In the 9th century, Mahavira declared that n - 0 equals n.  Explain why this makes sense!?
449990
Think about the number bar... if you start at 'n' and then move 0 distance from it toward the left, you are still at 'n'.\n\nFor 0-n, you would start at 0, and move 'n' distance toward the left, so 0-n = -n ... the only case where 0-n=n, is when n=0.
283236
20/10 = 2 because 2 x 10 = 20\nn/0 = n because n x 0 = n? No!!!\nSo n/0=n is wrong.
613146
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
937215
okay so what N means is that you are plugging in any integer value for n. integers, as a reminder, a whole, countable numbers like 1, 6, 88, 45798.\nSo teh first four integers (values of n) are obvious 1, 2, 3, 4\nso the sequence is\n(2*1 -5), (2 *2 -5), (2 *3-5), (2 *4-5)\n-3, -1, 1, 3.\n\nHere's one way to prove that it's arithmetic:\nYou can graph them as continuous variables along an x y coordinate system. That is, graph the points (1, -3), (2, -1), (3, 1), and (4, 3). If you connect them, you'll get a straight line. It's no surprise that the equation of that line (if you choose to find it) is y=2x-5. Because the points lie on a straight line, it's arithmetic (as opposed to geometric).
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
272752
One way to understand this is to think through the meaning of a divided by b. It is a number such that when you multiply by b, you get a. \n\nThus 1 divided by 0 is undefined, because there is no number such that when you multiply it by 0, you get 1.\n\nIn the same way 0/1 is equal to 0, as 1 times 0 is 0.
74602
The Fibonacci numbers form a sequence defined recursively by:\n\nF(n)= 0 if n = 0;\n      1 if n = 1;\n      F(n-1)+F(n-2) if n > 1.\n\nIn other words: one starts with 0 and 1, and then produces the next Fibonacci number by adding the two previous Fibonacci numbers. The first Fibonacci numbers for n = 0, 1, … are\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, …\n\nThe earliest known reference to Fibonacci numbers is contained in a book on meters by an Indian mathematician named Pingala called Chhandah-shastra (500 BC). As documented by Donald Knuth in The Art of Computer Programming, this sequence was described by the Indian mathematicians Gopala and Hemachandra in 1150, who were investigating the possible ways of exactly bin packing items of length 1 and 2.\n\nAs was pointed out by Johannes Kepler, the ratio of consecutive Fibonacci numbers, that is:\n\nF(n+1)/F(n),\n\nconverges to the golden ratio φ (phi).
1096569
The modular (and clock) arithmetic answers are correct, but I think they miss the point a bit, since in any modular arithmetic you still have 2+2 = 4 mod n. Let's suppose we are using the natural numbers with the usual addition.\n\nThe recurring decimal answers are just stupid, 1 = 0.9999... recurring (as shown in fact by the first answer here), so all of these answers are either irrelevant or wrong.\n\nThe '2' is just a symbol arguments possibly have some merit, but you should be careful. In Indianna in 1872 (or thereabouts) it was ruled by State legislature, that the value of pi is 4. Obviously this creates problems, if you go to Indianna you won't find many buildings that were designed in 1872. (Their mistake of course was that they didn't alter the representation of all the other numbers they where using accordingly).\n\nI'm not really a logician so I don't know the answer to this off the top of my head, I'm pretty sure that we don't have a satisfying answer here (sorry if we have but I missed it). \n\nPerhaps a good place to start looking for a serious discussion on this kind of topic would be to understand what a Peano system is (wiki it, I'm not explaining that here, 'tis an axiomatic description of the natural numbers). I've also heard that Bertrand Russell gave a proof that 1+1 =2 in Principa Mathematica, although I haven't read that and I don't know precisely where you can find it.
252229
Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
443911
I do not think you can.\nSo Bhaskara tried to solve the problem by writing n/0 = ∞. At first sight we might be tempted to believe that Bhaskara has it correct, but of course he does not. If this were true then 0 times ∞ must be equal to every number n, so all numbers are equal. The Indian mathematicians could not bring themselves to the point of admitting that one could not divide by zero. Bhaskara did correctly state other properties of zero, however, such as 02 = 0, and √0 = 0.
380829
Actually the Hindus started using the decimal numbering system, 0 to 9, 1500 years ago.  after 9 you add 1 infront to make 10, 12, 13, etc.
1137788
Nope, 0/0 could be 1 but it could also be anything else.   Division is the opposite of multiplication and 0 times anything at all, even infinity, is still zero.
449075
It can, sometimes:\nyou know n x 0 = 0\nso, if n x 0 = n\nthen 0 = n.
1060197
Grappling with the axiomatic logic is not suitable for most of\nus and these readings are difficult.  Let me attempt an inadequate simplifi-\ncation.  I would say a number "belongs to a concept" and is an\nextension of that concept and then statements about numbers correspond to\n"identities of concept".\n\nIn the sense of counting, one apple and one orange represent identical\nconcepts even if apples are not oranges, so we write 1 = 1.  Now 1 + 1 or\nfor that matter 1 + 1 + 1 + 1 + 1 + 1 (repeating)  are represented in the\nsense of identity by other symbols which, if arabic numerals are used, we\nwrite as 2 or 5.  The symbols are just compact notation for representation\nof an identical "concept".  I would say that number is not anything\nphysical nor is it subjective.  This may not be very satisfying but deep\nphilosophical questions about numbers may always remain unanswered in very\nsatisfying ways.
293196
Easy answer: It's defined that way to make a lot of things work right.  For example, let's say I ask you how many ways are there to choose a group of 9 people out of 9.  If you assume that order doesn't matter (i.e. a combination), then the answer is given by 9!/(9! 0!).  But there is only one way to choose a gorup of 9 out of 9 (choose everyone!), so this expression should equal 1.  the only way that can happen is if 0! = 1.\n\nThere is another more detailed reason using gamma functions, but I won't go there...\n\n\nSummary: Because it makes all your nice probability formulas true when 0! shows up.
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+