rktpro Posted April 15, 2012 Share Posted April 15, 2012 I came across a formula derived by Ankur Tiwari, which he says enables division by zero. The website claims This formula enables us to divide in a unique way without using denominator. This formula is based on the principle that, If the value of X divided by Y (X/Y) is A than by using this formula we can find out A without dividing X by Y directly, that means without dividing X by Y we can find out its value. This is the reason why ‘Bhartiya New Rule for Fraction’ is capable of diving by Zero. The interesting points in regard of this formula are :- 1.‘Bhartiya New Rule for Fraction’ is based on present phenomenon and rules of mathematics. 2. It is very simple and easy formula. 3.Greatest benefit of ‘Bhartiya New Rule for Fraction’ is that it is capable of dividing by Zero and giving its value as an integer. 4.‘Bhartiya New Rule for Fraction’ can be used to find out the value of four not defined trigonometric ratios tan90, cosec0, sec90, cot0. So that these values can be utilized in the field of astronomy and other fields related to mathematics. 5.If in place of simple division (X/Y), ‘Bhartiya New Rule for Fraction’ is used in any digital electronic device as its processing command for division in processor, it will results in permanently elimination of ‘divide by Zero’ error from that device. How would it affect mathematics? Is there a fallacy in there? Let us discuss... Link to comment Share on other sites More sharing options...

khaled Posted April 15, 2012 Share Posted April 15, 2012 (edited) The above photo was taken from his website, which shows the formula he use .. but is it valid analytically ? does the above formula equal X / Y ? Edited April 15, 2012 by khaled Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted April 15, 2012 Share Posted April 15, 2012 The above formula approximates X / Y by taking the average of the division at points to the left and to the right of Y. It's a very informal limit. 1 Link to comment Share on other sites More sharing options...

Bignose Posted April 15, 2012 Share Posted April 15, 2012 And it is clearly going to be dependent upon the 'calculating device'. i.e if you program the formula using single precision real numbers, you'll get a different number than if you used double or quad precision reals. I think that this is awfully unsatisfactory for defining a value for [math]\cot 0[/math] for example. Depending on if you used your calculator or Excel, you'll get a different answer. I also am not so sure why it is so hard for people to accept that the answer is undefined. Link to comment Share on other sites More sharing options...

khaled Posted April 15, 2012 Share Posted April 15, 2012 (edited) I've implemented his method, it works for most integers, especially positive numbers I've noticed that 0/X, X/0, and 0/0 all result in 0 The code: #include <stdio.h> // assume our virtual machine support 8 decimal places #define DBL_8_MIN 0.00000001 /* min 8-decimal-places positive number */ double f (int y) { return ((double)y - (double)DBL_8_MIN); } double g (int y) { return ((double)y + (double)DBL_8_MIN); } double bhartiya (int x, int y) { return (x/2.0) * (1.0/f(y) + 1.0/g(y)); } int main () { int x = 1; int y = 1; int u = 0; int v = 0; while (x < 999) { y = 1; while (y < 999) { u = bhartiya (x, y); v = ( x / y ); if (u != v) { printf(" %d/%d = %d .. but Bhartiya gives %d with y'=%.8lf & y''=%.8lf \n",x,y,v,u); } y++; } x++; } return 0; } Edited April 15, 2012 by khaled Link to comment Share on other sites More sharing options...

The Observer Posted April 16, 2012 Share Posted April 16, 2012 Have a look at the graph of 1/x and tell me what 1 divided by zero is. Link to comment Share on other sites More sharing options...

doG Posted April 16, 2012 Share Posted April 16, 2012 I've implemented his method, it works for most integers, especially positive numbers I've noticed that 0/X, X/0, and 0/0 all result in 0 IMO I'd say it doesn't work since X/0 and 0/0 are undefined by definition. Do we really want to teach kids that these equal 0 because that's exactly what they'll think if their calculator says so. What next? Do we start teaching that the limits of these equations approach 0 as the denominator approaches 0? What effect would this have on engineering? 1 Link to comment Share on other sites More sharing options...

khaled Posted April 16, 2012 Share Posted April 16, 2012 IMO I'd say it doesn't work since X/0 and 0/0 are undefined by definition. Do we really want to teach kids that these equal 0 because that's exactly what they'll think if their calculator says so. What next? Do we start teaching that the limits of these equations approach 0 as the denominator approaches 0? What effect would this have on engineering? Well, I didn't say his method is correct .. I just stated what his method result Link to comment Share on other sites More sharing options...

Bignose Posted April 17, 2012 Share Posted April 17, 2012 IMO I'd say it doesn't work since X/0 and 0/0 are undefined by definition. Do we really want to teach kids that these equal 0 because that's exactly what they'll think if their calculator says so. What next? Do we start teaching that the limits of these equations approach 0 as the denominator approaches 0? What effect would this have on engineering? I cannot agree more. This can be downright dangerous. It is all too easy in an engineering calculation to end up accidentally dividing by zero. If that gets set to zero, it is was too easy to miss a potentially costly or critical error. 3 Link to comment Share on other sites More sharing options...

AnkuR Tiwari Posted April 19, 2012 Share Posted April 19, 2012 (edited) Hello Friend, I am very glad to see that you all are discussing on the topics of My research. Thank You Very much. Kindly have a proper look over the complete detailed presentation of Bhartiya New Rule for Fraction. Use the link give below: https://docs.google....=1#slide=id.p13 Regards, AnkuR Tiwari, Personal info removed Edited April 19, 2012 by hypervalent_iodine Personal information removed. Link to comment Share on other sites More sharing options...

doG Posted April 19, 2012 Share Posted April 19, 2012 I am very glad to see that you all are discussing on the topics of My research. Thank You Very much. Kindly have a proper look over the complete detailed presentation of Bhartiya New Rule for Fraction. ... I see no need to put any time into reviewing such an absurd idea. The Bhartiya New Rule for Fraction does not solve any problems, it creates problems. It would create continuity in discontinuous functions and break traditional methods of analysis used for engineering. Consider for example Tan 90° = sin 90°/cos 90° = 1/0 which is both undefined and discontinuous. How would simply calling this value 0 with some flawed application of algebra be a benefit to analysis? 4 Link to comment Share on other sites More sharing options...

rktpro Posted April 19, 2012 Author Share Posted April 19, 2012 AnkuR Tiwari, I am really surprised that you arrived here. I hope you would take the criticism positively. Link to comment Share on other sites More sharing options...

khaled Posted April 19, 2012 Share Posted April 19, 2012 I was messing with 1/0 .. what do you think ? -1 Link to comment Share on other sites More sharing options...

mississippichem Posted April 19, 2012 Share Posted April 19, 2012 I was messing with 1/0 .. what do you think ? [math] \frac{1}{0} \neq \infty [/math] because [math] 0 \cdot \infty \neq 1 [/math], QED 4 Link to comment Share on other sites More sharing options...

Visionem Ex Illuminatio Posted April 20, 2012 Share Posted April 20, 2012 [math] \frac{1}{0} \neq \infty [/math] because [math] 0 \cdot \infty \neq 1 [/math], QED [math]0\times n=n-n[/math] by the iterative definition of multiplication. True for all numbers. Assume the existence of a valid, numerical infinity, as well as obvious properties like [math]n\pm\infty=\pm\infty[/math] positive/negative respectively. True for all real numbers. Now we want to find the value of [math]0 \times \infty[/math]. Let's call it [math]x[/math] [math]0\times \infty=\infty - \infty[/math] [math]x=\infty-\infty[/math] [math]x+\infty=\infty[/math] [math]x\in\mathbb{R}[/math], since we inductively established the postulate [math]n+\infty=\infty, x\in\mathbb{R}[/math]. In conclusion, [math]0\times\infty=\mathbb{R}[/math] indeterminate. Demonstratur ut putantur. Of course, this is only valid by establishing new grounds. Infinity itself isn't even a number in most cases, so why not go this far? Still, it's all meaningless without formally accepting infinity and solving the consequential contradictions we run into. And that would be pointless, since it wouldn't really help us in any way. 3 Link to comment Share on other sites More sharing options...

mississippichem Posted April 20, 2012 Share Posted April 20, 2012 [math]0\times n=n-n[/math] by the iterative definition of multiplication. True for all numbers. Assume the existence of a valid, numerical infinity, as well as obvious properties like [math]n\pm\infty=\pm\infty[/math] positive/negative respectively. True for all real numbers. Now we want to find the value of [math]0 \times \infty[/math]. Let's call it [math]x[/math] [math]0\times \infty=\infty - \infty[/math] [math]x=\infty-\infty[/math] [math]x+\infty=\infty[/math] [math]x\in\mathbb{R}[/math], since we inductively established the postulate [math]n+\infty=\infty, x\in\mathbb{R}[/math]. In conclusion, [math]0\times\infty=\mathbb{R}[/math] indeterminate. Demonstratur ut putantur. Of course, this is only valid by establishing new grounds. Infinity itself isn't even a number in most cases, so why not go this far? Still, it's all meaningless without formally accepting infinity and solving the consequential contradictions we run into. And that would be pointless, since it wouldn't really help us in any way. +1 Not a mathematician so sorry if this is stupid. Am I correct in thinking that we can define 1/0 in some scenarios but it just so happens that such a definition leads to nothing of interest? That is, it may work for some things but will lead to contradictions? Good post by the way. Welcome. Link to comment Share on other sites More sharing options...

imatfaal Posted April 20, 2012 Share Posted April 20, 2012 Demonstratur ut putantur. ? It is shown that it is settled? Welcome by the way - and I cannot turn the italic off Link to comment Share on other sites More sharing options...

John Posted April 20, 2012 Share Posted April 20, 2012 (edited) [math]0\times n=n-n[/math] by the iterative definition of multiplication. True for all numbers. Assume the existence of a valid, numerical infinity, as well as obvious properties like [math]n\pm\infty=\pm\infty[/math] positive/negative respectively. True for all real numbers. If infinity is supposed to be some real number, then [math]n\pm\infty = \pm\infty \implies n = 0[/math]. Now we want to find the value of [math]0 \times \infty[/math]. Let's call it [math]x[/math] If [math]\infty[/math] and [math]x[/math] are real numbers, then [math]x = 0 \times \infty \implies x = 0[/math]. [math]0\times \infty=\infty - \infty[/math] [math]x=\infty-\infty[/math] [math]x+\infty=\infty[/math] [math]x\in\mathbb{R}[/math], since we inductively established the postulate [math]n+\infty=\infty, x\in\mathbb{R}[/math]. Mhm. Notationally, [math]n+\infty=\infty, x\in\mathbb{R}[/math] strikes me as a bit odd, since the fact that x is a real number doesn't have much bearing on equations not involving x. In conclusion, [math]0\times\infty=\mathbb{R}[/math] indeterminate. Given the assumptions you've made so far, again, [math]0 \times \infty = 0[/math]. Demonstratur ut putantur. Of course, this is only valid by establishing new grounds. Infinity itself isn't even a number in most cases, so why not go this far? Still, it's all meaningless without formally accepting infinity and solving the consequential contradictions we run into. And that would be pointless, since it wouldn't really help us in any way. I've not studied very advanced mathematics, and so someone better educated may find problems with my reasoning above. In any case, as far as I know, systems in which division by zero is defined have slightly different definitions of the division operation. You, and maybe mississippichem based on his reply, might find http://en.wikipedia....l_number_system interesting. Edit: This post was made assuming that by "assume the existence of a valid, numerical infinity," you mean "include infinity in the set of real numbers." Edited April 20, 2012 by John Link to comment Share on other sites More sharing options...

Visionem Ex Illuminatio Posted April 20, 2012 Share Posted April 20, 2012 (edited) John (edit: sorry for the broken tex quotes below, not sure why it does that!) If infinity is supposed to be some real number, then . As I said, let's assume that there exists a valid, numerical value for infinity (but not call it necessarily a real number, more like a special value in an extension of the RNS). What happens when we add 1 to infinity? Or 2? Or 3? Well, of course this would make no sense in the, should I say, standard view of mathematics, but let's say it still remains infinite. Because infinity plus 1 should still be infinite. And we can apply this to all real numbers (again + and - infinity are excluded and treated as valid values exceptional properties). If and are real numbers, then . I defined [math]n\pm\infty=\pm\infty, -\infty<n<\infty[/math] where the positive/negative are independently respective. For the finite real numbers, any number times 0 is zero, but infinity's different in this case, since remember that [math]0\times n=n-n[/math] making [math]0\times\infty=\infty-\infty[/math], which would be satisfied by any real number (again, excluding the special values of +/- infinity). Mhm. Notationally, strikes me as a bit odd, since the fact that x is a real number doesn't have much bearing on equations not involving x. Yes, sorry for that. I was referencing to the property I defined in the beginning... I meant [math]n\in\mathbb{R}[/math]. And I don't find this to in any way demote the credibility of my (quite unorthodox) proof; I hope you don't see it like that. Given the assumptions you've made so far, again, . 0 is a satisfying value, but as I showed above, it could also be 1, -1, googol, e, pi ... in an indeterminate form, like 0/0. Edit: This post was made assuming that by "assume the existence of a valid, numerical infinity," you mean "include infinity in the set of real numbers." When I made that statement, I was referring to the idea that "infinity is merely a concept, and makes no sense in arithmetic, like 5+justice". I just meant that we should accept it (for the purpose of this proof) with a valid mathematical meaning. I didn't necessarily mean to include it with the real numbers in the sense that it would have the same universal properties of real numbers. I do hope that it would be seen as a real number in this proof (so it would seem comparable to the finite reals and not abstract), but of course with different rules. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Demonstratur ut putantur. ? It is shown that it is settled? Welcome by the way - and I cannot turn the italic off Demonstrated as speculated. I really don't remember where I got that from But it seemed to fit. Sorry for your italics problem. I hope you get better. Lol. Thank you +1 Not a mathematician so sorry if this is stupid. Am I correct in thinking that we can define 1/0 in some scenarios but it just so happens that such a definition leads to nothing of interest? That is, it may work for some things but will lead to contradictions? Good post by the way. Welcome. Thanks I'm pretty sure it's not about "lack of interest". I think any mathematician would be exceedingly interested to expand infinity to areas of math where it is forbidden (in a figurative sense). And yes, it's the contradictions. [math]0\times1=0\times2[/math], by "simplifying out" the zero you would get 1 = 2. Which, since 0 times any real number is 0, would make every number in the real number system equal to each other, causing a collapse of mathematics so disastrous as to cause an annoying internet meme of nuke-like explosions, bottomless pits in the ground, and black holes. Really, I don't think that argument is very convincing, since pretty much anything involving zero and an arbitrary value also involves indeterminate forms. It's a given that you will get different values equal to each other if you try reducing the equation. It's as valid as this... [math]x^2=9[/math] If I square root both sides, I'll get roots of 3 and -3. Does that mean I could say "Well, that would make 3=-3. Meaning every real number is equal to its negative. That doesn't make sense, so square rooting is undefined.[/math] In fact, this argument can be applied to any equation where two different values share a common function. That does not make them equal though, most especially in the case of indeterminate forms. As for usefulness, I'm certain we'll never use infinity in our everyday lives. Infinity itself now is useful enough for what we have in limits, sums, fractals, etc. However, if we allow it to be a tangible value it might help in describing certain things. E.g. An infinite amount of nothing is something. An infinite amount of 1-dimensional lines (no area) makes up a 2-dimensional plane (an actual finite area). Edited April 21, 2012 by Visionem Ex Illuminatio Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted April 21, 2012 Share Posted April 21, 2012 I think John's point was that if you start making things up outside of their standard definitions, then you could you could get your proof to say almost anything you wanted, which makes it fairly useless in terms of its ability act as a valid counter-proof for something. Link to comment Share on other sites More sharing options...

John Posted April 21, 2012 Share Posted April 21, 2012 John (edit: sorry for the broken tex quotes below, not sure why it does that!) No worries. Also, as others have said, welcome to the forums. As I said, let's assume that there exists a valid, numerical value for infinity (but not call it necessarily a real number, more like a special value in an extension of the RNS). What happens when we add 1 to infinity? Or 2? Or 3? Well, of course this would make no sense in the, should I say, standard view of mathematics, but let's say it still remains infinite. Because infinity plus 1 should still be infinite. And we can apply this to all real numbers (again + and - infinity are excluded and treated as valid values exceptional properties). I defined [math]n\pm\infty=\pm\infty, -\infty<n<\infty[/math] where the positive/negative are independently respective. Why bother with all this? For the finite real numbers, any number times 0 is zero, but infinity's different in this case, since remember that [math]0\times n=n-n[/math] making [math]0\times\infty=\infty-\infty[/math], which would be satisfied by any real number (again, excluding the special values of +/- infinity). This strikes me as self-contradictory, as you've essentially said a) infinity is different from the real numbers, b) infinity follows the same pattern as a real number n, then c) but infinity won't follow this pattern. Yes, sorry for that. I was referencing to the property I defined in the beginning... I meant [math]n\in\mathbb{R}[/math]. And I don't find this to in any way demote the credibility of my (quite unorthodox) proof; I hope you don't see it like that. Not at all. I pointed it out in case you were intending some special meaning that I didn't see. 0 is a satisfying value, but as I showed above, it could also be 1, -1, googol, e, pi ... in an indeterminate form, like 0/0. I see how you came to the conclusion you did based on your clarification of your intentions, but what do you mean here, i.e. what does something like "e in an indeterminate form" mean? When I made that statement, I was referring to the idea that "infinity is merely a concept, and makes no sense in arithmetic, like 5+justice". I just meant that we should accept it (for the purpose of this proof) with a valid mathematical meaning. I didn't necessarily mean to include it with the real numbers in the sense that it would have the same universal properties of real numbers. I do hope that it would be seen as a real number in this proof (so it would seem comparable to the finite reals and not abstract), but of course with different rules. Are you saying you were basically going for a proof by contradiction to show that [math]0\times\infty\neq1[/math]? I'm pretty sure it's not about "lack of interest". I think any mathematician would be exceedingly interested to expand infinity to areas of math where it is forbidden (in a figurative sense). As mentioned previously, there are areas in which division by zero can be defined, but in these cases one has to be careful with exactly how the operation is defined, and in what cases it's valid. And yes, it's the contradictions. [math]0\times1=0\times2[/math], by "simplifying out" the zero you would get 1 = 2. Which, since 0 times any real number is 0, would make every number in the real number system equal to each other, causing a collapse of mathematics so disastrous as to cause an annoying internet meme of nuke-like explosions, bottomless pits in the ground, and black holes. Really, I don't think that argument is very convincing, since pretty much anything involving zero and an arbitrary value also involves indeterminate forms. It's a given that you will get different values equal to each other if you try reducing the equation. It's as valid as this... [math]x^2=9[/math] If I square root both sides, I'll get roots of 3 and -3. Does that mean I could say "Well, that would make 3=-3. Meaning every real number is equal to its negative. That doesn't make sense, so square rooting is undefined. It can be shown that squaring either 3 or -3 will result in 9. It's obvious, however, that 1 does not equal 2. "Simplifying out the 0" in this case amounts to dividing both sides of the equation by 0, which is not valid as the operation is undefined. In fact, this argument can be applied to any equation where two different values share a common function. That does not make them equal though, most especially in the case of indeterminate forms. The argument can be applied, yes, but it's an invalid argument, which of course is the whole point. 1 Link to comment Share on other sites More sharing options...

doG Posted April 21, 2012 Share Posted April 21, 2012 (edited) As I said, let's assume that there exists a valid, numerical value for infinity.... In reality there is no valid numerical value for infinity so making that assumption to support any proof is flawed from the start. That assumption renders the rest of your proof as worthless. Edited April 21, 2012 by doG 1 Link to comment Share on other sites More sharing options...

Visionem Ex Illuminatio Posted April 21, 2012 Share Posted April 21, 2012 (edited) In reality there is no valid numerical value for infinity so making that assumption to support any proof is flawed from the start. That assumption renders the rest of your proof as worthless. Well this is of course speculative territory. And I'm also not sure why "reality" plays any part into this, since mathematics is abstract. Infinity does actually work in some fields of standard math, the weirdest involving real projective geometry. [math]\sqrt{-1}[/math] also used to have no valid numerical value, but now it is standard mathematics. Not that this is necessarily going to become universally standard, but the point is when we're heading into forbidden land, we have to make new laws so we don't break any rules and run into problems. Edited April 21, 2012 by Visionem Ex Illuminatio Link to comment Share on other sites More sharing options...

doG Posted April 21, 2012 Share Posted April 21, 2012 (edited) As I said, let's assume that there exists a valid, numerical value for infinity (but not call it necessarily a real number, more like a special value in an extension of the RNS). What happens when we add 1 to infinity? Or 2? Or 3? Well this is of course speculative territory. And I'm also not sure why "reality" plays any part into this, since mathematics is abstract. Infinity does actually work in some fields of standard math, the weirdest involving real projective geometry. Any 'value' you call infinity that you can add any number to was not really infinity to begin with .... Infinity Infinity, most often denoted as , is an unbounded quantity that is greater than every real number... Trying to assign a numerical value for infinity that you can add to breaks the definition that it is already greater than any number. IMO it is pointless to discuss any value of infinity that is less than infinity in order to define division by zero. Edited April 21, 2012 by doG Link to comment Share on other sites More sharing options...

Visionem Ex Illuminatio Posted April 21, 2012 Share Posted April 21, 2012 (edited) Any 'value' you call infinity that you can add any number to was not really infinity to begin with .... What? Not sure what you're saying. Whatever the statement, why not? You have to understand that in this case we're defining a new value with its own rules. You can't negate the proof with rules which only apply to the real numbers. Trying to assign a numerical value for infinity that you can add to breaks the definition that it is already greater than any number. IMO it is pointless to discuss any value of infinity that is less than infinity in order to define division by zero. How does it break the definition? Infinity is still [math]\infty>n, n\in\mathbb{R}[/math]. Again, we're not including [math]\pm\infty[/math] in the RNS, because 1) for convenience I won't change the definition of the RNS (thus changing the cardinality of the set and completely overturning Cantor's work on its head), and 2) it has special properties not universally granted to the real numbers. However, I do want you to think it's "real" in the sense that it's not an imaginary, complex, hyperreal, etc. number. It still has the same essence as 1, 2, pi, e, whatever. I never defined this value as less than infinity. It is still infinite, and it is still greater than all real numbers. However, it has the interesting indeterminate property that no matter what real number you add or subtract to it, it remains infinity (it does not change, greater or lesser). This is solely based on the inductive idea that any finite addition to infinity would still keep infinity, infinity. [math]1+\infty=\infty[/math] is what I defined. Of course, as I said, this is an indeterminate form. You cannot reduce it by subtracting infinity from both sides. Again [math]1+\infty\not>\infty[/math] but equal to infinity. BTW, though infinity doesn't work well in elementary arithmetic of the real numbers, there are standard, accepted indeterminate forms of expressions which involve it. (also zero, which is strange, since the multiplicative inverse of zero is infinity). [math]\dfrac{0}{0}, 0\times\infty, \infty-\infty, 0^0, 1^\infty[/math] Edited April 21, 2012 by Visionem Ex Illuminatio Link to comment Share on other sites More sharing options...

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