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skullpatrol

Why can you never divide by zero?

  • one year ago
  • one year ago

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  1. vac33
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    Because the answer would be undefined.

    • one year ago
  2. skullpatrol
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    What do you mean?

    • one year ago
  3. sthop456
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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]

    • one year ago
  4. vac33
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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.

    • one year ago
  5. zepp
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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.

    • one year ago
  6. sirm3d
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    |dw:1353147793508:dw|

    • one year ago
  7. skullpatrol
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    @sirm3d neither

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

    • one year ago
  9. scarydoor
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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.

    • one year ago
  10. scarydoor
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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)

    • one year ago
  11. scarydoor
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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.

    • one year ago
  12. sirm3d
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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

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

    • one year ago
  14. scarydoor
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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.

    • one year ago
  15. scarydoor
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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.

    • one year ago
  16. sirm3d
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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 \[1=\frac{ 5 }{ 4 }= \text{any other number}\]

    • one year ago
  17. scarydoor
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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...

    • one year ago
  18. skullpatrol
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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?

    • one year ago
  19. scarydoor
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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... )

    • one year ago
  20. sirm3d
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    @hartnn share your thoughts. =)

    • one year ago
  21. scarydoor
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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.

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

    • one year ago
  23. hartnn
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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

    • one year ago
  24. sirm3d
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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.

    • one year ago
  25. skullpatrol
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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

    • one year ago
  26. scarydoor
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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.

    • one year ago
  27. scarydoor
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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.

    • one year ago
  28. scarydoor
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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...)

    • one year ago
  29. sirm3d
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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

    • one year ago
  30. scarydoor
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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

    • one year ago
  31. scarydoor
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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).

    • one year ago
  32. skullpatrol
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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.

    • one year ago
  33. scarydoor
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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.

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

    • one year ago
  35. scarydoor
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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?

    • one year ago
  36. skullpatrol
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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.

    • one year ago
  37. scarydoor
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    I don't understand what you're saying.

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

    • one year ago
  39. skullpatrol
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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

    • one year ago
  40. scarydoor
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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).

    • one year ago
  41. scarydoor
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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..

    • one year ago
  42. skullpatrol
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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.

    • one year ago
  43. scarydoor
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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.

    • one year ago
  44. scarydoor
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    Are you convinced?

    • one year ago
  45. skullpatrol
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    no

    • one year ago
  46. skullpatrol
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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.

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

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

    • one year ago
  49. scarydoor
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    That's right.

    • one year ago
  50. skullpatrol
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    Now transform that original equation by multiplying both sides by 0.

    • one year ago
  51. scarydoor
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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.

    • one year ago
  52. scarydoor
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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.

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

    • one year ago
  54. scarydoor
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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.

    • one year ago
  55. scarydoor
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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.

    • one year ago
  56. scarydoor
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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.

    • one year ago
  57. scarydoor
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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.

    • one year ago
  58. scarydoor
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    Does that make sense?

    • one year ago
  59. scarydoor
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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).

    • one year ago
  60. skullpatrol
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    I agree :)

    • one year ago
  61. scarydoor
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    haha, good.

    • one year ago
  62. scarydoor
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    algebra is fun.

    • one year ago
  63. skullpatrol
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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.

    • one year ago
  64. scarydoor
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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.

    • one year ago
  65. scarydoor
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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.

    • one year ago
  66. scarydoor
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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.

    • one year ago
  67. scarydoor
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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...)

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

    • one year ago
  69. skullpatrol
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    Hi @helder_edwin

    • one year ago
  70. helder_edwin
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    Hi

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

    • one year ago
  72. helder_edwin
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    I read everything u people wrote.

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

    • one year ago
  74. helder_edwin
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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.

    • one year ago
  75. skullpatrol
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    Agreed.

    • one year ago
  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. \[ \large \frac{a}{b}=a\cdot b^{-1} \]

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

    • one year ago
  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.

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

    • one year ago
  80. skullpatrol
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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.

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

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

    • one year ago
  83. scarydoor
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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..."

    • one year ago
  84. helder_edwin
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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"

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

    • one year ago
  86. helder_edwin
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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.

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

    • one year ago
  88. skullpatrol
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    yes

    • one year ago
  89. skullpatrol
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    a=bc defines a/b=c.

    • one year ago
  90. helder_edwin
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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.

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

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

    • one year ago
  93. helder_edwin
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    u don't.

    • one year ago
  94. skullpatrol
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    "u don't." why not?

    • one year ago
  95. scarydoor
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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.

    • one year ago
  96. helder_edwin
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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".

    • one year ago
  97. skullpatrol
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    But you multiplied both sides by 0

    • one year ago
  98. skullpatrol
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    And then left :(

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

    • one year ago
  100. scarydoor
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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.

    • one year ago
  101. scarydoor
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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...

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

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

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

    • one year ago
  105. helder_edwin
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    *contradiction

    • one year ago
  106. skullpatrol
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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.

    • one year ago
  107. helder_edwin
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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.

    • one year ago
  108. skullpatrol
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    Thank you for confirming that Sir.

    • one year ago
  109. helder_edwin
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    @skullpatrol ???

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

    • one year ago
  111. helder_edwin
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    it is a proof by contradiction!!

    • one year ago
  112. scarydoor
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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)

    • one year ago
  113. scarydoor
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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.

    • one year ago
  114. scarydoor
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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.

    • one year ago
  115. skullpatrol
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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.

    • one year ago
  116. scarydoor
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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.

    • one year ago
  117. scarydoor
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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.

    • one year ago
  118. helder_edwin
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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.

    • one year ago
  119. skullpatrol
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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

    • one year ago
  120. helder_edwin
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    yes. u can't divide by zero.

    • one year ago
  121. scarydoor
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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.

    • one year ago
  122. skullpatrol
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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.

    • one year ago
  123. skullpatrol
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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.

    • one year ago
  124. scarydoor
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    yep. read a book on abstract algebra maybe.

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

    • one year ago
  126. scarydoor
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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.

    • one year ago
  127. helder_edwin
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    the discussion was very interesting. i hope we were helpful.

    • one year ago
  128. skullpatrol
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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 :(

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

    • one year ago
  130. scarydoor
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    http://en.wikipedia.org/wiki/Proof_by_contradiction

    • one year ago
  131. skullpatrol
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    Thank you for your patience and guidance.

    • one year ago
  132. DarkGhost
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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

    • one year ago
  133. jayz657
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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

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

    • one year ago
  135. jayz657
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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

    • one year ago
  136. skullpatrol
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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.

    • one year ago
  137. jayz657
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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

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

    • one year ago
  139. jayz657
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    it becomes undefined since its not possible to do so

    • one year ago
  140. jayz657
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    undefined is basically we dont know what happens

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

    • one year ago
  142. jayz657
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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

    • one year ago
  143. skullpatrol
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    But I asked "why"?

    • one year ago
  144. jayz657
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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

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

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

    • one year ago
  147. jayz657
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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

    • one year ago
  148. skullpatrol
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    The apples are divided by 0?

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

    • one year ago
  150. skullpatrol
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    I asked why can you NEVER divide by 0

    • one year ago
  151. jayz657
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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

    • one year ago
  152. jayz657
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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

    • one year ago
  153. skullpatrol
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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.

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

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

    • one year ago
  156. jayz657
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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

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

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

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

    • one year ago
  160. jayz657
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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

    • one year ago
  161. jayz657
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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

    • one year ago
  162. jayz657
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    dont know* (typo)

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

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

    • one year ago
  165. jayz657
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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

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

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

    • one year ago
  168. mayankdevnani
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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

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

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

    • one year ago
  171. skullpatrol
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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."

    • one year ago
  172. skullpatrol
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    Are you talking about amount of apple per friend?

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

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

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

    • one year ago
  176. StupidGenius
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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?

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

    • one year ago
  178. satellite73
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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

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

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

    • one year ago
  181. skullpatrol
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    @satellite73

    • one year ago
  182. skullpatrol
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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

    • one year ago
  183. geoffb
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    Chuck Norris can divide by zero.

    • one year ago
  184. skullpatrol
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    That^ is all this thread needed

    • one year ago
  185. StupidGenius
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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?

    • one year ago
  186. scarydoor
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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.

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

    • one year ago
  188. StupidGenius
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    yes i know some real analysis

    • one year ago
  189. Vogelfrei
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    http://en.wikipedia.org/wiki/Division_by_zero

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

    • one year ago
  191. StupidGenius
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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 )

    • one year ago
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