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It's approximate value, its infinite I know it but I want to know atleast the value upto $7$ decimal values.

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marked as duplicate by Yiorgos S. Smyrlis, Nick Peterson, Zev Chonoles, Najib Idrissi, Michael Grant Feb 13 '14 at 14:37

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The number $1/0$ does not exist. Division is only defined for non-zero numbers. – 5xum Feb 13 '14 at 13:59
It is not defined. Like to ask how much is 5+banana. – LeeNeverGup Feb 13 '14 at 14:00
From a mathematical point of view this is not very interesting, but you may run into issues like this if you try to do limited precision computations. Example that will work on many softwares using floating point numbers: 1/(0.9-0.3-0.3-0.3) – Dennis Jaheruddin Feb 13 '14 at 14:44

Division by zero is undefined. So $\dfrac 10$ does not exist.

You are mixing up limits with numbers. That is, we can meaningfully speak of $\lim_{x \to 0} \dfrac 1x$, but $$\lim_{x \to 0^+} \frac 1x \neq \frac 10$$

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perfect answer..+1 – Apurv Feb 13 '14 at 14:15
Thanks, @Apurv! – amWhy Feb 13 '14 at 14:16
@amWhy. Division by $0$ is forbidden by law. – Claude Leibovici Feb 13 '14 at 14:36
Actually, this isn't quite right. We can meaningfully speak of $\displaystyle\lim_{x\rightarrow 0^{+}}\frac1x$, meaning approaching from the right (so $x>0$), and we can meaningfully speak of $\displaystyle\lim_{x\rightarrow 0^{-}}\frac1x$, meaning approaching from the left (so $x<0$). But these limits are different, $+\infty$ and $-\infty$ respectively, so we cannot just define the limit. We need to be more careful. – user1729 Feb 13 '14 at 15:27
indeed, @user1729. – amWhy Feb 13 '14 at 15:30

I admit that this question troubled me also as a kid.. But I found this explanation satisfactory:

Saying $c=\dfrac {a}{b}$ means calculating the number of chocolates each donkey will receive if $a$ chocolates are divided among $b$ donkeys. But, if the number of donkeys itself is zero, we cannot divide the $a$ chocolates in any definite way, or the question of division would be ruled out, or division would become a meaningless word in the whole context. In other words, division would become undefined. Hence $c$ would become undefined. Therefore division by zero is not defined.

P.S. For a more mathematical answer, see amWhy's answer...

For further explanations, see this and this.

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+ Nice personal touch, Apurv! – amWhy Feb 13 '14 at 14:37
@amWhy, thanks... – Apurv Feb 13 '14 at 14:38

Saying that $1/0$ is "infinite" is a sloppy way to hide to children the hard truth :


The effect is the same with Santa Claus : when you are young your parents say you that he lives "far there ... in the northern countries", that he travel only by night, that he come in your house through chimneys ... all this in order to avoid explanations regarding the fact that you are never able to see and touch it.

We may say that Santa Claus lives in a "point at infinity".

The crude truth (and now you have discovered it) is that the division is not everywhere defined ... nobody one, in two o three millenia of hard study of math has been able to find it.

Of course, I'm joking !

The "exact" answer is that there is no reason why all "operations" must be performable in all possible cases.

In case of division, to divide by $1$ is already not to divide at all; so, we cannot imagine a way of dividing that is "less dividing" that not to divide at all ...

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LOL! Santa Claus! Are you serious this site is about Math? Anyways excellent answer..+1 – Apurv Feb 13 '14 at 14:14
OMG, you mean Santa doesn't exist? I thought this site was safe for children. – Carry on Smiling Feb 13 '14 at 14:25
@user4140, Santa Claus exists, but he lives in a point at infinity. – Apurv Feb 13 '14 at 14:29
@ClaudeLeibovici - You are right; we need at least a proof that the existence of Santa Claus implies a contradiction ... I suppose that this will be enough for the guys of this site. – Mauro ALLEGRANZA Feb 13 '14 at 14:34
@neofoxmulder, correct.. I found one such proof-See this – Apurv Feb 13 '14 at 15:38

$\frac{1}{b}$ is the number such that $\frac{1}{b}\cdot b=1$ (this is the definition for any inverse)

However $x\cdot 0= 0$ so $0$ does not have an inverse. This implies $\mathbb R$ does not form what mathematicians call a group under multiplication.

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You might specify that $\mathbb{R}$ with multiplication isn't a group. – Thomas Feb 13 '14 at 14:35
I agree, that could confuse people, since $\mathbb{R}$ with addition is a group. – Joachim Feb 13 '14 at 15:07

First the set of real numbers $\mathbb{R} = (-\infty, \infty)$ does not contain $\infty$. $\infty$ is not a number.

When you have two numbers $a$ and $b$, we have multiplication of the two numbers. We can, in fact, multiply any two numbers. We can also add and subtract any two numbers. When I perform any of these operations, then I always end up with a (real) number.

This is different for division. You can't divide any two given numbers. You can't divide by zero. $\frac{7}{0}$ is not a number nor is it a well-defined expression.

One way to think about this is that the equation $$ a = bx $$ should have the solution $\frac{a}{b}$. But if $b =0$ (and $a\neq 0$), then the equation becomes $$ a = x\cdot 0 $$ But if $a\neq 0$, then we don't have a solution since the product of any number with zero is zero. (This isn't really the right way to say it since you can point out that when $a = b = 0$ then there is a solution, but maybe a way to think about it.)

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Why the downvote? – Thomas Feb 13 '14 at 15:20
then I'll convert it to an upvote...+1 – Apurv Feb 13 '14 at 15:23

I certainly do not want to provide ammunition to jokers but the OP did not ask for an 'exact' answer so in this sense it is meaningful to answer

$$ \lim_{x \to 0^+} \ \frac{1}{x} \ = \ + \infty $$

$$ \lim_{x \to 0^-} \ \frac{1}{x} \ = \ - \infty $$

To simply say 'it does not exist' does not give the maximum amount of information available , IMHO.

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The nice thing about mathematics is that questions like these do have concrete answers. The answer is that $1/0$ isn't defined. So I would make sure to point that out as a first answer. Otherwise it might give the impression that you believe the expressions is defined. After that you could point out the thing about limits. – Thomas Feb 13 '14 at 14:49
... ... so, due to the fact that division is the inverse operation respect to multiplication, the stipulation must "elect" a number $\alpha$ such that $\alpha \times 0 =1$... and this is the problem we have not yet solved. Please, remember that the expression "lim ... $= ∞$ is like $\forall x ...$: in that context, you cannot separate the different components of the symbol and read it as if $∞$ is a number like $1$. 2/2 – Mauro ALLEGRANZA Feb 13 '14 at 15:01
Mixing up $\lim_{x\to 0} \frac 1x$ with $\frac 10$ is a huge hurdle folks need to get past. Note that I addressed (in my answer, long before yours) that it IS meaningful to speak of such a limit, but NOT meaningful to draw the conclusion $\exists x: x = \frac 10$. – amWhy Feb 13 '14 at 15:11
If you find the OPs answer meaningful, then why not provide him/her with a value approximated to 7 digits? – amWhy Feb 13 '14 at 15:15
Not a problem, neofoxmulder. I agree that dismissing the post outright is not at all helpful to the OP, hence my post, etc. I'm a bit of an unorthodox mathematician, in the sense that I find too many mathematicians eager to bite students' heads off, rather than trying to meet students where they are at, and trying to help them move forward. So I do sympathize with you. – amWhy Feb 13 '14 at 15:35

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