can any one prove that 1+1=0?
599876
no can do it equals 2
680886
You can't because it does!\n\nFirst assume x = 0.999... repeating\n\nx = 0.999...\nx = 0.333... + 0.333... + 0.333...\nx = 1/3 + 1/3 + 1/3\nx = 3/3\nx = 1\n\nSo 0.999... = 1\n\nThis is a valid proof with no faulty logic, division by zero etc.\n\n0.999... *does* equal 1\n\nAnother way to prove this is as follows:\n\nx = 0.999....\n10x = 9.999...\n10x - x = 9.999... - 0.999...\n9x = 9\nx = 1\n\nFor the person that says subtract from 1 and you get 0.0000000001, you don't.  You have to have an infinite number of zeros, not a finite set.  You could write this number as 10^-n where n = infinity.  This is also 1 / 10^n where n = infinity.  The lim as n->infinity of 10^-n is zero, similarly the lim as n->infinity of 1 / 10^n is also zero.  So you are adding zero.  This therefore another proof that 0.999... *equals* 1.\n\nAnd if you *still* don't believe me, read the Wikipedia entry below.
556722
look at my question\n\nBidDogARRRRR...\n1 minute ago\nDoes anyone know 0=1 and anything = anything else?\na = 1; b = 0\n1. a = b + 1 \n2. (a-b)a = (a-b)(b+1)\n3. a2 - ab = ab + a - b2 - b \n4. a2 - ab -a = ab + a -a - b2 - b\n5. a(a - b - 1) = b(a - b - 1) \n6. a = b\n7. b + 1 = b \n8. Therefore, 1 = 0
529993
If (1+2+4+8+16+...) is an infinite series you are basically saying that (infinity = 1 + 2*infinity)  which in itself is true because (infinity = infinity + 1) and (infinity = 2 * infinity).  So what you have basically proven is that (S = infinity = infinity) You can't subtract infinity from itself and obtain 1.  You get 0 which means both sides of the equation are equal.\n\nIf you make the series finite, then you can see how the logic doesn't hold to determine that S=-1.\nS = 1+2+4+8+16+32\nS = 1+2(1+2+4+8+16)\nS = 1+2(S-32)\nS = 1+2S-64\n-S = -63\nS = 63
494525
Think of it in terms of logarithms, since it is the inverse of exponents.\n\nStatement to prove:  1^infinity = 1\n\nRewritten in logarthmic form:  log[1](1) = infinity\n(Statement above is read: log base 1 of 1 equals infinity)\n\nWell that can't make sense, right?  It is a law of logs and of exponents that log[b](1) = 0 for any positive base b.  This is the same as saying b^0 = 1, for any b.  Also, log[b](b) = 1, for any positive b.  This makes since in b^1 = b.  Both of these laws would conflict this reasoning.  In fact, when b = 1, they conflict with one another.  Therefore, lets look at it with a variable instead.\n\nProve:  1^infinity = X\n\nRewrite in logartithmic form: log[1](X) = infinity\n\nNow, what value of X will give you infinity?\nCan you determine that?\n\nI would say it is indeterminate :)
66821
You can't divide anything by zero, so the answer mathematicians give is that 1/0 is "undefined". \n\nIt is very tempting to think that 1/0 = infinity but there is a consequence...\n\nIf 1/0 = infinity, and 0 times infinity = 1, then what is 2/0? Infinity? \n\nIf we say "yes", then we have to say that 0 times infinity = 1 AND also = 2 (and by implication also 3, 4, 5...). \n\nBut remember that anything times 0 is 0. It only takes the simplest of arithmetic to imagine an empty apple box, and to say "you can ask me for as many as you like - even infinity - but I can only give you zero apples"
846663
(x+3)(x-1)\n\n x= -3\nx=1
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
889713
3|(2x-1)/5| + 3 = 1?\nThats impossible. |anything| >= 0, so the LHS is >= 3. Did you mean something else?
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
510030
OK, found the error:\n\nThe sum you have defined is sum(i=1 to n)i(n-i+1)  <== Note "+1"\n\n1. Let n = 1.\nThen 1(1) = 1\nand (1/6)(1)(1+1)(1+2) = 1.\n\n2. Let this be true for n.\nThen sum(i=1 to n)i(n-i) = (1/6)(n)(n+1)(n+2)\n\nSo for sum(i=1 to n+1)(i(n+1-i+1))  <== Note "+1"\n= sum(i=1 to n+1)(i(n-i+1)) + sum(i=1 to n+1)(i)\n= sum(i=1 to n)(i(n-i+1) + 0 + sum(i=1 to n+1)(i)\n\n2a) = n(n+1)(n+2)/6 + (n+1)(n+2)/2\n= (n+1)(n+2)/6 [ n + 3 ]\n= (n+1)(n+2)(n+3) / 6  ==> proof since this is what you expect from the theorem.\n\nEdit: Missing the +1 in the original summation is the issue.
872873
There is only one value that works: \nx=1\n\nj
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
1067729
The sum of (-1)^n from 0 to infinite...1-1+1-1+1-1....\nGrouping this sum like this first: (1-1)+(1-1)+(1-1)...=0+0+0+...=0\nOR Use the following grouping: 1+(-1+1)+(-1+1)+...=1+0+0+0...=1\n\nThis is a fun problem, and shows that when using summation, be careful.
806805
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
978758
There's a mathematical riddle that claims to "prove" 1 = 2, but if you examine it closely enough, it turns out to be dividing by 0.  Sorry I csn't remember the riddle.
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.