If n x 0 = 0  Why can't n x 0 = n?
449075
It can, sometimes:\nyou know n x 0 = 0\nso, if n x 0 = n\nthen 0 = n.
111021
If you are sloving for X then anything greater than 2 will work. basically it is 6 time what (X) will be greater than 12. Well 6x2=12 so that won't work.
55379
I would say, \n\nonly in the case when n=0 or 1, then this equation stands good! If n is not equal to zero or one, then everytime the answer would definitely be above 1 (even when u use negative integers)
189562
This is not simply a quirk of our usual number system: in any number system that satisfies certain basic properties such as distributivity and associativity, 0 multiplied by anything will give 0. So if you start with any non-zero number x, there cannot be any number which when multiplied by 0 gives you x, so there can be no answer to the question "what is x divided by 0". Mathematicians say that "division by 0 is undefined", meaning there is no way to define an answer to the question in any reasonable or consistent manner.
187658
it's defined \n\n0^0=0
541628
One of the factors will be x-x=0.\nSo, the answer is 0.\nThis is because 0 times any number is 0.
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.)
404608
that is incomplete because you are missing one parenthesis. But x=0 is a correct answer.
905755
you must include x=0.
680886
You can't because it does!\n\nFirst assume x = 0.999... repeating\n\nx = 0.999...\nx = 0.333... + 0.333... + 0.333...\nx = 1/3 + 1/3 + 1/3\nx = 3/3\nx = 1\n\nSo 0.999... = 1\n\nThis is a valid proof with no faulty logic, division by zero etc.\n\n0.999... *does* equal 1\n\nAnother way to prove this is as follows:\n\nx = 0.999....\n10x = 9.999...\n10x - x = 9.999... - 0.999...\n9x = 9\nx = 1\n\nFor the person that says subtract from 1 and you get 0.0000000001, you don't.  You have to have an infinite number of zeros, not a finite set.  You could write this number as 10^-n where n = infinity.  This is also 1 / 10^n where n = infinity.  The lim as n->infinity of 10^-n is zero, similarly the lim as n->infinity of 1 / 10^n is also zero.  So you are adding zero.  This therefore another proof that 0.999... *equals* 1.\n\nAnd if you *still* don't believe me, read the Wikipedia entry below.
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
148466
You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,
556722
look at my question\n\nBidDogARRRRR...\n1 minute ago\nDoes anyone know 0=1 and anything = anything else?\na = 1; b = 0\n1. a = b + 1 \n2. (a-b)a = (a-b)(b+1)\n3. a2 - ab = ab + a - b2 - b \n4. a2 - ab -a = ab + a -a - b2 - b\n5. a(a - b - 1) = b(a - b - 1) \n6. a = b\n7. b + 1 = b \n8. Therefore, 1 = 0
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.
2302
(1) = x\n(2) = / \n(3) = x
835878
There's a better way of looking at it.\n\nFor every real number x in [0,1) you define D(x, n) as the nth decimal of x.  This is defined for every natural number n.\n\nFor x=1/3, we have D(1/3, n) = 3 for every n.  The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3.  It is EXACTLY the number 1/3.  Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...\n\nFor 0, we have D(0,n) = 0 for every n.  There is never a '1' decimal.  If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p.  I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1.  That number is greater than zero and smaller than the number that you gave me.  \n\nThe 0.5 = 0.49999... thing is just a quirk of decimal notation.  We can formally prove the equality as follows:\n\nLet a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.\nLet b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.\n\nProof by contradiction that a=b: Let's suppose on the contrary that a > b.  This means there must be a number x between a and b.\n\nLet's figure out the digits of x.  Since x is less that a=0.5, its first decimal must be a 4.  D(x,1) = 4.\n\nThe second decimal of x must be 9 since x > b `> 0.49.\nThe third decimal of x must be 9 since x > b >0.499.\nAnd so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.\n\nTherefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b.  Contradiction, ergo a = b.\n\nThat's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0.  In the latter case, it is possible to find a number that is between the two.