Is there a simple proof for 0! = 1?
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.
853846
Do not write x = 0 because think about it\n\nif x = 0, then that would mean that |0+4| + 4 = 0\n\nthat means 8 = 0... is that true? NO!\n\nSince it is imposible for |x - 4| to equal -4 (since absolute value is ALWAYS possitive or 0) there is no solution =)
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
72221
This is just a sneaky way of getting back to your question\non "0!".\n    This is the real answer.\n The definition of n! is a recursive one. It has a recursive\npart and a base. The recursive part is:\n                n!=n*(n-1)!\nThe base is:    1!= 1\nSubstituting 1 for n in the recursive part we get\n                1!=1*(1-1)!   The left side is defined as 1 and simplifying the right side we get\n                1=1*0!  and of course 1 times anything is\nitself so we get  1=0!  \n     Just thought you should know.
637914
1=1. Simple yet important.
287019
1 + 1 = 2\none plus one = oneone
999313
0^n = 0, where n is not equal to zero\nn^0 = 1, where n is not equal to zero\n0^0 = undefined. why?\n\nExpress the exponent 0 into n - n (n is not equal to zero)\n0^0\n= 0^(n - n)\n\nUse the property b^(x + y) = b^x · b^y\n= 0^n · 0^-n\n\nUse the properties 0^n = 1 and b^-x = 1/b^x\n= 1 · 1/0^n\n= 1/0\n= undefined\n\n^_^
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.
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
467241
Add zero to anything and you get the same number you started with. It does not matter at which number you start with add 0 to it, it will still be zero. For example, you can start with 0 and add any number (n), lets say, you will still get that number n. Thus the notation for the previous example would be [0+n]. Ex. [0+1]=1 and again in the notation that you are asking about [0+n]=n, in the example above i just "substituted" 1, or just let [n] that represents any number for instances equal one. \n\nThe same goes for [n+0]=n, let [n] equal 1. add 1 to zero. What do you get? You should get one. see they (n+0 and 0+n )are the same thing.
574280
a number to the zero exponent is always 1 as a answer
41284
Wait, all the answers are wrong\n\nx^2+4>4\nThen x^2-4x+4>0\nthen (x-2)^2>0\nBut hey, (x-2)^>0 is always true expect when x=2 which equals zero. An square is always positive or 0.\nSo the answer will be any value but 2 or R-{2}\nthese are the evidence:\nPut them in original non equation\nx=0 then 4>0 which is true\nx=-1 then 1+4>-4 so 5>-4 which is true\nx=3 then 9+4>12 so 13>12\n but at x=2 then 8>8 which is incorrect
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
981541
No, it follows from 1+1 = 2\n\nWhich follows from 1+0 = 1\n\nBut that can be considered an amazing "fact".  In reality, it's an axiom - a definition.  One of the only true defintions that form the basis of mathematics.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
1185063
(0! +0! + 0! +0! +0!)!
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?