# 1 divided by zero

## 1 divided by zero=?

• ### Other(explain):

• Total voters
13
• Poll closed .

## Main Question or Discussion Point

OK recently in class some of my classmates were debating whethter 1 divided by zero is undefined or infinity. I saw both points but im wondering what everyone else out there thinks. So 1/0= undefined or infinity or does it depend on other stuff?

matt grime
Homework Helper
Oh dear God, not again......

Icebreaker
Dividing by zero is undefined. There are ways around that using limits; however, in those cases, you are not dividing by zero, per se.

Alkatran
Homework Helper
lim(x->0)(1/x) = undefined
lim(x->0+)(1/x) = infinity
lim(x->0-)(1/x) = -infinity

Take your pick. It all depends on the question.

dextercioby
Homework Helper
The interesting part is that he posted a poll...He probably meant division.All operations involving infinities & dividing through 0 are undefined...

Daniel.

arildno
Homework Helper
Gold Member
Dearly Missed
I voted other: 0/1=0.

Icebreaker
This reminds me; if you ever see a book called "Zero: The Biography of a Dangerous Idea" and want to read it, be careful. The author frequently uses expressions like "dividing by zero/infinity". One of the passages goes like this: "[...]dividing anything by infinity gives zero; dividing anything by zero gives infinity. We can see the relationship these two numbers have throughout[...]"

http://www.chapters.indigo.ca/item.asp?Item=978014029647&Catalog=Books&N=35&Lang=en&Section=books&zxac=1

I am told, however, that the historical account of zero of the book is rather fascinating.

dextercioby
Homework Helper
Don't read the book.Division under the reals has the property

$$a:b=c\Rightarrow a=b\cdot c$$

You'll see that in the case of 0 or infinity,it doesn't work...(multivaluedness).

Daniel.

HallsofIvy
Homework Helper
Define your terms. If you are working in the standard real number system, division of any number is "undefined"-literally. However, some people use "infinity" as a shorthand for "undefined": saying $\lim_{x->0} \frac{1}{x}= \infty$ is precisely the same as saying "the limit does not exist".

matt grime said:
Oh dear God, not again......
Yes, inevitably.
We must make sense out of the expression EyEx(x/0 = y).
Is it tautologous or is it contradictory?

i.e. is (1/0) a number?, or, is (0/0) a number?, or, is (2/0) a number? etc..

dextercioby
Homework Helper
As Halls said,how do you define your concepts:the concept of "number"...?

Daniel.

arivero
Gold Member
When x=0, the expression 1/x="multiplicative inverse of x"="the number y such that x times y equals unity" is a lot less ambiguous that 0/x, "the number y such that x times y equals zero".

dextercioby
Homework Helper
Yes,but when u see the overall picture,they're both faulty.

Daniel.

i think, this question can take place on a math-FAQ.

people who think about division with zero seem to forget the meaning of division.
steps of division are, you subtract the divisior from the divdend as long as there's something remaining dividend, ie it hits 0. for each exact subtraction, add 1 to the counter (ie, division). (i'm not covering the details for fractional parts, since my purpose in telling this is to remind the definition)

so, applying this definition to 1/0: subtract 0 from 1 as long as you have a non-0, and add 1 to the counter each time. now, if you do this for 1/0, what's your counter?

similar idea can be applied to 0/0: subtract 0 from 0 until as long as you have a non-0. now, indefinitiny approcahes at this point. no matter how you repeat this (and remember that you're adding a 1 to the result in each step), it'll be valid against the definition of division. 0/0 = 5, 0/0 = 9, 0/0 = anything.

an application can be: $$x.0 = 0$$, this equation true for every value of x, ie, it's independent from the contents of x, as long as it's a valid number. dividing two sides by 0, you get $$x.\frac{0}{0} = \frac{0}{0}$$. an equation independent from the value of x. it's pretty weird at the first glance, but if you remember what division is, the explanation reveals itself.

i'd like to emphasize that division by zero in not undefined, it's indefinite (or maybe uncertain is a better word), and there're problem-dependent methods for getting rid of the indefinition.

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arildno
Homework Helper
Gold Member
Dearly Missed
fdarkangel:
Euclid's algorithm has nothing to do with the fact that the multiplicative inverse of 0 is undefined.

I saw an interesting "proof" that illustrates the problems of dividing by zero:

a=b
a²=ab
a²-b²=ab-b²
(a-b)(a+b)=b(a-b)
(a-b)(a+b)/(a-b)=b(a-b)/(a-b)
a+b=b
2b=b
2=1

The reason it comes out wrong is because (a-b)=0

Kinda Fun.

:D

The Rev

jcsd
Gold Member
We've had this question many times before:

It cna easily be shown in any ring that:

a*0 = a(b - b) = a(b - b) + 0 = a(b - b) + ab - (ab) = a(b - b + b) - (ab) = ab - (ab) = 0

i.e. a*0 = 0 for all 'a';

define divison by:

a = b/c iff a*c = b

let a = 1/0 therefore a*0 = 1, but a*0 = 0 for all 'a' so there is no 'a' which cna be identifed with 1/0.

jcsd said:
We've had this question many times before:

It cna easily be shown in any ring that:

a*0 = a(b - b) = a(b - b) + 0 = a(b - b) + ab - (ab) = a(b - b + b) - (ab) = ab - (ab) = 0

i.e. a*0 = 0 for all 'a';

define divison by:

a = b/c iff a*c = b

let a = 1/0 therefore a*0 = 1, but a*0 = 0 for all 'a' so there is no 'a' which cna be identifed with 1/0.
Of course.

jcsd
Gold Member
Owen Holden said:
Of course.

The thing to note about the proof is that it applies to all rings.

arildno said:
fdarkangel:
Euclid's algorithm has nothing to do with the fact that the multiplicative inverse of 0 is undefined.
are you sure you've read my post?

jcsd said:
Yes.
That which is equal to (1/0) does not exist.

1/x =df (the y: x*y=1), includes the definition of (1/0).

If x=0 then 1/0 = (the y:0*y=1)

But, there is no y such that: y*0=1.

That is, Ay(y*0=0).

Therefore, (the y:y*0=1) does not exist.

jcsd
Gold Member
you mean to say there does not exist any element of x in any element of ...... (a certain algebra, the most general class of al;gebras you probaly wnat to consider are called the rings) which meets this definition of 1/0. General, unqaulifed staements like "1/0 doesn't exist" lack meaning in a matehmatical context.

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jcsd said:
you mean to say there does not exist any element of x in any element of ...... (a certain algebra, the most general class of algebras which meet the two requiremnts you have stipulated are called rings) which meets this definition of 1/0. General, unqaulifed staements like "1/0 doesn't exist" lack meaning in a matehmatical context.
I don't agree.

E!x =df EF(Fx).

~E!(the x:~(x=x)), is a theorem.
All objects described by contradicory predications do not exist!
They have no properties at all.

~E!x <-> ~EF(Fx).

Alkatran
Homework Helper
1/0 = 1 or 2 or 1.5 or 1.2 or 1 + 5i or sqr(2) or ...

Why are we doing this, again?

jcsd