- #1

zeromodz

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Agree or disagree? and why?

Me and my peers are disputing this.

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- Thread starter zeromodz
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- #1

zeromodz

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Agree or disagree? and why?

Me and my peers are disputing this.

- #2

mathman

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As it stands 1/0 is "undefined".

"Approach" is used when talking about 1/x, as x -> 0. However strict pedantry requires one to say "becomes infinite" rather than "approaches infinity".

"Approach" is used when talking about 1/x, as x -> 0. However strict pedantry requires one to say "becomes infinite" rather than "approaches infinity".

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- #3

zeromodz

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As it stands 1/0 is "undefined".

"Approach" is used when talking about 1/x, as x -> 0. However strict pedantry requires one to say "becomes infinite" rather than "approaches infinity".

Yes, we must say its undefined because we cannot deal with a number that is not constant like infinity is, but I want to know if you agree with me that it becomes infinite, but we must say it is undefined because we cannot deal with such an outcome. If you disagree and think its something else, please back it up with some logic.

- #4

Hurkyl

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In the real number system and the extended real number system, it's nonsense.

In the projective real number system, it's equal to [itex]\infty[/itex].

- #5

Studiot

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The trick is to realize that infinity is not a single object like, say, the number 27. It is certainly not a real number.

Zero is a slightly unusual real number.

It was Cantor who first proved that there are many infinities. Some are 'greater' than others.

The knowledge he derived allow us to evaluate such expressions as

[tex]\frac{\infty }{\infty }[/tex] and [tex]\frac{0}{0}[/tex]

By L'Hopitals rule or other methods and come up with real number answers.

Zero is a slightly unusual real number.

It was Cantor who first proved that there are many infinities. Some are 'greater' than others.

The knowledge he derived allow us to evaluate such expressions as

[tex]\frac{\infty }{\infty }[/tex] and [tex]\frac{0}{0}[/tex]

By L'Hopitals rule or other methods and come up with real number answers.

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- #6

Algr

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Is this equation always true?

5 * (x/x) = 5

5 * (x/x) = 5

- #7

Anonymous217

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@Algr: Because there is an x as the denominator, you must set a restriction on what x can be. In other words, there is a domain where x is all real numbers except 0.

- #8

Hurkyl

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If we are talking about real numbers, then the answer is either:Is this equation always true?

5 * (x/x) = 5

- The equation is true -- because
*x*is a variable restricted to some domain of nonzero reals - The equation is nonsense -- because the domain
*x*varies over includes zero

(Note that "The equation is false" is not the same thing as "this equation is nonsense")

Another possibility opens up if you are working with partial functions, which works out to that equation being synonymous with the statement "x is nonzero".

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- #9

g_edgar

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In the system called "Riemann sphere" it is indeed true that [tex]1/0=\infty[/tex]. In the real number system this is not true, since there is no real number [tex]\infty[/tex] and operation [tex]1/0[/tex] is not defined there.

And, of course, there are many more numbers systems, too.

So ...

- #10

Mentallic

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Also, 1/x where x is infinitely small (but not zero) would equal infinite too. So, is 1/0 infinite or undefined?

- #11

CRGreathouse

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Another problem with 1/0 is the fact that it could be [itex]\pm \infty[/itex] in a sense, depending on which end you approach zero from.

In the extended reals, 1/0 is undefined (not +∞ or -∞).

Also, 1/x where x is infinitely small (but not zero) would equal infinite too. So, is 1/0 infinite or undefined?

In the hyperreals, if e is infinitesimal but nonzero, 1/e is infinite (the exact value depends, naturally, on e). But even there 1/0 is undefined.

In the projective line, just as in the Riemann sphere, 1/0 = ∞.

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- #12

Mentallic

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And of course the OP is talking about the real number system, else the entire group discussion would have been futile since they would have already learned the difference of 1/0 in each number system.

- #13

CRGreathouse

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And of course the OP is talking about the real number system, else the entire group discussion would have been futile since they would have already learned the difference of 1/0 in each number system.

(emphasis mine)

I'm not so sure. If someone posted "What is 3 divided by 2?", I wouldn't say, "3 can't be divided by 2 if you're working in the integers; 1 and -1 are the only units in Z". I would assume that the OP was working in a system where 3/2 made sense.

Similarly, the question "Is foo infinite?" doesn't make sense in a system, like the real numbers, without a concept of "infinite". I would sooner assume that they're working in a system cobbled together out of the real numbers and a misunderstanding of the calculus limits +∞ and -∞. This system seems to resemble the extended reals, so usually this is my first guess as to the best way to formalize such questions.

If that's how the question is understood ("In the extended reals, is 1/0 infinite?"), then the answer is no: it is undefined.

- #14

tauon

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Agree or disagree? and why?

Me and my peers are disputing this.

the simplest way to see why 1/0 is an undefinable fraction in the real numbers system is to assume the contrary: let's say

[tex]\frac{1}{0}=a[/tex]

then

[tex]0a=1[/tex].

how many [tex]a\in\mathbb{R}[/tex] do you know that satisfy that relation?

in general actually, division by 0 is undefined in any field, the argument for that is similar.

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- #15

Mentallic

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The thing is, if the OP was working in a system where 3/2 made sense then the question wouldn't have been asked in the first place. By the sounds of it, he was disputing between peers whether 1/0 is infinite or undefined. This is obviously indicating that they're using a number system where the norm was to simply state 1/0 is undefined, but 1/0 can also be seen naturally as infinite by the sloppy use of limits in calculus as you've said.I'm not so sure. If someone posted "What is 3 divided by 2?", I wouldn't say, "3 can't be divided by 2 if you're working in the integers; 1 and -1 are the only units in Z". I would assume that the OP was working in a system where 3/2 made sense.

But yes, I see the necessity to ask which number system one is intending to work with.

How about [itex]a=\infty[/itex]? The OP has already said this:the simplest way to see why 1/0 is an undefinable fraction in the real numbers system is to assume the contrary: let's say

[tex]\frac{1}{0}=a[/tex]

then

[tex]0a=1[/tex].

how many [tex]a\in\mathbb{R}[/tex] do you know that satisfy that relation?

in general actually, division by 0 is undefined in any field, the argument for that is similar.

so you haven't really swayed the discussion in favour of 1/0 being undefined.zeromodz said:There are an infinite amount of zero's that can go into 1

There are even examples in limits with calculus that show that a is undefined (1/0). Take the functions [itex]f(x)=x[/itex] and [itex]g(x)=csc(x)[/itex]. f(0)=0, g(0) is undefined. But f(0)g(0)=1 (well, as x tends to 0 of course). Now, if you let g(0) be the value a you mentioned then there you go, the answer is 1/0.

Now, is 1/0 infinite or undefined? :tongue:

- #16

zeromodz

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The thing is, if the OP was working in a system where 3/2 made sense then the question wouldn't have been asked in the first place. By the sounds of it, he was disputing between peers whether 1/0 is infinite or undefined. This is obviously indicating that they're using a number system where the norm was to simply state 1/0 is undefined, but 1/0 can also be seen naturally as infinite by the sloppy use of limits in calculus as you've said.

But yes, I see the necessity to ask which number system one is intending to work with.

How about [itex]a=\infty[/itex]? The OP has already said this:

so you haven't really swayed the discussion in favour of 1/0 being undefined.

There are even examples in limits with calculus that show that a is undefined (1/0). Take the functions [itex]f(x)=x[/itex] and [itex]g(x)=csc(x)[/itex]. f(0)=0, g(0) is undefined. But f(0)g(0)=1 (well, as x tends to 0 of course). Now, if you let g(0) be the value a you mentioned then there you go, the answer is 1/0.

Now, is 1/0 infinite or undefined? :tongue:

Okay, but undefined isn't an answer. Its just a cop out to say "hey we don't know what's going on here". I agree its okay to label it undefined, but when the process is actually done the answer will not be a number, it will turn into a concept that we call "infinity". Thats why I think its plausible to deal with saying its undefined.

Here is another argument to say its infinity

1 / ∞ = 0

Therefore

1 / 0 = ∞

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- #17

Hurkyl

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Er, "your question is nonsense" is a perfectly reasonable answer to a nonsensical question. :tongue: In real number arithmetic, "1/0" is the semantic equivalent of an English phrase like "the taste of ideas" -- those English words just don't combine in that fashion, nor do those mathematical symbols.Okay, but undefined isn't an answer.

(In a variant syntax based on partial functions instead of functions, 1/0 does make sense, but it doesn't have any value at all. e.g. the equation 1/0=x is identically false)

Where you get that idea?Its just a cop out to say "hey we don't know what's going on here".

Process? What process?but when the process is actually done

If your intuition tells you that, in the real numbers, 1/0 is infinite, then there is an error in your intuition, plain and simple. We have given you keywords that would let you go searching for further information on other number systems. (Another keyword: wheel, although AFAIK those are more of a curiosity than something people actually use)

However, if you're going to insist that all of mathematics must conform to your intuition the way it is right now, then you aren't going to learn anything.

- #18

Studiot

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But yes, I see the necessity to ask which number system one is intending to work with.

Why do they all have to be in the same number system?

There are many processes, both finite and infinite, in mathematics where the result of a process leads outside the origin set.

So why can the process of taking a reciprocal not lead outside the origin set in the case of zero?

Edit

The statement 'the result of this process is undefined' is really a short way of saying 'the result of this process is not a member of the origin set.' In some instances it has been worthwhile establishing what the result of the process is - even if it has meant creating a new mathematical object.

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- #19

Hurkyl

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Process? What process?So why can the process of taking a reciprocal not lead outside the origin set in the case of zero?

Zero is not in the domain of the real number reciprocal, so obviously one cannot take the real number reciprocal of zero to get something that isn't real.

- #20

Studiot

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- #21

Hurkyl

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Sure -- you can work with whatever functions you like. But if 0 is in the domain of that function, it's not the real number reciprocal function, and theorems about the real number reciprocal function will typically not apply to the function you want to work with.Mathematics does not enfoce any particular choice,

- #22

Studiot

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For instance I have started with the domain and asked what functions are of interest?

Rather than starting with the function and asking what results are of interest?

- #23

CRGreathouse

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Here is another argument to say its infinity

1 / ∞ = 0

Therefore

1 / 0 = ∞

Your argument rests on the assumptions:

1. 1 / ∞ = 0

2. a/b = c implies a/c = b

where #2 probably comes from

3. a/b = c if and only if a = bc

4. a = b if and only if a/c = b/c

5. ab/b = a

#3 through #5 (and thus #2) are valid rules for nonzero real numbers, but you have ∞ so you're not working with real numbers. But then, why assume that #3 through #5 still hold?

There are ways to extend the real numbers to cover division by zero. Define "/" as the preimage of multiplication: that is, a "/" b is the set {c: b * c = a}. Then for a,b in R and b not zero, a "/" b = {a/b}; for a in R and a ≠ 0 = b, a "/" b is the empty set {}; for a = b = 0, a "/" b is the set of real numbers R. If you'd like you can try to extend this to include ∞, however you choose to define that.

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- #24

pootette

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Are the theorems for non real numbers invalid for real numbers (and vice-versa)?

- #25

CRGreathouse

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Are the theorems for non real numbers invalid for real numbers (and vice-versa)?

A theorem that says, "If x is a real number, ..." does not let you conclude "If x is not a real number, ...". Similarly, a theorem that says, "If x is not a real number, ..." does not let you conclude "If x is a real number, ...".

But you might have a theorem of the form, "If x is an extended real number, ..." which does let you conclude "If x is a real number, ..." since all real numbers are extended real numbers.

- #26

JSuarez

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Are the theorems for non real numbers invalid for real numbers (and vice-versa)?

Every non-constant polynomial has a root.

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- #27

zeromodz

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If your intuition tells you that, in the real numbers, 1/0 is infinite, then there is an error in your intuition, plain and simple. We have given you keywords that would let you go searching for further information on other number systems. (Another keyword: wheel, although AFAIK those are more of a curiosity than something people actually use)

However, if you're going to insist that all of mathematics must conform to your intuition the way it is right now, then you aren't going to learn anything.

Fine, that may be the case where my intuition is wrong, then please tell me how many zero's can go into one?

I am not trying to argue, I just want an answer. You can use the Riemann sphere or any extended complex number system.

- #28

CRGreathouse

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Fine, that may be the case where my intuition is wrong, then please tell me how many zero's can go into one?

I am not trying to argue, I just want an answer. You can use the Riemann sphere or any extended complex number system.

In the Riemann sphere, 1/0 = ∞. ("∞" in the Riemann sphere is a point with greater magnitude than any other point, but with no particular argument. Similarly, "0" in the Riemann sphere is a point with lesser magnitude than any other point but with no particular argument. All other points on the Riemann sphere have a unique argument (mod 2π) and infinitely many numbers with greater (or lesser) magnitude.)

I don't know what the extended complex number system is. Presumably the same as the Riemann sphere, but giving magnitudes to infinite values? In that case 1/0 would be undefined there.

- #29

Hurkyl

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How many zeros "go into" one? As in the following equation?tell me how many zero's can go into one?

[tex]\stackrel{n\ \text{times}}{\overbrace{0 + 0 + \ldots + 0}} = 1[/tex]

In that case, this equation has no solution for

- #30

Studiot

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Zero is not in the domain of the real number reciprocal, so obviously one cannot take the real number reciprocal of zero to get something that isn't real.

I'm sorry but this is not an answer to my question which was (expanded a little)

If I take a real nonzero number a and divide it by the real number zero why should I expect the result to be a real number?

After all we move from the integers to the rationals to the reals by asking almost the same question.

I suggest that in the answer this question lies the way to understanding for zeromodz.

- #31

CRGreathouse

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If I take a real nonzero number a and divide it by the real number zero why should I expect the result to be a real number?

If you take the integer 2 and divide it by the integer 3, there is no integer that is the answer. But that doesn't mean that there's a unique answer outside the integers. In Q the answer is 2/3, but in Z/7Z the answer is 3.

That's why it's so important to know which system you're working in.

- #32

Studiot

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But that doesn't mean that there's a unique answer outside the integers.

I agree, although adding uniqueness is another burden.

However that did not answer my question.

It doesn't mean there is or isn't; it just draws attention to a gap in the system of mathematics (numbers in this case) as do roots, surds and trancendental numbers, vector cross products, some Fourier analyses... the list goes on and on.

As a result of that attention it has been found possible and convenient to develop new mathematics to deal with some such gaps, but not all known gaps have been plugged this way.

1/0 is one such unplugged gap. So we say it is undefined.

- #33

Hurkyl

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No, leaving 1/0 undefined is (these days anyways) a deliberate design decision.1/0 is one such unplugged gap. So we say it is undefined.

Ponder this question: why would you want to divide 1/0?

Most of the other number systems that people use -- e.g. the real numbers, the extended real numbers, the projective complexes, modular arithmetic -- are used because they are good for some purpose. The projective complex line, for example, are especially well suited for studying rational functions of one variable.

OTOH, I believe wheels were defined specifically for the purpose of defining a good arithmetic system where +,-,*,/ are defined for any pair of numbers

*: Technically, / is a unary operator in a wheel: /x is the "reciprocal" of x, and 1/x is just 1 times the reciprocal of x.

- #34

Studiot

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Ponder this question: why would you want to divide 1/0?

As a result of that attention it has been found possible and convenient...but not all...

No offence meant but that is what I said, right from the outset. It's the same thing from another point of view. 'We have chosen not to' is the opposite side of the same coin from 'we have chosen to'.

It's the same coin of reasons that we don't chose to define division as a fundamental operation betwen numbers and you used addition to discuss division in your post#29.

I was also trying to introduce a point of view that shows our current reticence in defining 1/0 is not just a whim but in keeping without current systems of mathematics in a wider sense to help make zeromodz more comfortable with this. Would you claim our system of maths is perfect?

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- #35

Mentallic

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prove that if two complex numbers, z and w, are equal to each other, then their real parts and imaginary parts are equal.

i.e. z=w, let z=x+iy, w=a+ib

therefore, x+iy=a+ib

prove x=a, y=b

Doing a proof by contradiction: let's say that b[itex]\neq[/itex]y. Then we have x-a=i(b-y) and then [tex]i=\frac{x-a}{b-y}[/tex] under the assumption that b[itex]\neq[/itex]y. But the RHS is a real number, while the LHS is an imaginary number by definition, so obviously b=y (and thus a=x).

Notice that in this case, rather than having a dispute about 1/0=[itex]\infty[/itex] or not, we have 1/0 being imaginary. Clearly 1/0 is undefined in this number system.

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