This one will do fine. If you don’t know the answer, it’s no big deal, there’s no way I would think less of you.

scherado wrote:

OlivierK wrote:

scherado wrote:Another thing I remember from college-days is that multiplication is addition by a different name: 7 x 3 = 7 + 7 + 7.

For division, we have subtraction by a different name: For 6/2, we subtract 2 from 6, take the reduced value and repeat UNTIL we either reach 0 or there's not enough remaining; count the number of iterations:

6 - 2 = 4
4 - 2 = 2
2 - 2 = 0

3 times we subtracted 2: 6/2 = 3

Now, do the same for 3/0:

3 - 0 = 3
3 - 0 = 3 ...

Infinite iterations.

There's that pesky infinity, again. (Not whistling Dixie.)

You think you reach 0 after subtracting 0 from 6 an infinite number of times?

How precious.

Do you want to reconsider...when you've sobered up?

No need to reconsider.

You're counting interations taken to reach zero.

In your first example you come up with the (correct) count of 3 iterations to reach zero.

In your second example you come up with the (incorrect) count of infinity iterations to reach zero.

The fact that "infinity" is a braindead answer to the second part is not my problem.

OlivierK wrote:

scherado wrote:

OlivierK wrote:

scherado wrote: ...

Now, do the same for 3/0:

3 - 0 = 3
3 - 0 = 3 ...

Infinite iterations.

There's that pesky infinity, again. (Not whistling Dixie.)

You think you reach 0 after subtracting 0 from 6 an infinite number of times?

How precious.

Do you want to reconsider...when you've sobered up?

No need to reconsider.

You're counting interations taken to reach zero.

In your first example you come up with the (correct) count of 3 iterations to reach zero.

In your second example you come up with the (incorrect) count of infinity iterations to reach zero.

The fact that "infinity" is a braindead answer to the second part is not my problem.

You most certainly to have a problem: Your problem is that, STILL, you won't accept the entire point of this thread. To wit, (you): "you come up with the (incorrect) count of infinity"

INFINITY IS NOT A VALUE, but I repeat myself.

And this: "you come up with the (incorrect) count of infinity iterations to reach zero."

The example NEVER REACHES ZERO, which is the point of that example.

Do you want to reconsider, sober or not?

How to tell if a troll has invaded your online forum:

Issues challenges to which he cannot recognize correct answers and fails to even make the attempt to answer challenges given back.

check

scott1328 wrote:How to tell if a troll has invaded your online forum:

Issues challenges to which he cannot recognize correct answers and fails to even make the attempt to answer challenges given back.

check

I thought a troll is an offensive offense.

Trolls, at least the inept ones we've had on this forum lately, are people with more available time than wits.

scherado wrote:

OlivierK wrote:

scherado wrote:

OlivierK wrote:
You think you reach 0 after subtracting 0 from 6 an infinite number of times?

How precious.

Do you want to reconsider...when you've sobered up?

No need to reconsider.

You're counting interations taken to reach zero.

In your first example you come up with the (correct) count of 3 iterations to reach zero.

In your second example you come up with the (incorrect) count of infinity iterations to reach zero.

The fact that "infinity" is a braindead answer to the second part is not my problem.

You most certainly to have a problem: Your problem is that, STILL, you won't accept the entire point of this thread. To wit, (you): "you come up with the (incorrect) count of infinity"

INFINITY IS NOT A VALUE, but I repeat myself.

And this: "you come up with the (incorrect) count of infinity iterations to reach zero."

The example NEVER REACHES ZERO, which is the point of that example.

Do you want to reconsider, sober or not?

Nope. If you agree that infinite subtractions of zero from 6 never get you to zero, then you have your answer right there about why division by zero is undefined. Infinity doesn't have anything to do with it, unless someone claims that's the answer to division by zero. I certainly don't, and now you say you don't, either. So: well done you, I guess, you got there in the end. Do you want a cookie?

OlivierK wrote:... Infinity doesn't have anything to do with it, unless someone claims that's the answer to division by zero. ...

One can't divide by zero because infinity is not a value.

scherado wrote:

OlivierK wrote:... Infinity doesn't have anything to do with it, unless someone claims that's the answer to division by zero. ...

One can't divide by zero because infinity is not a value.

How many times do you have to be told that "infinity" has nothing to do with the matter?
you are merely wrong.

scherado wrote:

OlivierK wrote:... Infinity doesn't have anything to do with it, unless someone claims that's the answer to division by zero. ...

One can't divide by zero because infinity is not a value.

No cookie for you.

You might as well say that you can't divide by zero because the Pope is not a dog.

But just because I could do with a giggle, can you explain how division by zero would be well-defined if infinity was a value?

Isn’t there a math symbol for infinity?

Well, yes there is, and it has a term called “lemniscate“. I think is it a two dimensional representation of a möbius loop, but it’s been a “few” years since my last math class.

Does an ♾ represent a value? If it didn’t it wouldn’t need be represented in formal math.

RS

theropod wrote:Isn’t there a math symbol for infinity?

Well, yes there is, and it has a term called “lemniscate“. I think is it a two dimensional representation of a möbius loop, but it’s been a “few” years since my last math class.

Does an ♾ represent a value? If it didn’t it wouldn’t need be represented in formal math.

RS

Let this be your xmas glyph from me:


∞

On my mac keyboard, it's in the extended ascii, and I type it with opt-5 right into this text box.

\infty to those who speak LaTeX.

Also: When wibbling comes to town, Octave is your friend. See? I can divide by zero! Me NaN says I can.

The problem is that infinity isn’t a number. It’s math, there has to be a number, right?

Must have blown people’s minds when zero was invented.

felltoearth wrote:The problem is that infinity isn’t a number. It’s math, there has to be a number, right?

Must have blown people’s minds when zero was invented.

Infinity: "I'm not a real number, but I play one on TV."

Cito di Pense wrote:

felltoearth wrote:The problem is that infinity isn’t a number. It’s math, there has to be a number, right?

Must have blown people’s minds when zero was invented.

Infinity: "I'm not a real number, but I play one on TV."

And wants to direct.

scott1328 wrote:

scherado wrote:Can anyone tell me--after it has been revealed already--the reason we can't divide by zero? In my 5 words or Matt_B's 6 words.

This thread has been over for a few days now. I thought it would take much longer to get a CORRECT answer: Give Matt_B the keys to the forum.

I'll have to start a new one.

Your 5 word simplistic answer is wrong. And your presumption that there is but one answer is wrong.

Why don't you answer my question? why is it incorrect to use L'Hopital's rule to evaluate the limit as x approaches 0 of sin(x)/x ?

Since the troll was banned. I will do the work he couldn't/wouldn't do (I do understand that the majority of those who responded to the troll already know the answer to this "challenge")

L'Hopital's rule states that if the limit as x approaches some value c of f(x) = 0 and g(x)=0, then the limit as x approaches c of the quotient of those two functions [ f(x)/g(x) ] can be found finding the limit as x approaches c of f'(x)/g'(x).

So it would seem that applying L'Hopital's rule to the quotient of sin(x)/x would be useful in evaluating the limit: since sin(0)=0 so we have the case 0/0. The rule says that we should be able to evaluate the limit by finding the derivative of sin(x) which turns out to be cos(x) and the derivative of x which turns out to be the constant 1. Now, evaluating the limit : the limit as x approaches 0 of cos(x)/1 is 1.

However, where the the trouble comes in, is in how you actually prove that the derivative of sin(x) does in fact equal cos(x). To find the derivative of sin(x) relies on knowing how to evaluate the limit of sin(x)/x. And therein lies the circularity.

scott1328 wrote:L'Hopital's rule states that if the limit as x approaches some value c of f(x) = 0 and g(x)=0, then the limit as x approaches c of the quotient of those two functions [ f(x)/g(x) ] can be found finding the limit as x approaches c of f'(x)/g'(x).

So it would seem that applying L'Hopital's rule to the quotient of sin(x)/x would be useful in evaluating the limit: since sin(0)=0 so we have the case 0/0. The rule says that we should be able to evaluate the limit by finding the derivative of sin(x) which turns out to be cos(x) and the derivative of x which turns out to be the constant 1. Now, evaluating the limit : the limit as x approaches 0 of cos(x)/1 is 1.

However, where the the trouble comes in, is in how you actually prove that the derivative of sin(x) does in fact equal cos(x). To find the derivative of sin(x) relies on knowing how to evaluate the limit of sin(x)/x. And therein lies the circularity.

Why does L'Hôpital's rule work at all? The reason I brought it up in the first place was to see if scherado knew anything about Taylor series expansions. The expansion does involve taking derivatives (of functions that are continuously differentiable, etc.) but maybe Euler's formula suggests another way to get the series expansions of functions like sin and cos without the circularity you note. Recognition of the series for sin and cos first involves elementary (?) analysis of the complex exponential (at the elementary level, we get Euler's formula). Maybe this is just a circularity on a less-trivial level, but at least we've gotten beyond a rote application of the rule "just take derivatives when we have the right kind of indeterminate form". If we have not obtained those series by circularly taking the derivatives of trig functions, then that limit can be evaluated without dividing by zero, which is still not allowed. L'Hôpital's rule works (and we can apply it repeatedly) because it leads to canceling common factors in numerator and denominator, or removing vanishing terms from a series.

Absolutely you can define the trig functions(and their hyperbolic counterparts) in terms of their MacLauren series expansions. In which case finding the derivatives of sin and cos are trivial matters. Motivating such a definition is problematic. This introduces a pedagogical difficulty however. Such definitions of sine and cosine are going to seem entirely arbitrary to first and second year calculus students. And would likely impede understanding rather than aid it.

scott1328 wrote:Absolutely you can define the trig functions(and their hyperbolic counterparts) in terms of their MacLauren series expansions. In which case finding the derivatives of sin and cos are trivial matters. Motivating such a definition is problematic. This introduces a pedagogical difficulty however. Such definitions of sine and cosine are going to seem entirely arbitrary to first and second year calculus students. And would likely impede understanding rather than aid it.

Agreed. There is much that first- and second-year calculus students find arbitrary. There are professors who know how to introduce it from this angle. Hopefully, the student discovers hints of the generality in seeking the interval (or radius) in which a function can be analyzed. It's of benefit far outside the bounds of pure mathematics.