can you prove that 1 = 0?
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
797702
Since there is no = sign in that equation, I assumne it should be equal to zero.  If that is the case then m= -1/8
206555
is the question x=x^2?\nif so 1 is the only number that can always satisfy that
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
44798
Any number raised to the power of zero is 1. Why? Start with a simple problem: Divide one number raised to the power of n by the same number raised to the power of n. (In the example below, I use ** to mean "raised to the power." And I use the number 3 just as an illustration. It works with any number, and if you wanted to make this a general proof, you'd probably substitute "x" for the 3.)\n\n3**n / 3**n \n\nWe all know that any number divided by itself is 1, right? So 3**n divided by 3**n is 1.\n\nAnd to divide one number that has an exponent by the same number with an exponent, you simply subtract the exponents.\n\nSo:\n\n3**n / 3**n = 3**(n-n) = 3**0\n\nSo 3**n / 3**n equals 3**0\n\nAND we know 3**n / 3**n also equals 1.\n\nSo, 3**0 equals 1.\n\nHope this helps!
166763
1+1=2 is an axiom - something that we consider being true without proving it.
930650
cos(0) = 1. so you can plug in x = 0 to get the limit as 0*1 = 0
709175
If you need a positive result, then y will only be a real number when x - 1 is >= 0.\n\nx - 1 >= 0\nx >= 1\n\nSo x must be greater or equal to 1.
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
163896
yes the previous equation mentioned would "prove" it, but the problem occurs as you're dividing by zero.  you can also make 1=2, with a similar process, involving the division of zero.  if you don't get what i mean, the previous explanation has this:\n\n(x-y)(x+y)=y(x-y)\nx+y=y\n\nwhat happens at this step?  you are dividing by (x-y).  since x=y is given, then (x-y)=0, thus you are dividing by 0, which is impossible in math.
89742
x and y are real numbers...if x+y=0,then how do i prove that x=-y?\nJust transfer the y variable to the right side of the first equation.\nx=0-y.\nAnd then you can just take out the zero.\nAnd you get x=-y
904022
Surprisingly, yes.  This bugged the heck out of me in middle school, but it's a consequence of the way infinity works.\n\nThe classic proof is:\n\n    x = 0.999999....\n10x = 9.999999....\n\nSubtract the first line from the second, and you get\n\n9x = 9, therefore x = 1.
600176
This is a Frequently Asked Question. Following is my previous answer:\n\nThis hypothesis is false:\nSuppose 0/0 = 1. Since 1 is the multiplicative identity, this implies (0/0)*a = a for all real numbers a. But (0/0)*a = (0*a)/0 = 0/0 = 1. This contradicts our hypothesis that 0/0 is the multiplicative identity, so the hypothesis must be false (proof by contradiction). QED\n0/0 is indeterminate as a limiting form. See http://mathworld.wolfram.com/Indeterminate.html .\nAs an algebraic expression over the real field, it is undefined: http://mathworld.wolfram.com/Undefined.html .\nThis is a Frequently Asked Question; the following is my previous answer:\n\nThis question may be asked so frequently (along with the infamous 0.999...) because no concrete axioms or definitions are ever given for the real number system until one studies higher algebra or real analysis; it is assumed that everyone knows what a real number is until then. The proliferation of these questions show that this assumption is no good.\nFor the sake of conciseness, I will not define what a real number is here, as only the algebraic structure of the real numbers is necessary to answer this question. The algebraic structure of the real numbers is what we call a field. A field is an algebraic structure that follows a finite set of axioms. These axioms define what statements are true and false in this system based on the laws of logic. You can find the field axioms here: http://mathworld.wolfram.com/Field.html .\nEach line is preceded by "For all real numbers a, b and c, the following is true...". Note that the proper form of the additive identity and multiplicative identity is that it prescribes the existence of some number 1 and some number 0 such that the equations hold. These same numbers are then used to prescribe the existence of additive and multiplicative inverses.\nDivision a/b is just shorthand for a*(b^(-1)). So in your question, you are looking for 0*0^(-1). It is easy to show that there is no such number as 0^(-1) in any field. Suppose 0^(-1) exists.\nThen 0*0^(-1) = 1. But multiplying both sides by 0^(-1) gives 0 = 0^(-1), which means 0*0 = 0 = 1. This is obviously false.\nSuppose 0 = 1. 0*a = a for all numbers a. But 0*a = (1 + (-1))*a = a + (-a) = 0. Thus 0 is not equivalent to 1 and our sole assumption that 0^(-1) exists must be false. Thus the expression 0*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs for trying to define 0/0 using limiting behavior, note that you are evaluating a limit over a plane of the function f(x,y) = x/y. As we approach (0,0) along the x-axis, this limit is 0, but as we approach (0,0) along the y-axis, the limit diverges. Thus, the limit of f(x,y) at (0,0) does not exist, so f(0,0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nThe expression 0/0 is said to be indeterminate as a limiting form because one can have many functions that approach the form 0/0 that have convergent limits of different values. For example, the limit as x approaches 0 of [sin(x)]/x yields the indeterminate form 0/0, but L'Hopital's rule reveals the limit to be 1. Similarly, multiplying the above limiting form by any real number will yield that number as the limit, even though the form is still 0/0. The expressions listed on this page: http://mathworld.wolfram.com/Indeterminate.html are called indeterminate because they do not have a fixed behavior when studied as limits of expressions.
1010309
Let a = b = 1.\n\na = b     [Given]\na ^2 = ab     [Multiplying a on both sides]\na^2 - b^2 = ab - b^2     [Subtracting b^2 on both sides]\n(a + b)(a - b) = b(a - b)     [Factoring both sides]\n(a + b) = b     [Dividing both sides by (a - b)]\n(1 + 1) = 1     [Substitution]\n2 = 1\n\nThe flaw in this so-called "proof" is that you can't divide both sides of an equation by (a - b), because with a = b, a - b = 0, and you can't divide a number by zero.
1142259
proof :\n\n1 -1 = 1 - 1\nlhs: 1 - 1 \nrhs: 1 -1 \n\nthaking 1 common  from lhs v hav \n1 (1 - 1 )\n\nexpanding rhs by the formula a^2 - b^2 = (a+b)(a-b)\n\n(1-1)(1+1)\n\nthus we hav \n\n1 (1 - 1 ) = ( 1 - 1 ) (1 + 1 )\ndividing througout by ( 1 - 1 )\n\nwe get 1 = 1 + 1\n\n\n\nDoes this help u??
148878
1power0 = 1\n1power1 = 1\nso that 0=1
606683
The factor (1/infinity) equals to zero.\n\nHence the term\n\n(infinity) - (1/infinity) = (infinity) - 0\n\nTherefore,\n\nX = infinity