How you will going to prove that 0=1?
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.
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
89742
x and y are real numbers...if x+y=0,then how do i prove that x=-y?\nJust transfer the y variable to the right side of the first equation.\nx=0-y.\nAnd then you can just take out the zero.\nAnd you get x=-y
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
287019
1 + 1 = 2\none plus one = oneone
144204
-10\n\nand to prove it -10 x -10 x -10 = -1000
200150
for all of those, all u do is set once variable at a time equal to zero. for instance..\n\n-x+0=2\n-x=2  <--- multiply x and 2 by -1\nx=2\n\nthen..\n\n0+y=2\ny=2
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.
654517
x^2-2x+1\n(x-1)(x-1)\nso zero at x=1\n\nx^2-x=6\nx^2-x-6=0\n(x-3)(x+2)=0\nzero at x= 3, and x=-2\n\nx^2-3x-4\n(x-4)(x+1)\nzero at x=4  or x=-1\n\nx^2+8x+15\n(x+3)(x+5)\nzero at x=-5 or x=-3\n\nx^2-6x+5\n(x-5)(x-1)\nzero ar x=5, or x=1\n\nx^2+9x+20\n(x+4)(x+5)\nso zero at x=-4 or x=-5
66821
You can't divide anything by zero, so the answer mathematicians give is that 1/0 is "undefined". \n\nIt is very tempting to think that 1/0 = infinity but there is a consequence...\n\nIf 1/0 = infinity, and 0 times infinity = 1, then what is 2/0? Infinity? \n\nIf we say "yes", then we have to say that 0 times infinity = 1 AND also = 2 (and by implication also 3, 4, 5...). \n\nBut remember that anything times 0 is 0. It only takes the simplest of arithmetic to imagine an empty apple box, and to say "you can ask me for as many as you like - even infinity - but I can only give you zero apples"
678157
2x^2(1-2x)=0\n\nx=0\nx= (1/2)
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
882621
My old math teacher showed me this.\n\nSince A=B, then A-B=0, and you can't divide by 0
1032089
discovery of the number 0?
469247
A = A\n\nThis is one.
163896
yes the previous equation mentioned would "prove" it, but the problem occurs as you're dividing by zero.  you can also make 1=2, with a similar process, involving the division of zero.  if you don't get what i mean, the previous explanation has this:\n\n(x-y)(x+y)=y(x-y)\nx+y=y\n\nwhat happens at this step?  you are dividing by (x-y).  since x=y is given, then (x-y)=0, thus you are dividing by 0, which is impossible in math.
717483
1\nbecause u have mentioned that 1=5 so 5=1 and it cant = 9