point A is treated in exactly the same way.\n\nNow we have a triangle (P-L1-L2), with two vertices L1 and L2 lying on the line L to the right of A in the order A,L1,L2 from left to right. \n\nSince L1 lies to the right of A (or coincides with A), the angle between P-L1 and L1-L2 is greater than (or equal to) pi/2.\n\nTherefore t
722973
K=positive then U=negative K=U=A and B=equal.
164065
The trick with "proofs" like these are generally an obscured division by zero. One of the lines seems to work, though when you divide to cancel out some factors, the quantity you're dividing by could/does equal zero. It's a trick and doesn't say anything substantial about mathematics.
955983
The answer is 42.  Prove or disprove by doing your homework.
864410
The proposition is\n\nSQRT(a) + SQRT(b) = SQRT(a+b)\n\nIf the equality is true, then we can rais both sides to the second power, yielding\n\n(SQRT(a) + SQRT(b))^2 = (a+b)\n\nIf we put c=SQRT(a) and d=SQRT(b) to simplify the equation, we have \n\n(c + d)^2 = a+b\nand expanding\nc^2 + 2cd + d^2 = a+b\n\nBut c^2 = SQRT(a) ^2, and d^2 = SQRT(b) ^2, so in the end\nc^2 = a and d^2 = b\nso back substituting we have\na +2cd +b = a+b\nand 2cd must be equal to zero, and that is possible only if either c or d is zero, and that is possible only if either a or b is zero.