morgsboi Posted May 20, 2012 Posted May 20, 2012 If asked if -0 =0 one would say yes. But I don't think that is the case. It is quite simple. If you divide a number by 0 your answer would be infinity. So if you divided a number by -0 you must have an answer of -infinity. So somewhere in 0, it must have a value but how can it? There symbol represents that there is no value to it. Could it be that 0 doesn't exist? Well that is impossible for it not to exist because one could say, "What is the answer to 10 - 10"? Is 0 a paradox or is it infinity and an infinitesimal at the same time?
DJBruce Posted May 20, 2012 Posted May 20, 2012 If you divide a number by 0 your answer would be infinity. So if you divided a number by -0 you must have an answer of -infinity. Dividing by zero is not defined. Is 0 a paradox or is it infinity and an infinitesimal at the same time? If you are asking the mathematical question what is zero, and looking at the rational or real numbers as fields -- i.e.: a place where you can add, multiple, have multiplicative and additive inverse, have additive and multiplicative identities, and other "natural" properties -- then 0 is simply the additive identity. If you are asking the philosophical question whether zero exists, then I don't think mathematics really has an answer for you, other than the fact that many mathematical objects have nice properties if we assume they have a "zero".
Greg H. Posted May 20, 2012 Posted May 20, 2012 (edited) Dividing by zero is not defined. Actually, despite all those math classes we all took, in floating point calculations (based on the IEEE 754 standard) division by 0 is only undefined for [math]\frac{\pm0}{\pm0}[/math] and [math]\frac{\pm\infty}{\pm\infty}[/math] . It also (apparently) has some applications statistical math. You can read more at: Signed Zero (Wikipedia) Some of the salient bits: The IEEE 754 standard for floating point arithmetic (presently used by most computers and programming languages that support floating point numbers) requires both +0 and −0. The zeroes can be considered as a variant of the extended real number line such that 1/−0 = −∞ and 1/+0 = +∞,division by zero is only undefined for ±0/±0 and ±∞/±∞. Negatively signed zero echoes the mathematical analysis concept of approaching 0 from below as a one-sided limit, which may be denoted by x → 0−, x → 0−, or x → ↑0. The notation "−0" may be used informally to denote a small negative number that has been rounded to zero. The concept of negative zero also has some theoretical applications in statistical mechanics and other disciplines. It is claimed that the inclusion of signed zero in IEEE 754 makes it much easier to achieve numerical accuracy in some critical problems,[1] in particular when computing with complex elementary functions.[2] On the other hand, the concept of signed zero runs contrary to the general assumption made in most mathematical fields (and in most mathematics courses) that negative zero is the same thing as zero. Representations that allow negative zero can be a source of errors in programs, as software developers do not realize (or may forget) that, while the two zero representations behave as equal under numeric comparisons, they are different bit patterns and yield different results in some operations. Edited May 20, 2012 by Greg H. 1
timo Posted May 20, 2012 Posted May 20, 2012 Someone should inform the real numbers about this floating point representation standard so that they can adapt to it.
morgsboi Posted May 20, 2012 Author Posted May 20, 2012 Dividing by zero is not defined. Yes it is according to my simple calculation. infinity/ 1 = infinitesimal or 0.000recurring 1 So the inverse of that is infinity Proof of infinitesimals being rounded can be proved by this. x = 0.999recurring 10x = 9.999recurring 10x - x = 9x 9x = 9 9x/ 9 = 1 x = 1 It's something I have worked out myself but it has been confirmed. Apparently it was actually some of the early work of Sir Issac Newton!
Bignose Posted May 20, 2012 Posted May 20, 2012 Someone should inform the real numbers about this floating point representation standard so that they can adapt to it. Yes, it is very interesting that a role reversal has occurred. Floating Point was invented in an attempt to mimic the real numbers as best possible by a computer. But, now we see people taking the computer representation as sacrosanct. There is an early Simpson's episode that makes fun of this: Mrs. Crabapple: "Whose calculator can tell me what six times seven is?" Millhouse: "Oh! Oh! Oh! Low Battery!" The point being that we have to remember that a computer is a tool -- an imperfect one at that. Imperfect in that unless care is taken in the programming, round off errors will occur. It is also imperfect in that the programming could simply be in error. Just trusting what a computer outputs is generally a poor idea. Especially when trying to look at questions that mathematics has left indeterminate and incalculable. Just because a computer spits out an answer, doesn't mean that it is a right answer. And why anyone would think that an approximation of the real number line should actually answer some of the unanswerable questions about the real number line is beyond me. 1
Janus Posted May 20, 2012 Posted May 20, 2012 infinity/ 1 = infinitesimal or 0.000recurring 1 So the inverse of that is infinity Proof of infinitesimals being rounded can be proved by this. x = 0.999recurring 10x = 9.999recurring 10x - x = 9x 9x = 9 9x/ 9 = 1 x = 1 It's something I have worked out myself but it has been confirmed. Apparently it was actually some of the early work of Sir Issac Newton! 0.000...1 approaches zero as a limit, but cannot equal zero the way that .999... equals 1 It requires that there be a final "1" in the series and this always causes the number to be just a bit larger than zero. For the same reason, 0.999...9 does not equal 1, because 0.999...9, is not equal to 0.999... 0.999... does equal 1, but 0.999...9 does not. In turn: 1/0.000...1 does not equal infinity but instead 10000...0, some arbitrarily large finite number. Again, 10000...0 is not the same as 10000... 1/0 is undefined, no matter what you seem to think.
studiot Posted May 20, 2012 Posted May 20, 2012 (edited) 1/0 is undefined, no matter what you seem to think. Maybe but what about [math]\frac{{1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8}. - ....}}{{10 - 9 - 0.9 - 0.09 - ....}}[/math] or [math]\frac{{3*4*5*6*.....}}{{2*3*4*5*6*7.....}}[/math] Edited May 20, 2012 by studiot
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 Haha, I've just realized I've been beaten to it..... again! Look at the computer arithmetic section. http://en.wikipedia.org/wiki/Division_by_zero
uncool Posted May 21, 2012 Posted May 21, 2012 0.000...1 approaches zero as a limit, but cannot equal zero the way that .999... equals 1 0.000...1 doesn't exist in the real numbers. It simply doesn't make sense. =Uncool-
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 0.000...1 doesn't exist in the real numbers. It simply doesn't make sense. =Uncool- Yes, that is what's known as an infinitesimal.
John Posted May 21, 2012 Posted May 21, 2012 As a note, 0.999... = 1. As for the rest, as others have noted, division by zero isn't defined and standards for handling computer arithmetic are based on the limitations of computer hardware and don't overrule mathematical theory. Dividing by zero is not defined. Yes it is according to my simple calculation. infinity/ 1 = infinitesimal or 0.000recurring 1 So the inverse of that is infinity Proof of infinitesimals being rounded can be proved by this. x = 0.999recurring 10x = 9.999recurring 10x - x = 9x 9x = 9 9x/ 9 = 1 x = 1 It's something I have worked out myself but it has been confirmed. Apparently it was actually some of the early work of Sir Issac Newton! You can't treat infinity as just another number, because it isn't. As discussed in a recent thread, there are instances in which infinity is added to the real number line, but even then infinity isn't a number and has to be treated carefully. Relevant Wikis: http://en.wikipedia....ivision_by_zero http://en.wikipedia.org/wiki/.999...
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 As a note, 0.999... = 1. As for the rest, as others have noted, division by zero isn't defined and standards for handling computer arithmetic are based on the limitations of computer hardware and don't overrule mathematical theory. You can't treat infinity as just another number, because it isn't. As discussed in a recent thread, there are instances in which infinity is added to the real number line, but even then infinity isn't a number and has to be treated carefully. Okay, I'm only here to learn. Please point out where I went wrong in my mathematics.
timo Posted May 21, 2012 Posted May 21, 2012 The first error is in the statement "If you divide a number by 0 your answer would be infinity". You're welcome.
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 (edited) The first error is in the statement "If you divide a number by 0 your answer would be infinity". You're welcome. Please.... tell me why. If x = 0: x/x = 1 2x/x = 2 3x/ x = 3 x/0 = infinity x/-x = -1 2x/-x = -2 3x/-x = -3 x/-0 = -infinity Edited May 21, 2012 by morgsboi
John Cuthber Posted May 21, 2012 Posted May 21, 2012 Please.... tell me why. If x = 0: x/x = 1 2x/x = 2 3x/ x = 3 x/0 = infinity x/-x = -1 2x/-x = -2 3x/-x = -3 x/-0 = -infinity Why? What's the point of telling you again? You plainly didn't pay attention when DJBruce told you in the second post in this thread. You ignored it again when Janus pointed it out. Still, just in case the thrid repetition gets the message through to you: Division by zero is not defined. 1
Greg H. Posted May 21, 2012 Posted May 21, 2012 (edited) Please.... tell me why. If x = 0: x/x = 1 2x/x = 2 3x/ x = 3 x/0 = infinity x/-x = -1 2x/-x = -2 3x/-x = -3 x/-0 = -infinity In pure mathematics, because it gives you nonsense answers, such as the following: It's only useful is certain applications in computer languages to handle, as I pointed out before, floating point math correctly. Youc an read more about it here: http://en.wikipedia.org/wiki/Mathematical_fallacy Edited May 21, 2012 by Greg H.
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 Where is your evidence??? You can't expect it just to be that without evidence! In pure mathematics, because it gives you nonsense answers, such as the following: Just to point out, this is where you went wrong.
Greg H. Posted May 21, 2012 Posted May 21, 2012 Where is your evidence??? You can't expect it just to be that without evidence! First, use the quote function properly. It's annoying when you don't. Second, I am glad you were able to spot the flaw in the math problem. Did you understand the point I was trying to make? (I'm guessing not, but hope is a virtue, even when it chances to be misplaced.)
morgsboi Posted May 21, 2012 Author Posted May 21, 2012 First, use the quote function properly. It's annoying when you don't. Second, I am glad you were able to spot the flaw in the math problem. Did you understand the point I was trying to make? (I'm guessing not, but hope is a virtue, even when it chances to be misplaced.) I deleted a line too many. And I see where you are coming from but algebraically I put it a number of different ways which should work. Since I showed where your one went wrong, please show where mine has gone wrong.
Greg H. Posted May 21, 2012 Posted May 21, 2012 I deleted a line too many. And I see where you are coming from but algebraically I put it a number of different ways which should work. Since I showed where your one went wrong, please show where mine has gone wrong. I knew mine was wrong before we started - I was using it to illustrate the reason division by zero is considered undefined. The fact that you didn't grasp the point behind my example leads me to two possible conclusions. A) You're being deliberately obstinate. B) You really don't understand what we're telling you. While I hope the answer is B, your continued refusal to accept even simple examples that you're incorrect leads me to believe my hope is in vain.
studiot Posted May 21, 2012 Posted May 21, 2012 Since I don't agree with a blanket statement that division by zero is unavailable I showed how to do it in certain circumstances. So I invite all the naysayers to disprove this. morgsboi has point even if only a partial one.
Greg H. Posted May 21, 2012 Posted May 21, 2012 Since I don't agree with a blanket statement that division by zero is unavailable I showed how to do it in certain circumstances. So I invite all the naysayers to disprove this. morgsboi has point even if only a partial one. I don't disagree that it's available in certain circumstances. I even posted a link that showed where it has a practical application. However, that doesn't mean that it's generally available for use in all math, and outside of those proper contexts, writing anything in a div-zero format is going to get you some funny looks at best.
Daedalus Posted May 21, 2012 Posted May 21, 2012 As I stated in a thread started by Morgsboi that was moved to the speculations forum: This is why division by zero is undefined: Given that [math]z=\frac{y}{x}[/math] assigns to each point ([math]x[/math], [math]y[/math]) a real number [math]z[/math]. Then the limit[math], \ \lim_{(x, y) \to (a, b)} \frac{y}{x} \ ,[/math] can only exist if all paths in the domain, [math]\{(x,\, y) \in \mathbb{R}^2\}[/math], approach the same value. If we choose multiple lines (paths in the domain) of the form [math]y=m\, x[/math], then as [math]x[/math] approaches zero [math]y[/math] also approaches zero such that: [math]\lim_{(x, y) \to (0, 0)} \frac{y}{x} \ =\ \lim_{x \to 0} \frac{m\, x}{x} = m[/math] Notice how the value of the limit depends on the direction at which you approach zero - different values of [math]m[/math]. Since we get different values for the limit as [math]x[/math] and [math]y[/math] approach zero, the limit does not exist. Hence the reason why division by zero is undefined.
studiot Posted May 21, 2012 Posted May 21, 2012 (edited) As I stated in a thread started by Morgsboi that was moved to the speculations forum: Yes I saw that and I think the threads could profitably be combined. However you have avoided my point that in certain circumstances the ratios zero/zero and infinity/infinity can be evaluated. Further there was no stated requirement to remain within either the reals or the complex numbers. Both the limits in the series ratios I displayed exist and are zero or infinity. Edited May 21, 2012 by studiot
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