Ivan said if a number ends in 0 and has a factor of 2 and 5 is he correct, why?
930195
yes because any number ending in zero is even but is divisable by five
127821
"Dividing by nothing" doesn't mean you would do nothing.  It means, "how many piles of nothing (zero) would you have to add up to get to the original number?".  \n\nHow many piles of zero would you have to have in order to have 2?  The answer isn't even infinity, because you would still never have 2.  So it's not possible, and 2/0 can't be represented by any number.
912338
Neither, it is undefined, i.e. it has none of the properties of all rational or whole numbers.\n\nI might be wrong, but I think this is one of those trick questions that are so common in math courses.
364553
I agree July 25th.  The five stays the same, but the first numbers starts at 2, declines to zero and then starts back toward 2
766859
Alright. i=1-5,because the number on the bottom is the first number you will put in the expresion, and the number on top is the last number.and you put all the number inbetween in the expression too. So i will equal 1, 2, 3, 4, 5. Plug each one into the expression next to E. Now you have 5 answers. Sigma means something like "add" so you add all 5 answers together, and that is the final answer.
782057
look at a few examples and it will make sense\n\nfor example \n\n02 (where the 2 is small)   -> _ O    (the _ is where a number goes)  (2)
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.
885437
The last two digits are '25'. \n\n(This is the same for ANY integer power of 5 greater than 1).\n\nThe calculator won't give you the answer since it is not accurate enough.
921859
No, because the opposite of -5 is 5 which is positive; the opposite of 0 is 0 which is neither positive nor negative.\n\nThis is why the expression "-x" should never be read as "negative x" but rather "the opposite of x" since if x itself were negative, -x would be positive.
510570
Because the exponet of both is an even number which is 2\n\nIf you divide by 8 you might get a .5 result.
133903
You can put it as the last number, right after 2, because that would be the 13th number.\nOr it could be after the number 6. If you add all the numbers and divide by 13, you would get 6.
145210
They call ZERO an even number, who are they that mention like this?\n\nActually, odd numbers are start with 1, then 3, 5, etc., even numbers are start with 2, then 4, 6, etc.\n\nIf you check at the numbers system, then you can understand what meant by odd number, even number, negative number, natural number, whole number and others. \n\nSo, here the point is, ZERO actually is not a even number (the smaller even number start with 2), then there was no reason to tell you that why can call it an even number.
283236
20/10 = 2 because 2 x 10 = 20\nn/0 = n because n x 0 = n? No!!!\nSo n/0=n is wrong.
50645
if we speak of numbers and not of limits, the answer is no without exceptions: 0 is defined in group theory as the null element, so it is BY DEFINITION the element which multiplied by any other gives itself: always 0.\nInstead, what we usually assume as always true, but is not true in some finite algebraic models, is that if A*B=0 then either A=0 or B=0. This is not always true in all algebraic models (for exemple in modular algebra).
950893
X cannot have any prime factors other than 2 or 5. Thats because 2 and 5 are the only prime factors of 10, which is the base you are working in (and if you changed base, the same idea still works.)\n\nReplying to your extra questions: well, X is an integer, of course it terminates. As for 1/X, its a rational of course, but that has nothing to do with it terminating or not.
469511
To quote dilbert: "Early Civilisations never had a concept of zero" (hey, you wanted it thorough right?) And they started the number system at one. .... Hmm, pretty much the panel cuts out... \nRemember, the ones and zeros are really the positive and negative polaritys on a magnetic disk. That is the key to your computer informations. Do not take a magnet to your computer.
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.