math:is 0 a compositive or prime 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.
8564
It looks to me that the integer part means that you take the integer that comes before the given number if the given number is not an integer thus 0 would be the correct answer.
696843
Check MathForum.org:\n\nQuestion:\nWhat is the quantity 0^0 (zero to the zeroth power)?  Would it be \n0, since 0 times anything is 0, or would it be 1, since anything to the 0 power is 1?  Or is this undefined, too?\n\nAnswer:\nThis, too, is an indeterminate form.  Its logarithm is 0 * infinity.\n\nMore detail from the website:\nSay you wish to introduce a new element (or maybe two elements) infinity which you wish to append to the real number system.  That is not prohibited.  After all, that is how we got from natural numbers to integers (appending negative integers and zero), and from integers to rationals (appending ratios of integers), and from rationals to reals (appending limits of convergent sequences), and from reals to complexes (appending the square root of -1).  What you end up with is not the real number system, however.  Furthermore, if you wish to define the four operations + - * and / for this new system, you probably want them to be the same on real numbers, and just add on the definitions of things like infinity + r and r/infinity, for any real number r.  Some of these work fine.  It makes sense to define:\n\n        infinity + r = r + infinity = infinity\n  (-infinity) + r = r + (-infinity) = -infinity\n                infinity + infinity = infinity\n          (-infinity) + (-infinity) = -infinity\n                       infinity - r = infinity\n                    (-infinity) - r = -infinity\n                       r - infinity = -infinity\n                    r - (-infinity) = infinity\n             infinity - (-infinity) = infinity\n             (-infinity) - infinity = -infinity\n        infinity * r = r * infinity = infinity  for r > 0\n  (-infinity) * r = r * (-infinity) = -infinity  for r > 0\n        infinity * r = r * infinity = -infinity  for r < 0\n  (-infinity) * r = r * (-infinity) = infinity  for r < 0\n infinity * infinity = (-infinity) * (-infinity) = infinity\n infinity * (-infinity) = (-infinity) * infinity = -infinity\n                       infinity / r = infinity  for r > 0\n                    (-infinity) / r = -infinity  for r > 0\n                       infinity / r = -infinity  for r < 0\n                    (-infinity) / r = infinity  for r < 0\n                       r / infinity = 0\n                    r / (-infinity) = 0\n\nWhere we get into trouble is with defining the following:\n\n        infinity + (-infinity)\n     (-infinity) + infinity\n        infinity - infinity\n     (-infinity) - (-infinity)\n               0 * infinity\n        infinity * 0\n               0 * (-infinity)\n     (-infinity) * 0\n        infinity / infinity\n        infinity / (-infinity)\n     (-infinity) / infinity\n     (-infinity) / (-infinity)\n        infinity / 0 = infinity\n    (-infinity) / 0 = -infinity\n\nThese expressions are called "indeterminate forms."  These can all have a large range of different values, depending on exactly where the "infinity" parts came from.\n\nAs a result, the system you construct is not closed under addition, subtraction, multiplication, or division.\n\nOther indeterminate forms are 0^0 and 1^infinity.\n\nSo the answer, is 0^0 is undefined or indeterminate.
1073227
Zero is such a complicated number.   If you look at the way numbers are sequenced -3, -2, -1, 0, 1, 2, 3 etc. it would appear to be even.  There is a problem, however.  Zero came from the Arabic Hindu system and was considered a placeholder.\n\nZero is used as a place holder.  Numbering systems start with #1 and progress to 9.  1 starts over in the tens position and is followed by the placeholder 0 so we know one is in the tens position and not the single unit position.  The same is so for 20 and 30 and so on and so forth.  Hundreds, thousands, tens of thousands, hundreds of thousands, so on and so forth.\n\nNow back to the numbering sequence -3, -2, -1, 0, 1, 2, 3.  Zero is a placeholder and is neither odd or even.  It is a null value.  You have heard of zip, zilch, nada?  That is zero.\n\nZero was brought to the West by a mathematician named  Fibonacci some time in the 1200s.  It made numbers a lot easier to write.  Most numbers were written with a space where a zero would go today.  Made for lots of problems.  The numbers could only be understood in the context of the writing and could never be used by themselves.   1 2 3 4 5 6 7 8 9 1 11 12 13 14 15 16 17 18 19 2 21 22 23 etc.  These are writen in sequence so you know the 2nd 1 is really a ten.  But standing alone, it still looks like 1.\n\nThe computer age really stirred the pot.  Programing in the binary system put 0 before 1.   Auuuuuuuuuugh! as Charlie Brown would say.  The computer needed continuity in the old on/off programing in binary language.  When 0 followed 1 the computer wouldn't compute.\n \nWhen people count, it is assumed that zero (or nothing) comes first before we have something (1).  But we couldn't use it as a placeholder, because one is a single unit.  There is nothing to hold a place for.  10 is another situation.  Zero is holding an empty unit spot.\n\nYou can have the shovel now, I think I've dug a deep enough hole. (0)  Chuckle.
1139439
Prime numbers are numbers that are divisible by itself and produce only two factors..\n\nThe two factors are 1 and the number itself(other than 1)...\n\nBetween 1 to 10.. There are only 4 prime numbers..\n\n2 = 2 x 1\n3 = 3 x 1\n5 = 5 x 1\n7 = 7 x 1\n\nThe remaining numbers are not prime numbers as they do not fit the criteria...\n\n1 = 1 x 1\n4 = 2 x 2 = 4 x 1\n6 = 3 x 2 = 6 x 1\n8 = 4 x 2 = 8 x 1\n9 = 3 x 3 = 9 x 1\n10 = 5 x 2 = 10 x 1\n\n\nHope it helps to clear your suspicions...\n\n\nCheers.... (",)
943730
This is Goldbach's conjecture and has NOT been proven\ntrue or false.  Good luck solving it!!\n\nI looked at the above answers, adn they dont seem to understand the question.  You shouldl have said:\n\nAny even number greater than 2 CAN be written as the sum of two primes. Prove true or false.
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
120534
No, prime numbers are numbers that cannot be divided by anything but itself and 1. 6, 12, 24, and 36 can all be divided by 2
797722
AARGHHH!!  I withdraw my earlier attempt.  It must be dementia, I'm afraid.  It's still asking about 1000! (1000 factorial), of course, but a careful accounting is needed, not only of the effect of all numbers ending in: \n\nA. 1, 2, 3, ... 9 (I did this by a variety of ways, always coming up with a final non-zero digit of 6 for the product of all such sets), \n\nbut also of the following two other types of number, which I regrettably neglected the first time around:\n\nB. 10, 20, 30 ...90 (as often as sets of numbers like that occur), and finally, of types:\n\nC. 100, 200, 300, ... 900 (this set occurs but once).  The final, unique 4-digit number (1000) only adds three zeros at the end and doesn't affect the final non-zero digit.\n\nO.K., here goes:\n\nNumbers of Type A.\n\n1 x 2 x 3 x ... x 9 = 362880; final non-zero digit is 8.\n\nThere are 100 such "basic sets" of type A altogether.  \nMultiplying them together, set after set, we get the \nfollowing repeated cycle of non-zero ending digits \n(top row is # of basic sets used so far; bottom row is the corresponding final digit.)\n\n1   2   3   4   5   6   7   8   9 ...\n8   4   2   6   8   4   2   6   8 ...\n\nNotice that after every four basic sets have been used, we end up with 6 as the final non-zero digit.  Since there are exactly 100 such sets (a multiple of 4), we end up with that same last non-zero digit for the product of ALL numbers of type A.  (But we're not done yet.) \n\nNumbers of Type B.\n\nThere are exactly 10 sets of numbers of Type B, i.e. ending with  exactly one (but not two or three) zeros.\n\nMultiplying just these sets together, we'll have the following analogous results:\n\n1   2   3   4   5   6   7   8   9  10  \n8   4   2   6   8   4   2   6   8     4   \n\nSo, the last non-zero digit in these "10's products" is 4 for this lot.\n\nNumbers of Type C.\n\nWe have just one set of "100's".  For this, the last non-zero digit is 8 (it's just our "basic set" with a large number of zeros following it); only the last non-zero digit is of interest.\n\nSo, to summarize, the different sets we've divided our numbers into, when multiplied together, give us final non-zero digits as follows:\n\nType A:   6\nType B:   4\nType C:   8 \n\nFinally, then, the last non-zero digit of the product of everything is the last digit of:\n\n6 x 4 x 8, or 2.  (PHEW!!)\n\nWell, it's now the value of 2 (as reported by zanti3 from the 100 factorial website at:\n\nhttp://www.justinwhite.com/big-calc/1000.html)\n\n(Finish off this URL with 1000.html, instead of 1000... The Yahoo site would not reproduce the full URL from my draft.  Alternatively, just go to Yahoo or Google with 1000 factorial 1000 - this worked for me.) \n\nBoy; what a problem!  Evidently the jury is still out on the question of my impending dementia.  (See zanti3's kind words, below about my earlier fiasco.) \n\nThanks again to zanti3 for bringing the existence of a (1000 factorial) web site to my attention.\n\nLive long and prosper, everyone.
336691
Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
236680
A perfect square is the product of a whole number times itself, therefore, there is at least one whole number (its square root) which evenly factors into the number.  Since that is the case, the number can't be prime.\n\nOr you could argue the other way, a prime number is a number for which there is no other whole number (other than itself and 1) which can be a factor of it.  Since a perfect square has at least its square root as an factor, a prime cannot be a perfect square.
630729
2.Only even number to be prime
643237
2 is the only (and hence also the highest) even prime number.\n\nOther even numbers would be divisible by 2 besides itself and 1, so they cannot be prime.
1085354
This a variation on the question of whether or not there are an infinite number of twin primes (obviously there cannot be any primes between twin primes p,p+2). This is an open question. It is not even known that every even natural number is the difference of two primes even without the restriction that they be consecutive primes. Obviously, it's true for the even number 666 since you gave an example, but it is clearly a problem for sufficiently large even numbers.  The solution to this is related to the Reimann hypothesis, as are many significant number theory problems like Goldbach's conjecture: "every even number can be expressed as the sum of two primes in at least one way."  However, if Pn+1 and Pn are two consecutive primes such that (Pn+1)-Pn = 2k then there are rigorous upper bounds for the number of such consecutive primes less than a given number X, and this number is less than C(k)* X/(logX)^2, where C(k) is a computable constant depending on k.\n\nThere is an excellent book: The New Book of Prime Number Records (by Paulo Ribenboim) which is very topically comprehensive.\n\nA good link:\nhttp://mathworld.wolfram.com/topics/PrimeNumbers.html
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....
832736
No other two numbers besides ONE and itself can equal to ONE. for example\n\n7 is prime because only\n\n1*7 can equal to 7\n\n1 is prime because only\n\n1*1 can equal to one
81372
I don't mean to be rude, but none of the above answers are quite right, and some are actually quite wrong. 0/0 is not undefined, 0/0 is not always infinity, 0/0 never means "nothing" (in fact, it is used quite frequently in calculus), and although there is no general way to define 0/0, it most certainly can be defined given enough information in a problem.\n\nNormlparker had it closest, when discussing the nature of division. In other words, 10/2 means take 10 things and divide them into 2 equal groups; when you do, each of those groups will contain 5 things, so 10/2=5. Dividing by 0 doesn't conceptually make sense. You can't take 10 things and divide them into no groups. That answer is UNDEFINED. (Captial letters for vocabulary emphasis, not because I am yelling.)\n\nNow dividing 0 by 0 is a different issue, because you can surely take no things and make them into no groups. However, what's the size of those nonexistent groups? It's not infinity, nor is it undefined, as some previous posters have said. Dividing 0 by 0 is an INDETERMINATE value, and it's exactly as you said--it could equal just about anything.\n\nTake for instance the fraction (x^2+2x-3)/(x-1). If you plug x=1 into the fraction, you get 0/0, but the function is definitely not infinity when you get close to x=1. If you factor the numerator, the fraction becomes [(x+3)(x-1)]/(x-1). Cancel those common factors and you get x+3. In other words, this fraction (although undefined at x=1) gets very close to x+3=1+3=4. In this case, 0/0 is basically 4.\n\nTo be specific, a non-zero number divided by 0 approaches infinity (or negative infinity), but 0/0 can approach lots of things. However, ANYTHING divided by 0 is not a real number. \n\nWhen you took your equation (0)(0)=0 and divided both sides by one of the 0s on the left side, you actually violated a mathematical principle. Sure you can divide both sides of an equation by any number, as long as you are certain that number is not 0.\n\nSome of the most famous mathematical fallacies are based upon divideng by 0 in a sneaky way. You can prove lots of untrue things (such as 1=2) if you do that, and it's actually kind of fun. I used to do that once a year in my geometry classes to "prove" to my students all math was worthless, and as soon as they got excited, I showed them how I fooled them. They usually didn't care so much, but nerds like me think that kind of stuff is hilarious.\n\nHere are some other indeterminate values that can equal anything: (0)(infinity), 1^(infinity), (infinity)/(infinity).