comment: >
  
  What can be divided can also be multiplied.
  1x1=1
  1x2=2
  1x0=0
  0x0=...1?
  We have somehow created something from nothing.  We shouldn't be able to do that.
  Zero is an expression of nothingness.  There is no defined quantity of nothing within nothingness.  Nothingness can't be positively expressed because it is absence. It is defined by the inability to be positively expressed.
  Put differently: how many zeros fit in zero?  The answer is all of them.
context: If 0/0=1, then we have:
  2*0 = 0
  (2*0)/0 = 0/0
  2*(0/0) = 0/0
  2*1=1
  2=1
  Letting 0/0 = 1 would result in a lot of contradictions with the rest of mathematics. >
  
  So why is it fine to multiply a number by 0?
  Like, it seems like the major issue is allowing the use of 0 either way for a multiplication or division. I never got the whole n*0=0 thing because it’s the only subset of multiplications that isn’t reversible, unlike a*b=c and a=c/b or b=c/a. Like 0 fundamentally beaks the symmetry between multiplication and division, but we just accept it’s okay for multiplication for reasons. >
  
  Because multiply by zero DOES give you zero. You can have 2 groups of nothing. You can’t divide something by nothing.
  I’m not going to find the actual calculus behind it, but think of numbers very close to zero in both cases.
  2*.000001=a number very close to zero.
  2/.0000001=a big number, and the more decimals you add the closer to infinity you get. It would never converge on an actual answer.
  Edit: the way I think of it is dividing by zero is kind of like multiplying by infinity. Might not be the most accurate but it’s how I’ve always thought of it. >
  
  Like how do you physically group nothing, and now how do you “have” two groups of nothing, now add those groups of nothing together, and now split that big pile back into two groups and count the things you have. Congrats: that’s multiplication AND division with 0. 
  0xGroups=0, 0/0=Groups. Groups could be any number. It breaks down with both multiplication and division. 
  It works one way but not the other because we say it does, but that’s not a why.
  2/0 makes a bit more sense, but still seem arbitrary to say it’s fine for infinitly small numbers but not for infinitely big numbers. >
  
  The reason is because defining it causes contradictions and is nonsensical. Suppose you defined a number G and said it was a number (for all intents and purposes its infinity), this division cannot be reversed via multiplication without causing contradictions, as shown multiple times in this thread. If you want to say this is a number, then many general statements about numbers can no longer be made. For example if a > b, then a + x > b + x, this is no longer true if x is G.
  Forcing x/0 to be defined as a number essentially prevents you from doing most of common algebra. So it makes way more sense to just leave x/0 undefined and let the rest of the numbers operate as per usual. >
  
  Right but a big part of why n/0 causes contradictions is because n*0 is defined. Having a multiplication that you can't reverse with division is already a contradiction. Already it means that a*b=c but c/b!=a when a or b are 0, breaking equivalence. >
  
  
  Having a multiplication that you can't reverse with division is already a contradiction.
  
  That's simply not true. A contradiction occurs when two propositions that cannot simultaneously be true are found to be true. There is no contradiction in n*0 = 0 being defined.
  
  Already it means that a*b=c but c/b!=a when a or b are 0, breaking equivalence.
  
  If you establish than n/0 is undefined, then in this instance you have one exception. If you also want to asset that n0 is undefined, you don't add anything and just create a second exception. Logically you would choose the more powerful system with n0 defined and n/0 undefined rather than the weaker one with both undefined. >
  
  You can put an infinite number of 0s into 0. 0+0+0+0+0+...=0, how can there be only one 0 in 0 when I can fit any number of 0s into 0?
  Dividing by 0 doesn't work under the standard axioms. Having 0 in the numerator doesn't change that >
  
  !delta  that… that makes sense. Thank you! >
  
  Why didn’t you just Google this instead of arguing with regards for 5 years? >
  
  Reddit would be pretty barren if we all googled things like this instead of seeking discussion here. >
  
  I disagree, Reddit has many non Google questions and even more non questions entirely. Also who cares if Reddit is barren lol >
  
  I mean, I care because I enjoy using it. I enjoy reading people's questions and the replies. Some of the questions are basic or easily searchable but not something I would have thought to ask. I don't disagree about the content of other posts and didn't think I was alluding to that. >
  
  Hey it’s cool bro I don’t truly care if you ask basic questions here or anywhere else squirt em out all over the place. I think it’s weird but that’s a personal problem >
  
  Let's do a proof by contradiction.
  We start by assuming 0/0 = 1
  Then that means 0/0 + 0/0 = 1 + 1 = 2
  But we also know x/z + y/z = (x+y)/z
  So 0/0 + 0/0 = (0+0)/0 = 0/0
  But if 0/0 = 1; and 0/0 + 0/0 = 2; and 0/0 + 0/0 = 0/0
  Then that means 2 = 1 unless our assumption that 0/0 = 1 is false. >
  
  This guy !deltas >
  
  This delta has been rejected. The length of your comment suggests that you haven't properly explained how /u/CrackaBox changed your view (comment rule 4).
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  ^Delta System Explained ^| ^Deltaboards >
  
  !delta  This is a solid mathematical proof to show that if we assume 0/0=1 that an impossible outcome occurs. Therefore the assumption of 0/0=1 is false. >
  
  Sheeiit look what you get for trying to be a nice guy >
  
  Mathematician here.
  I think that the issue at core here is understanding the division sign. Division is the opposite operation of multiplication, same as subtraction is the opposite of addition. In other words, when I write a-b, I essentially mean: “it’s a number x such that b+x=a”. So, when you write 0/0, it should be a number x such that 0*x=0. Any real number x would work here. (Remark that if there is no 0 in the denominator, the answer for x is always unique.)
  So, in principle, you could have declared 0/0 to be any number, the definition of the division operation would hold completely; but this would break the nice properties of the previous operations. For starters
  1/0 = (1+0)/0 = 1/0 + 0/0 = 1/0 + 1
  In other words, 1/0 cannot be defined as a number. We used the distributive property; we could assume that it does not work specifically for 0/0, but what’s the purpose of declaring an operation result something and then doing it an exception of all rules? You do not create anything new arithmetically speaking.
  Worse than that, as other commentators mentioned,
  1 = 0/0 = (0+0)/0 = 0/0 + 0/0 = 2, and that is a much bigger problem, as it means that you can’t define 0/0 and keep numbers staying distinct at the same time.
  Weirdly enough, this can be generalized: whatever structure you have with “nice” addition and multiplication operations, the neutral addition element (0) can never be invertible with respect to multiplication.
  Edit: trickier exercise: why would we not set 0/0=0? >
  
  You explained the group of invertible elements in a field in passing. Great answer. >
  
  the invertible elements of a field do not form a ring >
  
  Apologies. You are correct. >
  
  they do make a multiplicative set >
  
  Sure. But not a ring though. Sum of invertibles need not be invertible. They indeed form a group and I have amended my comment to say the same. >
  
  actually i guess the units of the trivial ring do form a ring