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SolomonZelman

  • one year ago

I have already asked, but I want to ask your opinion about my statement (I will prove my statement)

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  1. SolomonZelman
    • one year ago
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    Divide by 0 doesn't exist because the result would split and converge to \(+\infty\) and \(-\infty\) all at once. (that makes no sense how a point can break into two, and certainly that both way it diverges, and this is why division by zero doesn't exist). (hold... I will post my reason)

  2. SolomonZelman
    • one year ago
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    lets see what happens when the divisor (the number you divide by) approaches 0 from the right. (I am using 1 is dividend for convenience, could be be any C) \(\large\color{black}{ \displaystyle 1\div 1=1 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{2}=2 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{3}=3 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{4}=4 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{5}=5 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{\rm n}=\rm n }\) \(\large\color{black}{ \displaystyle 1\div \lim_{{\rm n}\rightarrow0^+}\frac{1}{n}=\infty }\)

  3. SolomonZelman
    • one year ago
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    Now, as divisor approaches zero from the left side, \(\large\color{black}{ \displaystyle 1\div -1=-1 }\) \(\large\color{black}{ \displaystyle 1\div \frac{1}{2}=-2 }\) \(\large\color{black}{ \displaystyle 1\div \frac{-1}{3}=-3 }\) \(\large\color{black}{ \displaystyle 1\div \frac{-1}{4}=-4 }\) \(\large\color{black}{ \displaystyle 1\div \frac{-1}{5}=-5 }\) \(\large\color{black}{ \displaystyle 1\div \frac{-1}{\rm n}=\rm -n }\) \(\large\color{black}{ \displaystyle 1\div \lim_{{\rm n}\rightarrow0^+}\frac{-1}{n}=-\infty }\)

  4. SolomonZelman
    • one year ago
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    oh the last line should say \(\large\color{black}{ \displaystyle 1\div \lim_{{\rm n}\rightarrow0^-}\frac{1}{n}=\infty }\) (minus by 0)

  5. SolomonZelman
    • one year ago
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    Division by zero is a division by two sides limit, and thus \(\large\color{black}{ \displaystyle 1\div \lim_{{\rm n}\rightarrow0}\frac{1}{n}=\left\{ -\infty,~+\infty\right\} }\) so would be true for any number C, (at least if C isn't zero) \(\large\color{black}{ \displaystyle C\div \lim_{{\rm n}\rightarrow0}\frac{1}{n}=\left\{ -\infty,~+\infty\right\} }\)

  6. SolomonZelman
    • one year ago
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    ((division of a number by number gave two points. ))

  7. SolomonZelman
    • one year ago
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    i have a typo when I am showing the left sides limit. second line in that reply should be -1/2

  8. SolomonZelman
    • one year ago
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    (if you want I can remove the mass and repost it entirely)

  9. freckles
    • one year ago
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    \[\text{ do you mean to talk about } \frac{1}{\frac{1}{0}} \text{ or } \frac{1}{0}?\] And when I put the 0 on bottom just pretend that means n approaches 0.

  10. SolomonZelman
    • one year ago
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    Do you mean to ask if my definition of 0 is \(\large\color{black}{ \displaystyle \lim_{{\rm n}\rightarrow0}\frac{1}{n} }\) or \(\large\color{black}{ \displaystyle \lim_{{\rm n}\rightarrow0}{\rm n}}\) ?

  11. freckles
    • one year ago
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    \[\lim_{n \rightarrow 0}(1 \div \frac{1}{n})=\lim_{n \rightarrow 0} (1 \cdot n)=0\]

  12. SolomonZelman
    • one year ago
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    no, I am dividing by 1/n

  13. freckles
    • one year ago
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    why does 1/infty or 1/-infty = infty or -infty respectively?

  14. SolomonZelman
    • one year ago
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    My idea is that \(\large\color{black}{ \displaystyle {\rm C}\div0 =}\) \(\large\color{black}{ \displaystyle {\rm C}\div\lim_{n\rightarrow0}~\frac{1}{n} =\left\{{\rm a,~b}\right\}}\) (that is, it is equivalent to a set of 'a' and 'b', equivalent to two points.) point 'a' diverges to -infinity point 'b' diverges to +infinity.

  15. SolomonZelman
    • one year ago
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    u r right actually

  16. SolomonZelman
    • one year ago
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    should just be \(\large\color{black}{ \displaystyle\lim_{n\rightarrow0}~\frac{C}{n} =\left\{{\rm a,~b}\right\}}\)

  17. SolomonZelman
    • one year ago
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    that is how I should put it

  18. freckles
    • one year ago
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    ok I think I'm cool then you are basically using the graph of f(x)=1/x on (-inf.inf) to say why we shouldn't divide by 0 because it has a break at x=0 one side goes positive large while the other side goes negative large

  19. SolomonZelman
    • one year ago
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    yup, it is the graph of 1/x .....

  20. SolomonZelman
    • one year ago
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    I saw the idea of 1/(1/2) , 1/(1/3) , 1/(1/4) ... 1/(1/n) => infinity so then I though, about the negatives. the essence and a very good demonstration is the graph of 1/x (or any f(x)=C/x)

  21. freckles
    • one year ago
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    you know what in your work about each time you got closer to the end of one of your posts it looked like you were making the n larger (and not closer to 0) So maybe we could have wrote: \[\lim_{n \rightarrow \infty}(1 \div \frac{1}{n})\] or negative large \[\lim_{n \rightarrow -\infty}(1 \div \frac{1}{n})\]

  22. SolomonZelman
    • one year ago
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    Oh, yeah I made that technical error. Apologize.

  23. SolomonZelman
    • one year ago
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    I should have used infinity, or I could have said \(\large\color{black}{ \displaystyle C\div \color{orangered}{\lim_{n \rightarrow 0}\left(n \right)}}\)

  24. SolomonZelman
    • one year ago
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    or \(\large\color{black}{ \displaystyle 1\div \color{orangered}{\lim_{n \rightarrow 0}\left(n \right)}}\)

  25. ybarrap
    • one year ago
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    So division by 0 is undefined because \(\cfrac{1}{0}\) means there is an \(a\) such that $$ 0\times a=1 $$ But \(0\times \text{Anything}=0\) Would that be sufficient?

  26. SolomonZelman
    • one year ago
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    I was proposing it (division by 0) is undefined, because when you take any Real number C and divide by 0, you get a set of 2 values {a,b} where a diverges to -infinity and b diverges to + infinity. division by 0 gives a, where a diverges to -infinity = that is the left sided limit of 1/n division by 0 gives b, where b diverges to +infinity = that is the right sided limit of 1/n division by 0 (the two sides limit) gives you the set of a and b (wher a diverges to neg. inf. and b diverges to pos. inf.)

  27. SolomonZelman
    • one year ago
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    C / D = E but not C / D = {set of more than 1 value}

  28. SolomonZelman
    • one year ago
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    I am even disregarding the fact that a and b diverge

  29. ikram002p
    • one year ago
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    im founding this very interesting, good job !!

  30. SolomonZelman
    • one year ago
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    Tnx ikram...

  31. Hero
    • one year ago
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    People talk about dividing by zero as if it is such a mystery. Division implies breaking a whole into parts. The term "dividing by zero", when you really think about it, is an oxymoron. If you divide something by 3, you divide it into three parts. If you divide something by 2, you divide it into 2 parts. If you divide something zero times, you divide it into zero parts. The whole remains unchanged. In other words, you've done nothing. You haven't divided anything at all. The process of "division" never took place.

  32. SolomonZelman
    • one year ago
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    nice fading on the pic

  33. SolomonZelman
    • one year ago
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    Hero, that is like a pizza/cake reasoning. You can't split a pizza into 0 friends. But, you can't split a pizza into p/q friends or into any x friends if x is not a natural number.

  34. SolomonZelman
    • one year ago
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    acc to that reasoning division by non-integers is not applicable

  35. Hero
    • one year ago
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    If you take the common sense approach, you won't have to prove division by zero doesn't exist.

  36. SolomonZelman
    • one year ago
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    Well, I knew it doesn't exist ever since I learned that in 3rd grade in Russia.... but people reason it differently... I was just trying to offer that division by zero gives a result of two points ((and not only you have a division of a number by another another that gives you two answers that makes no sense, but also these elements of my answer are not tangible they are infinities))

  37. ikram002p
    • one year ago
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    well, as long as @SolomonZelman took the limit notation i see no harm using the word "division" in fact the idea of divide on zero with limit notations show u the tiny value u can divide by for example what if i wanna divide on such SMALL number near to zero ? and yes x/0 approaches to infinity its UNDEFINED value means you cant express it in number system terms that ppl deals with and count, but if ur a good mathematician you would use it and know the value behind it, unlike 0/0 this term is indeterminate which mean it can ANYTHING they ar such mathematical proofs shows its 2,3,4,.... or anything else

  38. SolomonZelman
    • one year ago
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    Yeah, elementary level: if C is a none-zero number, C / 0 = x checking: x times 0 = 0 (and can't equal to non-zero number C) So, C/0 = no solution (or undefined) if C is not zero. 0 / 0 = x checking x * 0 = 0 x can literally be any number. I am not sure if 0/0 is really invalid from that standpoint. ((( likewise, 0/x=0 and x can be any number. )))

  39. SolomonZelman
    • one year ago
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    Just an infinite number of solution doesn't necessarily indicate that there is no solution, or perhaps when I said 0/x , then there is no division, and that is why there is an infinite number of solutions? Perhaps, 0/x == u r not even dividing. just like saying 0=0 well, then why wouldn't 0/0=0 be like 0=0 ?

  40. Hero
    • one year ago
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    You divided by values that were close to zero and realized you could do that infinitely for both positive and negative values. But you never actually divided by zero. The term dividing by zero has no mathematical meaning. Dividing by zero won't result in two points. You don't know what will happen when you divide by zero because no one has yet to actually do it. When they say dividing by zero doesn't exist, there's a reason for it.

  41. SolomonZelman
    • one year ago
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    yes;) ... i think i got another reason

  42. anonymous
    • one year ago
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    The reason why division by zero is not defined is because if we allow it to be defined, many algebra rules are broken. Consider if we let \(1/0 = z\) and say \(c/0 = cz\). The object \(z\) would not obey most algebra rules.

  43. SolomonZelman
    • one year ago
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    just that I need to express in a better way. Will first start from this definiton \(\color{blue}{\rm Dividend \div Divisor = Quotient} \) and will name them, \(\rm A \div B = C \) A=Dividend B=Divisor C=Quotient \(\rm (this~~is~~for~~ convenience)\) What is dividing A by B mean? (lets choose a unit of a meter, and assume this same unit for all A B and C) ((Will just look at flat piece of size A, and a flat piece of size B.)) |dw:1434667396117:dw| and the C here, is the number of B-sized pieces that A can contain. There is certain amount/number C of B's in the A. (hope this phrase is not abstruse) If your B is 0, then A can contain as many piece as you want. And from there you get very large results when you divide by small numbers (by small B's). (just another thought)

  44. Hero
    • one year ago
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    Actually, it's not just because of the fact that it would break algebra rules. You don't even need to use algebra rules to show that division by zero is not possible. If you think of division as it relates to subtraction, then you can describe it this way: If you wanted to know how many times you could take 2 away from 12, the shortcut way to do it is to just divide. 12/2 = 6. You could take 2 from 12 six times before you have nothing left. To show this same thing using long subtraction, you have: 12 - 2 - 2 - 2 - 2 - 2 - 2 = 0 And you could do this with almost any division as long as the divisor isn't zero. Suppose we tried this same approach with zero as the divisor. We want to know how many times we can take zero from twelve before we have nothing left. In other words, 12/0 = what? Well, if we try long subtraction, here's what would happen: 12 - 0 - 0 - 0 - 0 - 0 - 0 - 0 ... We would keep subtracting zero from 12 infinitely but we would never get to zero. We'd remain at 12. Division is nothing but a shortcut for long subtraction. So when we say we can't divide a number by zero, what we're really saying is, We can subtract zero from a number as many times as we won't but we will never be able to determine exactly how many times it would take us to subtract before we have nothing left because we will always have the same amount we started with.

  45. SolomonZelman
    • one year ago
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    beautiful.....

  46. anonymous
    • one year ago
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    Division is an abstract concept that can be interpreted as repeated subtraction for cases when the divisor divides the dividend, but otherwise the interpretation falls short. You know that for 1/3 this would happen 1 - 3 - 3 - 3 ... and it will never get to zero, either. The next step is to consider values beyond the integers. The only way to really argue whether or not an operation or value deserves a definition is by considering how many algebra rules it breaks and how many it obeys.

  47. Hero
    • one year ago
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    Actually, you can demonstrate it for 1/3 as well when I apply the shorthand version of the long subtraction. In other words, each time I subtract 2, I multiply by 1 which signifies how many times I subtracted the number. The demonstration is here: 12 - 2(6) = 0 If I want to do the same for 1/3, then it would be 1 - 3(1/3) = 0 So it is possible to subtract 3 to get to zero, you just have to subtract 1/3 of 3 to get there.

  48. Hero
    • one year ago
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    What I showed immediately above is the way I meant to demonstrate it initially. To show the subtraction version of division it's dividend/divisor = quotient dividend MINUS divisor TIMES quotient = 0 The long hand version of the same thing I showed above is 1 - 1/3 - 1/3 - 1/3 = 0

  49. anonymous
    • one year ago
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    That's not really an actual definition. You're using subtraction and multiplication properties of equality to manipulate an equation. Also you had to create a class a numbers (non-integral rational numbers) for it to work.

  50. Hero
    • one year ago
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    All I was demonstrating was that division is a shortcut for long subtraction.

  51. anonymous
    • one year ago
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    And I agree that it makes sense when the divisor perfectly divides the dividend, but I disagree that it can be used to justify not defining zero division

  52. Hero
    • one year ago
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    I demonstrated also using long subtraction why zero division makes no sense. That's the first thing I did.

  53. SolomonZelman
    • one year ago
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    C / 0 = X C - 0 - 0 - ... - 0 = 0 \_____________/ X number times.

  54. SolomonZelman
    • one year ago
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    I like Hero's idea:)

  55. SolomonZelman
    • one year ago
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    well, for all non zero C's

  56. SolomonZelman
    • one year ago
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    and for C=0 you get X=anything

  57. SolomonZelman
    • one year ago
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    I want to know one thing. I previously asked it, but maybe not explicitly. 0/0=X 0 -0-0-0...-0 =0 \___________/ X times In this case, because X can be absolutely anything, does that make 0/0 undefined, or, 0/0 does make sense?

  58. anonymous
    • one year ago
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    I don't think it is satisfactory. It really just leads to the conclusion that \(12/0 = \infty\), which is fine in the context presented; however, it is still a definition.

  59. anonymous
    • one year ago
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    An indeterminate form is an operation where all values satisfy it given the right context.

  60. SolomonZelman
    • one year ago
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    12/0 is not infinity if you go by the subtraction idea. Or not necessarily infinity.

  61. SolomonZelman
    • one year ago
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    this is why division by doesn't exist, because then 1=0

  62. Hero
    • one year ago
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    @SolomonZelman, you made a mistake somewhere in your reasoning.

  63. anonymous
    • one year ago
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    There is no context in which 12/0 = 12

  64. SolomonZelman
    • one year ago
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    yes=X is mistake

  65. SolomonZelman
    • one year ago
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    oops

  66. ybarrap
    • one year ago
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    12/0=X equivalent to 12 - 0 -0-0 ... -0 = 0 \______________/ X times \(12 - X\times 0 = 0\) Not defined

  67. SolomonZelman
    • one year ago
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    tnx for correcting. 12/0=X 12 - 0 -0 -0 ... -0 = 0 \______________/ X times X= DNE goes quite well then.....

  68. Hero
    • one year ago
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    @wio, I understand why you are not really impressed with my demonstration. I explained an abstract concept using basic mathematics.

  69. SolomonZelman
    • one year ago
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    that was a shift. I assert that:) jk

  70. anonymous
    • one year ago
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    If we say \(12/0=x\), we don't get to be presumptuous about how \(x\) works. For example, you can't presume that the subtraction property of equality works for \(x\). Consider for example \(\sqrt{-1}=i\). If we have the inequality \(2<5\), can we multiply both sides by \(i\)?\[ 2i<5i \]We can't do this because it presumes \(i\) is positive. Just because \(i\) doesn't follow all the properties of real numbers doesn't make it undefined. It's a matter of how many algebra rules the concept forces us to qualify. If it doesn't provide much insight and makes add asterisks to a lot of rules, then it's better off not beind defined.

  71. SolomonZelman
    • one year ago
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    The subtraction approach is not like this. There is a huge difference. for any a>b you can't multiply both sides time i. And in this case, subtraction method only doesn't work when divisor is zero. This is not an evidence for invalidity of the method, but rather, it is a genuine evidence to why division by 0 doesn't exits.

  72. SolomonZelman
    • one year ago
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    This is my opinion, of course.

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