how is 1/0 gets infinity?
457166
This is a Frequently Asked Question. 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 1*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 1*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 of the function f(x) = 1/x. As we approach 0 from the negative x-axis, this limit decreases without bound, but as we approach 0 along the positive x-axis, the limit increases without bound. Thus, the limit of f(x) at 0 does not exist, so f(0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nWhile 1/0 is not defined in any number system, |1/0| is defined as positive infinity in extended number systems (which are not fields by the above argument) for various purposes. See http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html .
4300
zero divided by zero is not defined\none divided by zero is infinity\n\nIf you have a series like a_n/b_n whre both and b converge to zero as n goes to infinity, it might still have a limit, determined by what goes to zero faster a and b
820085
We say that division by zero is undefined, for it is not consistent with division by other numbers.\n\nHere's a little experiment for you to try on your calculator. Observe the output when you try the following set of calculations:\n       1/1\n       1/.1\n       1/.01\n       1/.001\n       1/.0001\n       1/.00001\n       etc... \n\nuntil your calculator can't go any further or you get tired. You should notice that the answers continue getting larger and larger. \nAnother way of thinking of it is to imagine filling a box with apples. Say a box can hold 100 apples. Now try filling it with apples that are half the size of these apples. You can put 200 in the box. Now imagine a special, magic apple that takes up no room at all. How many can you put in the box?\n\nWell, the answer is... there is no answer! That is why mathematicians refer to numbers that are divided by 0 as "undefined." Some people tend to think of them as being infinite, but this isn't exactly true. There simply is no answer.
2367
C = A - B = infinity - infinity = undefined :-), not 0.\n\nIf you have the specific series for A and B, then you could technically substitute them into C and figure out C for different {A,B}.\n\nThe short answer is this: not all infinity series are the same, so their difference is really not zero.
387091
Actually your question is the explaination of division operation of the number 1000000 to the number 0...\nThere is infinite zeros at any Real number... The number of a million is allready a Real number... Then we can say that the answer is infinite... Even if you like to ask another Real number the answer doesn't change... It is infinite for all Real number...
941593
1/3 *3 is one right? right\nand 1/3 is equal to .3333... right? right\nand .33333...*3 =.99999..... right? right\n\nwe have just proved that 1=.999999...\n\nthere is no rounding, it is a fact.\ncrazy but true.\n\nClick link for more examples and a longer explanation of exactly why this is true:\nhttp://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html\nthis will take you step by step as to why.\n\n(its .9999..... that is equal to one .\n.999 is not 1, is .999=.999)\nBut I know what you meant.
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
854009
lim as x approaches 0 from the positive of (ln x) is negative infinity.\nlim as x approaches 0 from the positive of (sqrt x) is 0.\n\nif the numerator goes to negative infinity and the denominator goes to 0, the limit is negative infinity. it might look like L'hospital's rule is required, but it is not.
490253
infinity/0 is infinity.  If you have any function f(x) and g(x) where f(x) goes to infinity as x -> k and g(x) goes to zero as x -> k, then 1/g(x) goes to infinity as x -> k, so\n\nf(x)*(1/g(x)) -> infinity as x -> k.\n\ninfinity*zero depends. If you look at 1/x and x^2, then as x -> 0, 1/x -> infinity and x^2 -> 0, but\n\n(1/x)*(x^2) = x, so its limit is 0 as x -> 0.\n\nIf you look at 1/x^2 and x, then as x -> 0, 1/x^2 -> infinity and x -> 0, but\n\n(1/x^2)*(x) = 1/x, so its limit is infinity as x -> 0.
1062482
Infinity is NOT a number; it is simply never ending.\n\nTo answer your question about infinity minus infinity, here is probably the simplest way to explain this: infinity minus infinity has three possible answers: 0, infiity, and some constant I will provide an example for each answer using one anology:\n\nImagine you have a bag of infinite marbles. What would happen if you removed ALL the marbles in the bag, you would have none in the bag (thus infinity minus infinity is zero).\n\nWhat happens if you remove all the even numbered marbles? Since there is an infinite number of even marbles, as well as an infinite number of odd numbered marbles, you have infinity minus infinity equaling infinity (all the infinite even marbles were removed, but there are still an infinite number of odd marbles in the bag).\n\nWhat happens if you want to have all but 3 marbles in the bag? You have to remove the all of them but 3. Thus infinity minus infinity equals a constant.\n\nTo answer your question about (0/0) - (0/0) This is simply undefined: you CANNOT devide by 0. An undefined value does not necessarily mean that it is infinite; there is a difference.\n\nInfinity divided by invinity will provide you the same answers as the subtraction problem. Zero raised to any power will always be 1 and infinity raised to the infinite power will more than likely be infinity.\n\n\nI hope this answers your question as to the concept of infinity.
12883
how do you count to infinity?
748218
Think of it like this...\n\n10/10 = 1\n\nthen try a number closer to zero...\n\n10/1 = 10   see the number gets bigger\n\ntry a number even closer to zero...\n\n10/.001 = 10,000\n\nSo the closer the denominator gets to zero, the closer the answer gets to infinity.
90067
There's no answer actualy. O cannot be divided by 0. \nThe answer infinity is just an assumption when \nnumber really near to 0 divided by another number near to 0.\n\nSuch as 0.0000000000000000001 / 0.0000001\nAlthough the example doesn't really show number near to 0, but if you change to 0.00000000000000000....01, you will see the same result
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....
504653
The set of all fractions {1/N} is equinumerous(the same size) to the positive integers N. They are both infinite sets.\n\nc does not really apply as it is a value, not dealing with set theory, but it's value is infinity, meaning it's larger than any value you can think of. If you are talking about the set {lim(x->0) 1/x}, then it only has one element, making it much smaller than a & b\n\n"a) and b) are of equal size, though if you take any fractions instead of just those between 0 and 1, you will get that a) is the largest. I attended a lecture about this a while ago, hence I can't remember the arguments, but the lecturer clearly concluded that there are more fractions than integers."\n\nThis is almost correct. If you just take all the fractions between 0 & 1 it will be a larger set than b. Since 1/N is equinumerous with N+, it follows that 2/N is equinumerous as well. All the fractions between 0 & 1 would include 1/N, 2/N, 3/N, etc. making it much larger than N+
664717
I'm not sure what you are asking, but yes, there are an infinite number of numbers between 1 and 0.  \n\nIn fact, the whole numbers are the smallest group of infinities on the number line.  The infinite set of irrational numbers far exceed the whole numbers.\n\nTherefore, if you were to have all of the numbers on a number line, and you were to throw a dart, the probability of you hitting a whole number with that dart is basically zero!
880821
This is an odd question because we are treating ininity like a number. Half of something going to infinity is still going to infinity. There are ways to count different infinities -- there are countable and uncountable ones -- but dividing by 2 does not make a difference.\n\nMy advice is ignore the half infinity for this reason and treat it as a particle in free space. This is common answer which can be found in all text books. The way to approach is take a finite square well and find the condition as we increase the boundries to infinity. They will be the same answer -- namely a free particle in space.