justification of 0!=1. why it become 1?
88833
Math is really a system of agreed upon rules.  The concept of factorial is DEFINED like this:\n\n0! = 1\nn! = n * (n-1)!\n\nWhoever came up with the idea could have defined it any way he or she wanted.  They could have said that it's not defined for zero (just like it's not defined for negative numbers).\n\nSuppose they had defined 0! = 0.  Then for any number n, n! would be 0.  That would not be a very interesting concept.
161780
There is a simple mistake in your logic.\nYou want to say you will add x , x times. As you want to add it x time you are making it constant.( Because when you are diffrentiating, x can't change)\nwhen you diffrentiating const, all will be 0.\nso you have proved 0=0.
832515
(1) any of the following may be the root of the expression,+1,-1,+2,-2,+3,-3,+6 or -6.\nNow putting the value of x=-1,see ifthe value of the expression becomes zero or not?If it becomes zero,then x=1 is one factor of the expression,if it does not become zero try with =1,then -2.+2 and so on.\nLuckily for us,with x=-1,the value of the expression becomes zero\nTherfore,x+1 is one factor\nx^4++x^3-7x^2-13x-6\n=x^3(x+1)-7x(x+1)-6(x+1)\n=(x+1)(x^3-7x-6)\n=(x+1){x^2(x+1)-x(x+1)-6(x+1)}\n=(x+1)(x+1)(x^2-x-6)\n=(x+1)^2(x-3)(x+2)\n(2)Will you please check up the second one once more and if there is any correction send it to me through Yahoo answers.
312174
It seems undefined since for x <= 0 and x != odd number, x^x doesn't work out so well.  If it's x->0+, then it's just 1.
149171
1,365\n\nor 15!/(11! * 4!)\n\nie n!/[k! * (n-k)!]
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
388398
i think the false statement is C. as the degree of any number is always 1. i dont understand the statement b tho :S
89444
The answer is "i" because there is no number, that when mutliplied by itself, will give you -1.  Think about it: the only possible combination that would make sense is 1 X 1, which equals 1...not -1.
997321
Division by 0 is undefined. \n\nIf you divide 4 by 2, you get 2. Reverse the process: 2 multiplied by 2 gives you 4.\n\nThe reason you can't divide by 0 is because you can't multiply 0 by any other number and get a number besides 0.\n\n4 divided by 0 gives you "undefined." That would mean that "undefined" times 0 should equal 4. But that's silly.
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
18903
There is no number that will multiply by zero to not equal zero because for example 5x5 would be the quantity of five added five times, 5x1 would be the quantity of five added one time, and 5x0 would be the quantity of five added zero times meaning there is nothing there or, if you prefer, that the quantity (Five) is not addded at all.\n\nBTW- even infinity multiplied by zero equals zero\nBTW- There is NO NUMBER!!! that will multiply by zero to to equal anything but Zero. Not Infinity or undefined!! 0x0=ZERO Infinity x 0=ZERO, ETC.
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
306774
Anything divided by 0 is actually undefined, not infinity or 0. When you think about how that works, it makes sense:\n\nDividing a numerator by smaller numbers makes your number larger. Divide 4 by 2, you get 2. Divide 4 by 1, you get 4. Divide 4 by 0.5, you get 8. And so on until you are dividing by a number so small it's pretty close to 0. The overall number gets larger and goes off into infinity.\n\nBut, if you divide a negative number in the same manner, your number gets smaller. -4 by 2 is -2, -4 by 1 is -4, -4 by 0.5 is -8, etc. As you approach 0, your numbers go off into negative infinity.\n\nYou'll have the same weird effect if you divide by negative numbers - a positive divided by an increasingly smaller negative will yield negative infinity, and a negative divided by an increasingly smaller negative will yield positive infinity.\n\nYou can't have it both ways - dividing by zero yields both a positive and negative infinity! So someone a long time ago decided that since a number can't really have two values, it will be undefined.
596523
If you are approaching zero from the left, then -1 divided by 0 is positive infinity.\nIf you are approaching zero from the right, then -1 divided by 0 is negative infinity.\n-1 divided by 0 is undefined because you get different results when coming in from the right side of 0 and the left side of 0.\nI learned this in Calc 1 (possibly Calc 2).\nThere are different levels (degrees) of infinity.
314173
(n+1)!=(n+1)(n)(n-1)(n-2)...\n  n!  =(n)(n-1)(n-2)...\n\n(n+1)!/n! = n+1\n\n\n(n+2)!=(n+2)(n+1)(n)(n-1)...\n  n!  =(n)(n-1)(n-2)...\n\n(n+2)!/n! = (n+2)(n+1)
523398
yes it is.\n\ni forget why, though.\n\nfactorials is basically taking the number that they give (i.e. 5!) and just multiplying it by every positive integer below it (5*4*3*2*1)\n\ni guess 0! is 1 because you never multiply 0, so it has to be 1?
812449
1/n! - 1/(n+1)! =  (n+1)!/[(n!)(n+1)!] - n!/[(n!)(n+1)!]\n                        = 1/[(n!)(n+1)!] X {(n!)(n+1-1)}\n                        = n*n!/[(n!)(n+1)!]                                 (divide by n)\n                        = n!/[(n-1)!(n+1)!]                  (divide by n!)\n                        =  1/[(n-1)!(n+1)]