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skullpatrol

  • 2 years ago

Why can you never divide by zero?

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  1. vac33
    • 2 years ago
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    Because the answer would be undefined.

  2. skullpatrol
    • 2 years ago
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    What do you mean?

  3. sthop456
    • 2 years ago
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    because you can then create a bunch of false proofs such as 2=1. Notice the last step of the following false proof is actually just division by zero: a = x [true for some a's and x's] a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a-2x = a+x-2x [subtract 2x from both sides] 2(a-x) = a+x-2x [2a-2x = 2(a-x)] 2(a-x) = a-x [x-2x = -x] 2 = 1 [divide both sides by a-x]

  4. vac33
    • 2 years ago
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    Because when you get a smaller and smaller number in the denominator the number gets larger and larger (assuming it is positive) for example 1/0.00001 is much larger than one. If its 1/0 it would be infinitely large following the same trend.

  5. zepp
    • 2 years ago
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    Let c (result) be \(\LARGE \frac{a}{b}\), if b was 0, then \(\large cb=a\), we need to find a number \(a\) that could be the answer to \(c*(0)\), and for all \(c\), a would always be 0. In this case, we leave this undefined.

  6. sirm3d
    • 2 years ago
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    |dw:1353147793508:dw|

  7. skullpatrol
    • 2 years ago
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    @sirm3d neither

  8. sirm3d
    • 2 years ago
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    @skullpatrol so there is no correct answer. any correct answer will be contradicted by the same examples above.

  9. scarydoor
    • 2 years ago
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    In abstract algebra, 1 is the multiplicative identity. That means that you can multiply any element by 1 and you get the number back. 4*1 = 4. 876*1=876 etc. It's the unique element that does that. You also want to have 'inverse elements'. Those are numbers that when you multiple them together, you get the identity. For example, the inverse element of 5 is 1/5, because 5 * 1/5 = 1, the identity. In multiplication, when you 'divide' by something, you are actually multiplying by the inverse element of the thing, with respect to multiplication. So when you say 4 / 5 you are actually saying 4 * 1/5. The question is, does 0 have a multiplicative inverse? What that's asking is, is there a number x such that 0 * x = 1? There can't be, because, well, from some other places, we sort of want 0 to be the element such that when you multiply anything by 0, then you get 0. So 0 has no inverse. If it did, then lots of the things from abstract algebra, which lay the foundations for arithmetic, would be violated and no longer work properly.

  10. scarydoor
    • 2 years ago
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    (above, where I said 4 * 1/5, what I mean is that you're actually multiplying 4 by the number defined as being 1/5. So in a sense there's actually no division. There's just multiplying, by the correct number)

  11. scarydoor
    • 2 years ago
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    So when you say you want to divide by zero, what you mean is that you want to multiply by the number that is the "multiplicative inverse" of 0. And the multiplicative inverse of zero is the number x such that 0 * x = 1, and x * 0 = 1. And it can be thought that if such a number exists, then it would wreck everything.

  12. sirm3d
    • 2 years ago
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    |dw:1353148249632:dw| if n is the correct answer, what is n then? it would be more sensible to agree that there is no solution than argue that one answer is "more correct" than the other

  13. skullpatrol
    • 2 years ago
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    How about 0*(1/0) = 1 @scarydoor ?

  14. scarydoor
    • 2 years ago
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    well, the 'symbols' like 1/5 and 1/0 still need to be translated back into decimal numbers of some kind. We can represent 1/5 in decimals. What value would 1/0 take? The only two possibilities are -infinity and +infinity, which aren't numbers.

  15. scarydoor
    • 2 years ago
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    Also, suppose you decide on some decimal value for 1/0 to take, called x (just some number). Then 1/0 = x. Then 1 = x * 0 = 0. So you have that 1 = 0, which means you're going to have some problems.

  16. sirm3d
    • 2 years ago
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    if 1/0 is a number, then \[\large 1=0\left( \frac{ 1 }{ 0 } \right)=\frac{ 0 }{ 0 }=\frac{ 0\times5 }{ 0 \times 4 }=\frac{ 0 }{ 0 }\left( \frac{ 5 }{ 4 } \right)=\frac{ 5 }{ 4 }\] so we must agree that \[1=\frac{ 5 }{ 4 }= \text{any other number}\]

  17. scarydoor
    • 2 years ago
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    Even if you had 1/0 = infinity, then you'd have 1 = infinity * 0. There are areas of maths where you have accepted rules to do arithmetic on the symbol infinity, and those rules say that infinity * 0 = 0. So you'd still have 1 = 0. But also, I'm not so sure you'd actually still be able to use the algebraic manipulations I used there, since if you have infinity in the system, then it's no longer really a structure of numbers. It's got some symbols in there too...

  18. skullpatrol
    • 2 years ago
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    If 1/0 = x. Then 1 = x * 0 , but how did you algebraically derive 1 = x*0 from 1/0 =x @scarydoor?

  19. scarydoor
    • 2 years ago
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    Well, we said 1/0 is the multiplicative identify of 0. That means (the definition of multiplicative identity) that 0 * 1/0 = 1. So 1/0 = x. (now multiply each side by 0, to get...) 0 * 1/0 = 0 * x (then 0 * 1/0 = 1, so...) 1 = 0 * x. (I wrote the right hand side the other way around, but, don't worry about that... )

  20. sirm3d
    • 2 years ago
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    @hartnn share your thoughts. =)

  21. scarydoor
    • 2 years ago
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    Also, this might seem weird and unconnected to normal arithmetic that you do. But this is really minimalist algebra, and was really just invented to justify the way that we think arithmetic should work (i.e., we want it to be "sensible"). These are really conservative conventions. Just abstract sounding.

  22. sirm3d
    • 2 years ago
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    @skullpatrol what about \[\large \frac{ 5 }{ 0 }\] is it five times as many as \[\large \frac{1}{0}\]

  23. hartnn
    • 2 years ago
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    logically speaking, dividing something with a number 'n' means making 'n' equal parts of that 'something. dividing by 0 means you are not making any parts...so such operation is not defined

  24. sirm3d
    • 2 years ago
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    \[0\left( \frac{ 5 }{ 0 } \right)=\frac{ 0 }{ 1 }\left( \frac{ 5 }{ 0 } \right)=\frac{ 0 }{ 0 }=1\] because \[0\left( \frac{ 1 }{ 0 } \right)=\frac{ 0 }{ 1 }\left( \frac{ 1 }{ 0 } \right)=\frac{ 0 }{ 0 }=1\] since this number must obey the multiplication rule for fractions.

  25. skullpatrol
    • 2 years ago
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    But, you have multiplied both sided of an equation by 0. You have used 0 as a multiplier in transforming an equation, which says the final equation will be satisfied by any number no matter what the equation is. @scarydoor

  26. scarydoor
    • 2 years ago
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    Also, try to divide 1 by 0. Divide it by increasingly smaller numbers. 1/(1/2) = 2. 1/(1/3) = 3. 1/(1/4)=4. In general, divide 1 by 1/n, where n is a natural number. 1/n gets ever closer to 0 as n gets increasingly larger. So 1/(1/n) = n. And as n goes to infinity (which means you get closer to the "true" value of 1/0) then 1/(1/n) = infinity. So, eventually you would guess that 1/0 might need to be infinity.

  27. scarydoor
    • 2 years ago
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    Oh yeah! Good point skullpatrol. Haha... Well, maybe that part of the explanation why it doesn't work needs some improvement. The other peoples' explanations might be better to look at then.

  28. scarydoor
    • 2 years ago
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    (also, above, where I was dividing by natural numbers, you could divide by -n, and then the limit comes to -infinity. So there are two implied values for 1/0, at least for that approach...)

  29. sirm3d
    • 2 years ago
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    IF you are right that it is possible to divide by zero, that this division by zero results in some quotient and remainder, can you give an answer that is unassailable by any sound argument? @skullpatrol

  30. scarydoor
    • 2 years ago
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    people have struggled with the idea of 0 and infinity for thousands of years. Lots of very smart people didn't know what to do with the complications. There are two good books about it. I read one (the one where reviewers say there is less maths, whichever that is...) and found it to be a very entertaining read on the topic: http://www.amazon.com/Zero-Biography-Dangerous-Charles-Seife/dp/0140296476/ref=sr_1_3?s=books&ie=UTF8&qid=1353150007&sr=1-3&keywords=zero http://www.amazon.com/The-Nothing-that-Is-Natural/dp/0195142373/ref=pd_vtp_b_2

  31. scarydoor
    • 2 years ago
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    You might be able to find one of them at a library. They will give you lots more intuition about the development of the idea of 0 (and infinity).

  32. skullpatrol
    • 2 years ago
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    All I'm saying is a lot of Professors give the argument: If a/0=c, then a=0*c. But as we have shown this argument is flawed by the fact that you are using 0 as a multiplier in transforming a/0=c into a=0*c.

  33. scarydoor
    • 2 years ago
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    hmn. actually.... coming to think about it, we can multiply each side by zero, right? The reason is that it "should" make both sides equal to zero. (and in fact you can multiply each side by the same number, for any number, and the equality will hold) So actually I'm going back on my reneg, and say that it was fine to multiply both sides by zero.

  34. scarydoor
    • 2 years ago
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    The problem is that we have a number (1/0) which we defined so that 0 * 1/0 = 1.

  35. scarydoor
    • 2 years ago
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    Can you explain why it shouldn't have been okay to multiply both sides by 0? Or why you think that shouldn't have been allowed? Because if you can be convinced that it was allowed, then that would confirm that we shouldn't be allowed to divide by zero, right?

  36. skullpatrol
    • 2 years ago
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    Also,they are implying that 0*(a/0) = 0*(1/0)*a and then 0*(1/0) = 1 but we know that 0 has no reciprocal because 0 times any number is 0 not 1.

  37. scarydoor
    • 2 years ago
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    I don't understand what you're saying.

  38. sirm3d
    • 2 years ago
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    the trouble with accepting 1/0 as a number is determining the value of a * (1/0)

  39. skullpatrol
    • 2 years ago
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    You have used 0 as a multiplier in transforming an equation, which says the final equation will be satisfied by any number no matter what the equation is. @scarydoor

  40. scarydoor
    • 2 years ago
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    if (1/0) = x, and x is the multiplicative inverse of 0, then... 0 * (1/0) = 0 * x (multiply both sides by 0. (if two numbers are equal, then to multiply them by the same number, they should still be equal) 1 = 0. (since 0 * 1/0 = 1 (left hand side) and 0 * any number = 0 (right hand side) 1=0 is absurd. so the assumption that (1/0) exists and is a real number, must have been wrong. (proof by contradiction).

  41. scarydoor
    • 2 years ago
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    by the same assumptions I made in the first line, you could say that... 1/0 = x (then multiply both sides by a to get...) a * 1/0 = a * x (post multiply both sides by 0...) a * (1/0 * 0) = (a * x) * 0 a * 1 = 0 a = 0. This is true for any number a..

  42. skullpatrol
    • 2 years ago
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    Any equality will be satisfied by any real number if you multiply both sides by 0. for example x=5 is only satisfied by 5 but 0*x=0*5 or 0*x=0 is now satisfied by any real number so it is not equivalent to the original equation x=5.

  43. scarydoor
    • 2 years ago
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    But to make that claim, you have assumed you can divide by 0! That's the bad link. you said that if you can say 0*x = 0*5, then x must be equal to 5. But how did you get rid of the 0? You must have tried to divide by 0. That's basically what I'm saying in the argument above.

  44. scarydoor
    • 2 years ago
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    Are you convinced?

  45. skullpatrol
    • 2 years ago
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    no

  46. skullpatrol
    • 2 years ago
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    I'm saying starting with the original equation x=5 and asking for what values is it satisfied for? The answer is 5, because when we substitute 5 into the original equation we get a true statement 5=5.

  47. scarydoor
    • 2 years ago
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    which original equation? I'm not sure where you think the problem is.

  48. skullpatrol
    • 2 years ago
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    Take any equation as your original equation try: 8z=40 the solution set to this equation is {5}.

  49. scarydoor
    • 2 years ago
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    That's right.

  50. skullpatrol
    • 2 years ago
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    Now transform that original equation by multiplying both sides by 0.

  51. scarydoor
    • 2 years ago
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    I think you're saying that if I now say that 0 * 8z = 0 * 40, then the solution for z is any number. And you think that's a problem because before we said that z = 5.

  52. scarydoor
    • 2 years ago
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    The problem is in direction of implication. I can say that if 8z = 40, then 0 * 8z = 0 * 40. But I can't say that if 0 * 8z = 0 * 40 then 8z = 40. Because I'm saying that 0 = 0 means that 8z = 40.

  53. skullpatrol
    • 2 years ago
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    Yes, that is correct. Zero is not allowed as a multiplier in transforming equations.

  54. scarydoor
    • 2 years ago
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    It's allowed as a transformation, but the implication is only one way. If it's not zero then I can get back again. but with zero I can't.

  55. scarydoor
    • 2 years ago
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    So, before I said that 1/0 = x is an equality. Then I multiply both sides by 0, I'm going in the correct direction for the implication... 1/0 * 0 = x * 0. This equality still holds. Now I use that 1/0 is the identiy of 0, and that x * 0 = 0... 1 = 0.

  56. scarydoor
    • 2 years ago
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    What you are trying to do is go in the other direction. But I don't need to go in the other direction, so I don't care that I can't.

  57. scarydoor
    • 2 years ago
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    think about this.... 6 = 3 * 2. That's correct. Now multiply both sides by 0. We get that 0 = 0. Can we conclude, from 0 = 0 that 6 = 3 * 2? We can't. Actually this all comes back to the fact that we can't divide by zero, and that's because 0 is the annihilator. If we could divide by zero, then we could go in the backwards direction too.

  58. scarydoor
    • 2 years ago
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    Does that make sense?

  59. scarydoor
    • 2 years ago
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    If you have x = y, then a * x = a * y for any number a. Now, so long as a is not zero, then 1/a * a * x = 1/a * a * y. Which means that 1 * x = 1 * y. x = y. So you can go both ways. Notice that the only reason you can't go backwards again if a is equal to zero is that you can't divide 1 by zero. (where a = 0, 1/a = 1/0).

  60. skullpatrol
    • 2 years ago
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    I agree :)

  61. scarydoor
    • 2 years ago
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    haha, good.

  62. scarydoor
    • 2 years ago
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    algebra is fun.

  63. skullpatrol
    • 2 years ago
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    But the definition of equivalent equations is equations that have the same solution set and multiplying both sides by 0 , because 0 is the annihilator. Annihilates all the information with any equation, so it is not allowed.

  64. scarydoor
    • 2 years ago
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    it's not allowed if you want to preserve all the information. But I didn't want to, so it didn't matter. I only wanted to produce a new true equation from an old true equation. The new true equation was all I cared about.

  65. scarydoor
    • 2 years ago
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    you're completely right though, that if you want to solve for some variable or something, but doing equality preserving arithmetic on them, then you don't want to multiply by zero. But in this case, notice that I didn't need to preserve everything.

  66. scarydoor
    • 2 years ago
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    Also, in most cases you wouldn't want to multiply by zero, because it destroys all information. BUT here, we said that there are two things with zero: it makes everything zero, and that 0 * 1/0 = 1. This second one isn't normally the case because we actually can't divide by zero. And so it's only in this "what if we could" scenario that we get some benefit out of multiplying both sides by zero. So I guess the reason that you feel strongly against multiplying by zero, is that in every other case, it's not a good thing to do. .... and the reason for that is because we can't divide by zero.

  67. scarydoor
    • 2 years ago
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    for example, if you have a x = b, and you want to find x. You would multiply by 1/a, to get x. You could also multiply by any other number, except zero, and still have hope of finding x. If you multiplied by 0, then you have destroyed all information (unless I could make the zeroes disappear by divide through by them...)

  68. scarydoor
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    (.... like I assumed you could. and then ended up with 1=0)

  69. skullpatrol
    • 2 years ago
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    Hi @helder_edwin

  70. helder_edwin
    • 2 years ago
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    Hi

  71. helder_edwin
    • 2 years ago
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    I don't believe there's much more to say.

  72. helder_edwin
    • 2 years ago
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    I read everything u people wrote.

  73. skullpatrol
    • 2 years ago
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    I believe the argument: If a/0=c, then a=0*c. Is flawed. @helder_edwin

  74. helder_edwin
    • 2 years ago
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    first of all \(1\neq0\) because otherwise we would have \[ \large a=1\cdot a=0\cdot a=0 \] for every real \(a\). so all the real numbers would be zero and that makes no sense and would render calculus useless.

  75. skullpatrol
    • 2 years ago
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    Agreed.

  76. helder_edwin
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    in order to define division we gotta have multiplicative inverses (i don't the word in english for "recíprocos") i.e. \[ \large \frac{a}{b}=a\cdot b^{-1} \]

  77. helder_edwin
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    provided b is not zero. that is, zero doesn't have multiplicative inverse.

  78. helder_edwin
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    suppose 0 actually has a multiplicative inverse, \(0^{-1}\). then \[ \large \frac{0}{0}=0\cdot0^{-1}=1 \] but also \(0\cdot0^{-1}=0\). so we should have 0=1.

  79. helder_edwin
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    therefore, zero does NOT have multiplicative inverse, so one cannot define division by zero.

  80. skullpatrol
    • 2 years ago
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    Therein lies the problem. I'm saying you can not write any number divided by 0 first. Because multiplication defines division.

  81. helder_edwin
    • 2 years ago
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    regarding the argument (proposition) \[ \large \frac{a}{0}=c\qquad\Rightarrow\qquad a=c\cdot0 \]

  82. skullpatrol
    • 2 years ago
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    Yes, how can you algebraically go from a/0=c --?--> to a=0*c

  83. scarydoor
    • 2 years ago
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    @skullpatrol it sounds like you're saying "I don't believe it's true that we can't divide by zero" so then the argument is "okay suppose we can divide by zero, then..." and then you're saying "but how can you do that? You can't divide by zero..."

  84. helder_edwin
    • 2 years ago
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    maybe you professor was trying to make the point like this: "suppose we could write for \(a\neq0\) \[ \large \frac{a}{0}=c \] then we could infer from this that \[ \large a=c\cdot0=0 \] which contradicts the assumption"

  85. skullpatrol
    • 2 years ago
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    How do you "infer"? I want to know the algebraic steps used, Sir.

  86. helder_edwin
    • 2 years ago
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    "now. even if we let \(a=0\) then we would have \[ \large \frac{0}{0}=c\Rightarrow 0=c\cdot0 \] and this last equation is satisfied by any number c. so we end up having \[ \large \frac{0}{0}=0=1=2=\pi=e^\pi=... \] which is absurd.

  87. scarydoor
    • 2 years ago
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    @skullpatrol do you know how a proof by contradiction works?

  88. skullpatrol
    • 2 years ago
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    yes

  89. skullpatrol
    • 2 years ago
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    a=bc defines a/b=c.

  90. helder_edwin
    • 2 years ago
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    that's the whole point. if u r supposing somehing is true then, all other relevant rules apply. including common algebra.

  91. skullpatrol
    • 2 years ago
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    So we are supposing if a/0=c, then how do you prove a=0*c.

  92. scarydoor
    • 2 years ago
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    You are also supposing that there exists a number 1/0 such that 1/0 * 0 = 1.

  93. helder_edwin
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    u don't.

  94. skullpatrol
    • 2 years ago
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    "u don't." why not?

  95. scarydoor
    • 2 years ago
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    a/0= a*1/0 = c, then a * 1/0 * 0 = c * 0. Then a * 1 = c * 0 = 0. ie a= 0.

  96. helder_edwin
    • 2 years ago
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    because it's algebraic manipulation: \[ \large a\cdot0^{-1}=\frac{a}{0}=c \] \[ \large a=a\cdot0^{-1}\cdot0=c\cdot0 \] there u have a "proof".

  97. skullpatrol
    • 2 years ago
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    But you multiplied both sides by 0

  98. skullpatrol
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    And then left :(

  99. scarydoor
    • 2 years ago
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    if you know that x = y, is it true that x * 0 = y * 0?

  100. scarydoor
    • 2 years ago
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    just finished a maths degree... you? I'm trying to tell you that it's okay to multiply both sides of an equation by zero. I've just taken all undergraduate courses on algebra that my university offers. I think I'm qualified to state that it's okay in this case :p.

  101. scarydoor
    • 2 years ago
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    (the above is a reply to the question by @skullpatrol patrol "what grade are you in?") Thanks for the implication that I don't appear very intelligent, buddy. Maybe I'll stop trying to help you understand it, eh...

  102. scarydoor
    • 2 years ago
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    the question seems to have been deleted just as I answered.

  103. helder_edwin
    • 2 years ago
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    sorry i left. i'm in a place with the worst internet service ever.

  104. helder_edwin
    • 2 years ago
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    we (myself and @scarydoor ) are multiplying by zero because we are using a proof by con

  105. helder_edwin
    • 2 years ago
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    *contradiction

  106. skullpatrol
    • 2 years ago
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    So it is OK to multiply both sides of an equation by 0 even though the final equation is satisfied by any real number.

  107. helder_edwin
    • 2 years ago
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    u never multiply an equation by zero, unless, as u pointed out, u want to loose the "information" the equation provides.

  108. skullpatrol
    • 2 years ago
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    Thank you for confirming that Sir.

  109. helder_edwin
    • 2 years ago
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    @skullpatrol ???

  110. skullpatrol
    • 2 years ago
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    So why are we allowed to do it here? If a/0=c, then a=0*c.

  111. helder_edwin
    • 2 years ago
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    it is a proof by contradiction!!

  112. scarydoor
    • 2 years ago
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    you are always allowed to do it. Just usually it doesn't do you any good. We only do it here because here, we have 1/0 * 0 = 1, because we assumed we could do that, because it's a proof by contradiction. In any other situation we know that dividing by zero destroys all information, and therefore is useless to us (even though it's correct, and the equality is preserved), so we don't do it in any other place. (even though we can. but it never helps us elsewhere)

  113. scarydoor
    • 2 years ago
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    The reason people tell you not to is not because it breaks arithmetic rules. The reason they tell you not to is because it won't help you do anything, where we haven't temporarily altered the rules of arithmetic.

  114. scarydoor
    • 2 years ago
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    That's why I asked you, if y=x, then does 0*y = 0*x? answer that, and then you'll see that there's nothing wrong with the proof. The only problem you find is that, normally you would find no practical value in multiplying by zero. Besides that, it doesn't break any rules to do so.

  115. skullpatrol
    • 2 years ago
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    But this (1/0)*0=1 is already a contradiction of the multiplicative property of 0, that is 0 times any number is 0, not 1.

  116. scarydoor
    • 2 years ago
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    but wasn't your theory that you can divide by zero? That's what you're assuming is true. If you can, then 1/0 * 0 = 1. That's basically the definition of what it means to be able to divide by zero. That's what we assume, and then derive that a = 0 for any number a.

  117. scarydoor
    • 2 years ago
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    Yes, it's a contradiction to the axiom that 0 times any number is 0. So that's one problem. What we show is that another problem, ignoring that first one, is that if you take it to be true then a = 0 for any number a.

  118. helder_edwin
    • 2 years ago
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    again. that's the whole point. u started with something u SUPPOSED to be true, and infered something absurd. ergo, your initial assumption was wrong.

  119. skullpatrol
    • 2 years ago
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    If a/0=c, <----u started with something u SUPPOSED to be true then a=0*c. <----inferred something absurd. ergo, your initial assumption was wrong.a/0=c

  120. helder_edwin
    • 2 years ago
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    yes. u can't divide by zero.

  121. scarydoor
    • 2 years ago
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    the whole of the assumption is: 0 is invertible, and so for any number a, a/0 is defined. ie a/0 = c for some number c. That's the assumption that is proven false.

  122. skullpatrol
    • 2 years ago
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    And it is proven to be false by using the false assumption that 0 is invertible and using 0 as a multipier in transforming the equation.

  123. skullpatrol
    • 2 years ago
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    If a/0=c, <----u started with something u SUPPOSED to be true 0*(a/0)=0*c 0*(1/0)*a=0*c 1*a=0*c then a=0*c.

  124. scarydoor
    • 2 years ago
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    yep. read a book on abstract algebra maybe.

  125. skullpatrol
    • 2 years ago
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    How can you prove something is false using a false assumption?

  126. scarydoor
    • 2 years ago
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    when I asked if you know how a proof by contradiction works, you said you do... You should go and read about it. I might not be in a high enough grade for you to take me seriously, afterall.

  127. helder_edwin
    • 2 years ago
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    the discussion was very interesting. i hope we were helpful.

  128. skullpatrol
    • 2 years ago
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    It just seems so obvious that 0 times any number is 0. And then to use 0*(1/0) = 1 feels completely wrong!!! In my opinion :(

  129. skullpatrol
    • 2 years ago
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    I guess the lesson to learn here is "1/0" is not any number.

  130. scarydoor
    • 2 years ago
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    http://en.wikipedia.org/wiki/Proof_by_contradiction

  131. skullpatrol
    • 2 years ago
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    Thank you for your patience and guidance.

  132. DarkGhost
    • 2 years ago
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    I'll just leave this here http://www.khanacademy.org/math/arithmetic/number-properties/v/why-dividing-by-zero-is-undefined

  133. jayz657
    • 2 years ago
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    for example if you have 2 apples and you are supposed to split it with 0 people, it is not possible because there is no one to split it with

  134. skullpatrol
    • 2 years ago
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    You could say the answer to that question is nobody gets the 2 apples.

  135. jayz657
    • 2 years ago
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    no i mean you are required to split it to 0 people, it is not possible since you cannot split it with 0 people 2/0 = undefined

  136. skullpatrol
    • 2 years ago
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    Here: 2/0 "=" undefined you are using the "=" for "is" What I'm saying is 2/0 means nobody gets the 2 apples.

  137. jayz657
    • 2 years ago
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    what im saying is that you cannot just say no one gets the 2 apples if you are required to divide among 0 people

  138. skullpatrol
    • 2 years ago
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    If there is nobody around and you are required to divide 2 among 0 people what happens to the apples?

  139. jayz657
    • 2 years ago
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    it becomes undefined since its not possible to do so

  140. jayz657
    • 2 years ago
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    undefined is basically we dont know what happens

  141. skullpatrol
    • 2 years ago
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    Have you fulfilled your own requirement: "if you are required to divide "

  142. jayz657
    • 2 years ago
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    your question was why can you not divide by zero? so the requirement was to divide n by 0 and i just made an example by using n=2 to show you that it is undefined when you do so

  143. skullpatrol
    • 2 years ago
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    But I asked "why"?

  144. jayz657
    • 2 years ago
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    i showed you why you cannot split 2 apples among 0 people since there is no one to split it with

  145. jayz657
    • 2 years ago
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    so what happens to the 2 apples? we dont know so we call that undefined

  146. skullpatrol
    • 2 years ago
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    But you said "it is undefined ***when you do so*** " so that means it ***can be done***?

  147. jayz657
    • 2 years ago
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    the apples are divided we just dont know how they are divided so we use the term undefined to answer it

  148. skullpatrol
    • 2 years ago
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    The apples are divided by 0?

  149. jayz657
    • 2 years ago
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    yea when you divide the apples by 0 we will get undefined since we dont know how they are divided

  150. skullpatrol
    • 2 years ago
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    I asked why can you NEVER divide by 0

  151. jayz657
    • 2 years ago
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    like i said your question was why can you never divide by 0 so i used an example of n = 2 we'll just use n then if you have n number of items and you want to split them among 0 people then you cannot because there is no one to split n items with, and we do not know how n is divided so we get an answer called undefined

  152. jayz657
    • 2 years ago
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    no matter what n is if you have no one to split it with, we will never know how n is divided

  153. skullpatrol
    • 2 years ago
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    That is what I said: no one gets the n. That is the answer. It is not undefined because nobody gets the apples is a well defined sentence.

  154. jayz657
    • 2 years ago
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    i didnt say no one gets the n i said we do not know how n is divided

  155. skullpatrol
    • 2 years ago
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    We do know how n is divided: "It is not divided" because nobody is there.

  156. jayz657
    • 2 years ago
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    it looks like you are making assumptions here you are saying since we dont know how n is divided you are assuming that n does not get divided instead, just because no one is there doesnt mean it does not get divided it means we dont know how it is divided so we say it is undefined

  157. skullpatrol
    • 2 years ago
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    If no one is there we do know nothing will happen to the apples.

  158. jayz657
    • 2 years ago
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    again you are making assumptions how can you be so sure nothing will happen to the apples?

  159. skullpatrol
    • 2 years ago
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    How can you be so sure we don't know?

  160. jayz657
    • 2 years ago
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    because it is more reasonable to say that we do not know rather that making an assumption and saying nothing happens

  161. jayz657
    • 2 years ago
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    you made a conclusion by saying nothing happens whereas i say we dont know, which means we havent come to a conclusion yet, so we just give it the term undefined to say we do know what happens

  162. jayz657
    • 2 years ago
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    dont know* (typo)

  163. skullpatrol
    • 2 years ago
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    So your answer is "we don't know" why we can never divide by 0.

  164. jayz657
    • 2 years ago
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    no as i said before we do not know how it is divided so we say its undefined

  165. jayz657
    • 2 years ago
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    since if we ever do divide by 0 do not know what happens so you are better off just not dividing by zero

  166. mayankdevnani
    • 2 years ago
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    @skullpatrol http://www.math.utah.edu/~pa/math/0by0.html

  167. sunnymony
    • 2 years ago
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    when u dived by zero so ans will be infinite.means undefind which has no meaning

  168. mayankdevnani
    • 2 years ago
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    You cannot divide by zero because there is no meaningful result to that operation. No element of any given set of numbers - real, imaginary or otherwise - would fulfil this equation. Infinity is not a defined number, it is merely a symbol meaning "exceeding all values". You cannot use it to perform calculations like these. ok @skullpatrol

  169. skullpatrol
    • 2 years ago
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    So each friend doesn't get any apple is the answer?

  170. skullpatrol
    • 2 years ago
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    3/6 = 1/2 each 2/6 = 1/3 each 1/6 = 1/6 each 0/6 = 0 apples each

  171. skullpatrol
    • 2 years ago
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    "If you have 3 apples, and six friends, you can divide each apple in half that way, each friend gets half an apple."

  172. skullpatrol
    • 2 years ago
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    Are you talking about amount of apple per friend?

  173. skullpatrol
    • 2 years ago
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    So you are counting the number of pieces each friend gets.

  174. skullpatrol
    • 2 years ago
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    6/3 =half each friend 6/2 =3rd each friend 6/1 =6th each friend

  175. zostale
    • 2 years ago
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    true...its undefined. Watch the khan academy tutorial on it. Pretty logical and interesting.

  176. StupidGenius
    • 2 years ago
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    i want say something, we can't divided by zero but if my problem want to divided by zero ?! we know the velocity equal to difference between density over difference between time? \[\frac{ v2-v1 }{ t2-t11 }\] question is what is the Instantaneous velocity? i.e find the velocity when t2 =t1?

  177. StupidGenius
    • 2 years ago
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    answer is we can't put t2=t1 directly but we but it by limit at t2 tents to t1

  178. satellite73
    • 2 years ago
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    you can't divide by zero for the same reason that it make no sense to ask how many miles per gallon i get on my bicycle \(a\div b=c\iff bc=a\) try solving \(a\div 0=b\iff 0\times c=a\) for \(c\) and non zero \(a\) for example try solving \[2\div 0=c\iff 0\times c=2\] for \(c\) and you see that you cannot

  179. satellite73
    • 2 years ago
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    typo there, i means try solving \[a\div 0=c\iff 0\times c=a\] for \(c\)

  180. skullpatrol
    • 2 years ago
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    hmm... shouldn't it be a=bc defines a/b=c?

  181. skullpatrol
    • 2 years ago
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    @satellite73

  182. skullpatrol
    • 2 years ago
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    That is a=bc ---> a/b=c not a/b=c---> a=bc which you are combining to give a/b=c<--->a=bc

  183. geoffb
    • 2 years ago
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    Chuck Norris can divide by zero.

  184. skullpatrol
    • 2 years ago
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    That^ is all this thread needed

  185. StupidGenius
    • 2 years ago
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    you right because we con divided by zero @satellite73 but if i have to divide by value tends to zero what shall i do?

  186. scarydoor
    • 2 years ago
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    @StupidGenius , if you're saying you're dividing 1 / n, and taking the limit as n gets closer to zero, then that is a limit that doesn't converge to any real number. It's just a case of a limit that doesn't converge to any real number. There are lots of them. That's all covered in the area of maths called real analysis.

  187. scarydoor
    • 2 years ago
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    first line, should have said: "dividing 1 by n (i.e. 1/n)"

  188. StupidGenius
    • 2 years ago
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    yes i know some real analysis

  189. Vogelfrei
    • 2 years ago
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    http://en.wikipedia.org/wiki/Division_by_zero

  190. radar
    • 2 years ago
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    Just remember dividing by zero is done at your own risk, I choose to avoid doing such a thing.

  191. StupidGenius
    • 2 years ago
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    real analysis study the series and sequence at infinite terms and it's study to seek if the valve gives infinite or number (divergence and convergence )

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