Need help with proofs!!!?
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The best way to attack proving problems in geometry is to answer and answer more proving related problem. No one can help to solve a problem. No one can help you speed up how you solve. It's you who can do that.\n\nMost of my classmates want to be good in algebra.\nBut they hate to solve algebraic proble,.\nI told them to try  answering all the questions and exercises in a book of algebra. At first, they did not believe what I have told them. Until they realize that they are becoming better and better in attacking problems in algebra.\n\nSo i think, it's up tp you if you will be good in proving geometric relations.\nI have also done it myself.\nAnd i can't believe believe that i can answer a problem with ease. It only took me a few second to answer a problem that my classmates took lots of minutes to answer.\n\nBelieve me. Solve and solve lots of problem. And you will soon realize that you will solve difficult problems with ease.
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First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
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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,
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(A)  Show base case is true:\n\n(2*1)^2 = (2*1)(2)(2*1 + 1) / 3\n4 = 4\n\nAssume the equality holds for n = k\nShow that it holds for n = k+1\n\n2^2 + 4^2 + 6^2 + ... + (2k)^2 + (2(k+1))^2\n= 2k (k+1) (2k + 1) / 3 + (2k+2)^2\n= 2k (k+1) (2k + 1) / 3 + (4k^2 + 8k + 4)\n= (2k^2 + 2k) (2k + 1) / 3 + (12k^2 + 24k + 12) / 3\n= (4k^3 + 6k^2 + 2k + 12k^2 + 24k + 12) / 3\n= (4k^3 + 18k^2 + 26 k + 12) / 3\n\n2(k+1)(k+2)(2k+3) / 3 \n= 2 (k^2 + 3k + 2)(2k + 3) / 3\n= 2 (2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6) / 3\n= (4k^3 + 18k^2 + 26k + 12) / 3\n\n(B)  Show base case is true\n5 (2!) ≥ 3^2\n10 ≥ 9\n\nAssume it is true for all n < k, Show it is true for n = k+1\n\n5(k+1)! ≥  5 (k+1) k! ≥  5 * 3 * 3^k ≥ 5 * 3^(k+1)
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just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
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1 -> b\n2 -> e\n3 -> c\n4 -> d\n5 -> a\n6 -> f
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1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
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This is a better link that the previous one posted here!! try it! \n\nhttp://pubs.logicalexpressions.com/Pub0009/LPMArticle.asp?ID=305
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Well, I agree with Rosie. You can do proofs really fast if you have a very clear understanding of all the required theorems & formulae. But for this you need to really really practice - and the time won't be far away when the only time that will be required will be the time you take to write the proof! I was very poor in this subject - but I was so determined that I would sit long hours and try to prove it on my own. Believe it or not I would spent 2 or 3 hours for trying to prove certain problems  i.e without taking anybody's help. It was very tough initially - sometimes i was not able to do it - i would go to sleep at night with the problem in mind - honestly i would always get it right the next morning. With gradual practice i could do it in a jiffy! But my dear don't forget H.W.Longfellow - "The heights of great men reached and kept, Were not obtained by sudden flight, But they, while their companions slept Were toiling upward in the night."
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Here's some checks of proof ( Click back then go on http://www.100wgezi,com     to get started)
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I will proof read it. my email it flores_debilee@yahoo.com. You can also check out my webpage @ www.groups.yahoo.com/group/3latedremnant
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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
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Here is a link to my geocities site.  You will find a pdf with a proof of this problem and other yahoo answers answers.\nhttp://www.geocities.com/euleratuo//index.html
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Here are a few sites you can check out:\nhttp://www.freemathhelp.com/\nhttp://www.mathgoodies.com/\n\nHope these sites can help you..
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Check out this site. It gives you a few examples. It was helpful for my cousin when she was leaning how to do them.\nhttp://www.purplemath.com/modules/solvelin.htm
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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?
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Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}