Undefined For Different Reasons

[harpazo]

Both a/0, where a CANNOT equal 0 and 0/0 are undefined, but for different reasons. Explain each reason.

For a/0, the answer is division by 0 is not possible. For 0/0, I know it has something to do with indeterminate form. I also know that 0/0 does not equal 0.

Without getting too technical, can you explain why 0/0 is undefined? Please, do not provide a calculus answer. I have not studied calculus.

 

[ksdhart2]

Without using calculus or limits, I'm not sure I can give a rigorous answer, but there's a fairly intuitive hand-wavy way of thinking about why 0/0 is undefined. Consider two of the most basic rules of division:

\(\displaystyle \frac{0}{x} = 0\) (Rule #1)

\(\displaystyle \frac{x}{x} = 1\) (Rule #2)

In reality, neither of these rules apply when \(x = 0\), but for this explanation, let's naively assume they are universal. What, then, would that make 0/0? Well, by Rule #1, the answer would have to be 0, since the numerator is 0. But, also, by rule #2, the answer would have to be 1, since the numerator and the denominator are the same. In some sense, we can think of this as implying that 0/0 is "trapped" between these two answers - it has to take on both values simultaneously but it can't so we say it's undefined.

Later on, when you do study calculus and limits, you'll find out more, and learn that 0/0 isn't just undefined but specifically what's called an "indeterminate form." Depending on how fast and on what path the numerator and denominator approach 0, a limit of the form of 0/0 can be any real number or it may not exist at all.

 

[pka]

Why not just say that division by zero is not defined (or allowed).

 

[HallsofIvy]

a/b= c is equivalent to a= bc.

a/0, where a is non-zero, is "undefined" because if a/0= c then a= 0c which is untrue.

0/0 is "undefined" (many texts say "undetermined" rather than "undefined" for this case) because if 0/0=c then 0= 0c which is true for any c.

 

[Dr.Peterson]

I second Halls' approach. Summarizing it, a/0 is undefined because there is no number c such that a = 0*c, while 0/0 is undefined because there are "too many" numbers c such that 0 = 0*c. In the first case, we can't find a possible answer; in the second case, we can't choose which answer to give, because every number has an equally good claim.

So we just say that division by 0 is always undefined -- that is, we can never define a unique value to assign such an expression.

 

[lookagain]

0/0 is "undefined" (many texts say "undetermined" rather than "undefined" for this case) because if 0/0=c then 0= 0c which is true for any c.

harpazo, many texts state 0/0 is additionally "indeterminate," because it cannot be limited to one value. It is not determined.

 

[Otis]

… Without getting too technical, can you explain why 0/0 is undefined? Please, do not provide a calculus answer …

Hi. I'm going with pka, on this one. At your level, a/0 and 0/0 are both undefined because we cannot divide by zero (in the Real number system).

Limits are a pre-calculus topic, so I'm not sure whether you want to get into limits now, but 0/0 does not mean zero divided by zero. We write 0/0 as a symbol (not a number) to represent a situation where both the numerator and denominator are approaching zero (in a limit of some rational function). The form 0/0 doesn't mean that the numerator and denominator are zero. It's a form that arises in limits.

If you're not interested in limits, yet, then just think of 0/0 as undefined because we cannot divide by zero. Later, if you study the concept of a limit and work with rational functions, you'll learn more about indeterminant forms.

?

 

[harpazo]

Why not just say that division by zero is not defined (or allowed).

This is not my own question. This question is from the Michael Sullivan College Algebra textbook, Section R.1, Real Numbers.

 

[harpazo]

Without using calculus or limits, I'm not sure I can give a rigorous answer, but there's a fairly intuitive hand-wavy way of thinking about why 0/0 is undefined. Consider two of the most basic rules of division:

\(\displaystyle \frac{0}{x} = 0\) (Rule #1)

\(\displaystyle \frac{x}{x} = 1\) (Rule #2)

In reality, neither of these rules apply when \(x = 0\), but for this explanation, let's naively assume they are universal. What, then, would that make 0/0? Well, by Rule #1, the answer would have to be 0, since the numerator is 0. But, also, by rule #2, the answer would have to be 1, since the numerator and the denominator are the same. In some sense, we can think of this as implying that 0/0 is "trapped" between these two answers - it has to take on both values simultaneously but it can't so we say it's undefined.

Later on, when you do study calculus and limits, you'll find out more, and learn that 0/0 isn't just undefined but specifically what's called an "indeterminate form." Depending on how fast and on what path the numerator and denominator approach 0, a limit of the form of 0/0 can be any real number or it may not exist at all.

A few things about me to consider.

1. I am reviewing college algebra on my own to reconnect with material learned over 20 years ago.

2. All questions posted come from Michael Sullivan's College Algebra Edition 9 textbook.

3. I am not a student in a formal classroom setting. I am a middle aged man not a young college guy trying to get others to do my work.

 

[harpazo]

I want to thank everyone for your input and ideas. As you know, in grammar school teachers wrongly teach that 0/0 is 0. Teachers in grades k-5 have a limited understanding of mathematics. Later on in my studies, I will come across the INDETERMINATE FORM.

 

[pka]

This is not my own question. This question is from the Michael Sullivan College Algebra textbook, Section R.1, Real Numbers.

I cannot find any of Sullivan's books. I wrote a very negative review for one of them, so the publisher never sent me another.
Here is my view: I see no reason for this sort of question in a College Algebra text. If it were a foundations of mathematics, an introduction to analysis, even an advanced calculus then those questions are expected. But not college algebra.

 

[Otis]

… Without getting too technical, can you explain why 0/0 is undefined? …

Why not just say that division by zero is not defined …

This is not my own question. This question is from … Michael Sullivan …

pka answered your question (above). Are you saying that you want to start studying limits? That is, you want to learn how to provide the explanation requested in the book? If so, then tell us whether the book covers the form 0/0 vs having mentioned it only in one of those 'for further study' exercises.

 

[harpazo]

pka answered your question (above). You haven't posted a question from the book. If you need help with an exercise in the book, please follow the guidelines.

Here is the question from Section R.1 again.

"Both a/0, where a CANNOT equal 0 and 0/0 are undefined, but for different reasons. Explain each reason."

This is the actual question in quotes.

 

[Otis]

Hi. You posted while I was editing my thread; please read it again. Why did you create this thread?

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[harpazo]

Hi. You posted while I was editing my thread; please read it again. Why did you create this thread?

\(\;\)

Again, this is a question in Section R.1 in Sullivan's College Algebra textbook Edition 9. Section R.1 is a review of Real Numbers. In fact, it is one of several review sections before actual college algebra begins in Chapter 1. It is a question that he expects students to research online to find the best, logical answer in the form of a paragraph.

 

[Otis]

… [the book] expects students to research online to find the best, logical answer in the form of a paragraph.

Is that what the book says?

 

[Dr.Peterson]

The question does not involve limits. I am quite sure it is intended to be answered in some way similar to posts #4 and #5.

I suspect that if we could see this book, we would find that it at least hints at an answer in the text. Harpazo, please read carefully through the section and let us know what it says there about division by zero, so we can be sure what we are talking about.

Actually, I found a copy of the book (some edition) online (probably illegal) by searching for the title and a snippet from the problem. I found that it says this in section R.1:


The exercise in question is:


So that's what they've said, and what the question is. It's asking for "Discussion and Writing", and is intended probably to give an instructor an opportunity to discuss this with the class. Harpazo, it's possible that this kind of problem will be too hard for you to learn from without a teacher (and without accidentally provoking arguments), so you might want to focus on problems for which answer can be given in the back of the book. This is not really essential to getting through the book (even though I can appreciate the desire to really understand things deeply).

 

[harpazo]

Is that what the book says?

See Dr. Peterson's latest reply for the actual question.

 

[harpazo]

The question does not involve limits. I am quite sure it is intended to be answered in some way similar to posts #4 and #5.

I suspect that if we could see this book, we would find that it at least hints at an answer in the text. Harpazo, please read carefully through the section and let us know what it says there about division by zero, so we can be sure what we are talking about.

Actually, I found a copy of the book (some edition) online (probably illegal) by searching for the title and a snippet from the problem. I found that it says this in section R.1:
View attachment 13123

The exercise in question is:
View attachment 13124

So that's what they've said, and what the question is. It's asking for "Discussion and Writing", and is intended probably to give an instructor an opportunity to discuss this with the class. Harpazo, it's possible that this kind of problem will be too hard for you to learn from without a teacher (and without accidentally provoking arguments), so you might want to focus on problems for which answer can be given in the back of the book. This is not really essential to getting through the book (even though I can appreciate the desire to really understand things deeply).

You nailed it, as usual.

 

[Otis]

… I know it has something to do with indeterminate form.

… Later on in my studies, I will come across the INDETERMINATE FORM.

That was going to be my next suggestion, and I had almost posted as Dr. Peterson did -- that you don't need to do any extra-curricular activites to begin reviewing algebra. It is review, correct? Start with Chapter 1.

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