can you prove that 1 = 0?
355004
1. a = b + 1 (Given, let a and b be two numbers such that a = b + 1)\n\n2. (a-b)*a = (a-b)*(b+1)\n\n3. a^2 - ab = ab + a - b^2 - b\n\n4. a^2 - ab - a = ab + a  - a - b^2 - b\n\n5. a*(a - b - 1) = b*(a - b - 1)\n\n6. a = b\n\n7. b + 1 = b\n\n8. Therefore, 1 = 0
717483
1\nbecause u have mentioned that 1=5 so 5=1 and it cant = 9
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
497395
1.\nBecause e to the power minus infinity is 0.
981541
No, it follows from 1+1 = 2\n\nWhich follows from 1+0 = 1\n\nBut that can be considered an amazing "fact".  In reality, it's an axiom - a definition.  One of the only true defintions that form the basis of mathematics.
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 :)
674498
1+1=2. That's it. The answer for everything else would be simply 42
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
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
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.
489740
let 0=0\n   then a^2-a^2=(a-a)(a+a)\n  then a(a-a)=(a-a)(a+a)\nthen a=2a\nthen 1=2\nthen 2=3
930650
cos(0) = 1. so you can plug in x = 0 to get the limit as 0*1 = 0
206555
is the question x=x^2?\nif so 1 is the only number that can always satisfy that
436555
1=0 when reference to derivatives in calculus or when in reference to places numbers are from 0 - 9 \nthen they are repeated. etc 1_0 1_1 1_2 .. you get the idea. \n Any way first place is  = 0
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
325892
= 1-{1-[1-(0)]}\n= 1-{1-[1]}\n= 1-{0}\n= 1
431708
0/0=(1-1)/(1-1)=(1+1)*(1-1)/(1-1)=> 0/0=2\nor this:\n0/0=(4-4)/(2-2)=(2+2)*(2-2)/(2-2)=> 0/0=4\nor, maybe, this one:\n0/0=(9-9)/(3-3)=(3+3)*(3-3)/(3-3)=> 0/0=6\n.........and so one......\n\nhowever, it's not immpossible, just undefinied....