How do you prove this?
817604
There is no shorter way than, first, to tabulate n(n+1) mod 5 for each possible value of n :-\n\nn = 0, n(n+1) = 0 * 1 = 0\nn = 1, n(n+1) = 1 * 2 = 2\nn = 2, n(n+1) = 2 * 3 = 6\nn = 3, n(n+1) = 3 * 4 = 12 = 2 mod 5\nn = 4, n(n+1) = 4 * 0 = 0\n\nNote that 0 and 2 can be halved, and we have not reduced 6 to 1 because we are going to halve it too.\n\nSo half of n(n+1) can only be 0, 1 or 3.  QED.
166307
first prove that they do\nthen i will prove that wrong
148466
You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,
493260
Knowing what a proof is, is the first step in answering this question. From what I am familiar with...\n\nA proof is: a set of logical steps acquired through deductive (therefore, not making any giant leaps in logic, unless by definition), and hence, empirically (from the evidence provided) resulting in a direct equivalence (being, among other types of equivalence, but primarily, in permutation,  multiplicative/additively & negatively/positively & even/odd... meta-mathematically) of states, that's shortest distance is (in absolute terms), either infinity, zero, and/or, also, one.\n\nReally, the attempted 'proof' of 2 + 2 = 5 is based on a distorted type of Trigonometry, which was in essence the source of today's Calculus (just try to draw Tangent or Secant without running into the idea of Calculus' derivative & integral, respectively), and actually is the result of any additive equavalence of any two numbers' to being alike to any number, (because measuring hypotenuse of a given sides is essentially multiplicative, hence partially irrational). \n\n(Which makes me wonder... is there a 2 * 2 = 5 equivalent? and the answer is a resounding, yes! But first the 'proof' as written by Charles Seife.)\n\nLet a & b each be equal to 1. Since a ^ b are equal,\n\nb^2 = ab ...(eq.1)\n\nSince a equals itself, it is obvious that \n\na^2 = a^2  ...(eq.2)\n\nSubtract equation 1 from equation 2. This yeilds\n\n(a^2) - (b^2) = (a^2)-ab ...(eq. 3)\n\nWe can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get\n\n(a+b)(a-b) = a (a-b) ...(eq.5)\n\nSo far, so good. Now divide both sides of the equation by (a-b) and we get\n\na + b = a  ...(eq.5)\n\nb = 0 ...(eq.6)\n\nBut we set b to 1 at the very beginning of this proof, so this means that\n\n1 = 0 ...(eq.7)\n\n ...Anyways, getting that far gives us the jist of the proof, later in the proof, Charles Seife goes on to prove that Winston Churchill was a carrot! if you want to know how that is possible, I recommend you read the book.\n\nFrom equation 7, add a number to either side and get it equal to any other number, one greater than itself.\n\nMultiplying equation 7 after adding to it, and one can get: any number is equal to any other number. \n\nHence, conceptually, any number is equal to zero, and, theoretically, that includes infinity. But that's also the reason why when you divide by zero, it is 'Undefined.' Which, consequentially, is what is happening in this equation... just subsistute 1 into equation 3 and one will see that we are dividing by zero in equation 5. \n\nThis is what lead to the invention of calculus. Really, from here this segways into Hilbert Space... but that is best left for another entry, hopefully, on the actual subject of quantazation.\n\nThat's all I have time for... \n\nTHIS PROOF IS BY DEFINITION INCORRECT, but it provides a good tool as of why we define things in mathematics the way we do.\n\nA good question to ask from here would be (based on my previous tangent):\n\nDoes 1/3 plus 1/3 plus 1/3 = 1?\nOr, does it equal just zero point nine repeating?
842660
What exactly are you trying to prove?
658128
There is no way to prove a negative.  You are asking me to prove that something doesn't exist.
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.
1065017
I'm writing up a proof right now that this is not possible.\n\nCheck out the pdf at:\nhttp://www.geocities.com/euleratuo/
658173
sorry the answere on that is ist too HARD to prove...
855974
neither exist... show me some proof.
252734
This is how I remember it...hope it helps.\n\nA Proof\nLet 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).
460547
This looks close enough to the techniques used to show that the roots are close to the critical line. The actual estimates are difficult, though, and nobody has been able to actually give a proof using these techniques (obviously). Go ahead and try it. Perhaps you can see through to a proof. If so, you will be famous. If not, you will have learned somthing.
317141
To prove that a statement is false, all you have to do is find ONE case where it is false.\nLet a = 1 and b = 1\n(a+b)^3 = (1+1)^3 = 2^3 = 8\na^3+b^3 = 1^3 + 1^3 = 1 + 1 = 2\n8 does not equal 2.\nThus, I have just proven that it is false.
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
927149
if u "prove" it then u might have a point
864701
Prove by induction.\n\nWhen k=1, then (n+a)^1 = n+a = Θn, true\n\nAssume true for k.\n\nFor k+1,\n(n+a)^(k+1) = (n+a) * (n+a)^k = Θn * Θ(n^k) = Θ(n^(k+1))\n\nTherefore it holds for all k>0
626246
ok prove it first --- I have a very HIGH bar set for proof.