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Few days ago, i was attending lecture of Introduction to Computers (ITC) in my University and there was one question. **

What is 1/0 or 1(divide-by) 0.

** I checked it on my phone and it says 1/0 is infinite and my Professor said that it is not defined there is not answer for 1/0. I am totally confused in both. I think this forum is the best place to get the answer.
I really want to know the answer because every time i try to calculate this following equation. I always think of 2 answers. Waiting for your precious replies!

Meer Faisal Ali
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    Your professor is right, it's not defined. – Thomas Nov 09 '18 at 19:56
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    If $\dfrac 1 0 =a$ for some $a$, then $1 = a \times 0$. But this is impossible; thus, there is no number $a$ such that $\dfrac 1 0 =a$. – Mauro ALLEGRANZA Nov 09 '18 at 19:57
  • It’s undefined. Some people say infinity because $\frac{1}{x}$ gets unboundedly large as $x$ approaches $0$. That’s not correct, since it doesn’t actually reach $0$. Giving a value of $n$ to $\frac{1}{0}$ would essentially mean $0\cdot n = 1$, which no value of $n$ satisfies. – KM101 Nov 09 '18 at 20:06
  • @KM101 what about the limit of $1/x$ as $x$ approaches zero? – Phil H Nov 09 '18 at 20:09
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    It’s undefined/doesn’t exist since it depends on whether it’s $x \to 0^+$ or $x \to 0^-$. – KM101 Nov 09 '18 at 20:11

2 Answers2

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Your professor is right.

Dividing by zero is not defined.

You may think of the limit of $\frac {1}{x} $ as $x$ approaches $0$ and that does not exist because depending on which side of $0$ you are you may get a very large positive or very large negative number.

For example $$\frac {1}{0.0001} = 10,000$$ while $$\frac {1}{-0.0001}=-10,000$$

Mohammad Riazi-Kermani
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The division by zero is not defined. I know it is awful but... Anyway, the definition of field provided a reason that was enough for me:

A field is a set $K$ endowed with two different operations $+$ and $cdot$ (if you know what a field is, skip this part): With respect to $+$, $K$ is an abelian group; with respect the other, $K$ is just a semigroup but, if $0$ is the identity element for the first one, then $K\setminus\{0\}$ is another abelian group. $+$ and $\cdot$ must satisfy other conditions too, but it does not matter for our purpose. The key is that the second operation makes $K$ a group only on when we consider $K\setminus\{0\}$; hence we are not allow to ask who is the inverse of $0$ with respect to $\cdot$ because it is not defined.

Then, $\mathbb R$ is a field with respect to the usual sum and usual multiplication. The consequence is that we are not allowed to ask for the inverse element of the identity element of the sum, i.e. we are not allowed to ask for the inverse of $0$ with respect to the multiplication.

The answer $1/0=\infty$ arises from the fact that $\lim_{x\to0} 1/x=\infty$. And thi is not actually true because depending if we take the limit by the right or by the left we take two different results.

Dog_69
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