Can it be proven that 0 is equal to 1?
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
659301
Zero divided by any number (except zero) is zero.
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.
194832
mate, zero divided by any no. is always zero
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
853846
Do not write x = 0 because think about it\n\nif x = 0, then that would mean that |0+4| + 4 = 0\n\nthat means 8 = 0... is that true? NO!\n\nSince it is imposible for |x - 4| to equal -4 (since absolute value is ALWAYS possitive or 0) there is no solution =)
717165
-1 because -1*-1=1 the reciprocal of a number times the original number ALWAYS equals 1.
155864
yes, it equals x(ln(x)-1)
536877
According to this article, (as well as my own knowledge) it is equal to 1.
574280
a number to the zero exponent is always 1 as a answer
930195
yes because any number ending in zero is even but is divisable by five
750843
you cannot divide by 0 = a-b\n&#92;
148878
1power0 = 1\n1power1 = 1\nso that 0=1
348468
This 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, 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. Tus the expression 0*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs to the limit argument, 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. 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.
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.
365423
This proof is trivial. The number -1 is defined as the number satisfying the equation 1 + (-1) = (-1) + 1 = 0. \nMultiply both sides by -1:\n(-1)*1 + (-1)*(-1) = 0\nby the distributive property.\nBy the fact that 1 is the multiplicative identity and the definition of additive inverse, we get:\n-1 + (-1)*(-1) = 1 + (-1)\nAdd 1 to both sides (additive inverse property):\n(-1)*(-1) = 1.\nQED.
55379
I would say, \n\nonly in the case when n=0 or 1, then this equation stands good! If n is not equal to zero or one, then everytime the answer would definitely be above 1 (even when u use negative integers)