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questions about hyperreals

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1 hour ago, uncool said:

I'm not sure I understand what you are trying to say. What's the difference between "rounding to the nearest real number" and truncation?

The result on the hyperreals is also exact. The "standard" function is specifically defined to give a result in the real numbers, not the hyperreal numbers.

There are two correspondences I can think of, but the one I think you are referring to is that the reals are a subset of the hyperreals (i.e. there is a natural embedding of the reals in the hyperreals). This isn't "super advanced algebraic structure theory"; it's just...a fact. I'm not sure what you think needs enlightening. 

I will need to find the references that led me to my comments before replying.

That should make them more sensible.

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8 hours ago, Conjurer said:

It was proved by Isaac Newton in the Principia Mathematica.

That cannot be entirely accurate. When Principia was written, Newton had long since passed on. Anyway, if that is where you got it from, then surely you can provide the relevant volume and page number? 

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11 hours ago, taeto said:

That cannot be entirely accurate. When Principia was written, Newton had long since passed on. Anyway, if that is where you got it from, then surely you can provide the relevant volume and page number? 

I don't believe he wrote it from the grave, and I don't see how that is possible unless it was planted by ancient alien astronauts.  Last I heard, it was considered to be a holy relic that couldn't be accessed by anyone, so people like you couldn't get their grubby hands on it.  The fact the mods would desire me to have to provide proof of this is absolutely disgusting, and a clear indication of the downfall of the public education system.

It is ridiculous that I would even have to provide this on a science forum, since it is just the basic proof of a derivative.  It is used to learn the first lesson in the first year of physics to determine instantaneous velocity of an accelerating body.  Do you believe that this process is somehow wrong?  If so, it would mean that all of physics is wrong.  It is basically the first thing that came about that even started physics in the first place.  You just need to take a first year physics or calculus class to learn this.

 

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9 minutes ago, Conjurer said:

It is ridiculous that I would even have to provide this on a science forum, since it is just the basic proof of a derivative.

Why do you think that you would have to provide the definition of a derivative, if that is what you mean by "proof of a derivative"? Nobody asked you about that. Instead you were asked to provide some kind of reference or evidence of your claim that Newton could make sense of division by zero. Maybe it would help if you just admit that you are simply making up nonsensical stuff all the time.

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17 minutes ago, taeto said:

Why do you think that you would have to provide the definition of a derivative, if that is what you mean by "proof of a derivative"? Nobody asked you about that. Instead you were asked to provide some kind of reference or evidence of your claim that Newton could make sense of division by zero. Maybe it would help if you just admit that you are simply making up nonsensical stuff all the time.

I guess you didn't watch the video and you are unwilling to learn.  The video describes how you can divide where the change in x and y becomes zero.  The equation for the derivative is the same equation they use in hyperreals.    

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In fairness to conjurer, Principia Mathematica is not enough of a reference by itself.

Russell and Whitehead wrote the treatise with that precise title, following on from Newton, in around 1907.

This was intended to be the pure mathematical encyclopedia of known mathematics at that time.

 

Newton's work was actually about Physical Science and entitled

Philosophae Naturalis Principia Mathematica

Both are commonly referred to as Principia or Principia Mathematica.


So before the crossed wires develop further let us all shake hands.

Edited by studiot

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17 minutes ago, studiot said:

Philosophae Naturalis Principia Mathematica

Fair enough. This is the first time I heard it referred to as just Principia Mathematica. Thanks for pointing it out Studiot, fist bumps and shake hands all around, I hope. My bad.

Then the claim stands, I suppose: Newton did know how to make sense of division by zero? Somewhere in one of the three volumes and on some page?

 

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25 minutes ago, Conjurer said:

I guess you didn't watch the video and you are unwilling to learn.  The video describes how you can divide where the change in x and y becomes zero.  The equation for the derivative is the same equation they use in hyperreals.    

Except in neither case is there any actual division by 0.

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3 minutes ago, taeto said:

Fair enough. This is the first time I heard it referred to as just Principia Mathematica. Thanks for pointing it out Studiot, fist bumps and shake hands all around, I hope. My bad.

Then the claim stands, I suppose: Newton did know how to make sense of division by zero? Somewhere in one of the three volumes and on some page?

It is a necessary step to finding the instantaneous value of anything in Newtonian Physics.  This is like asking if Newton knew how to do Newtonian Physics.

2 minutes ago, uncool said:

Except in neither case is there any actual division by 0.

I already explained this, and it is described in the video.  Khan academy is now an accredited university.

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6 minutes ago, Conjurer said:

I already explained this, and it is described in the video.

Please point out where you explained any actual division by 0, which is the specific distinction taeto is making.

7 minutes ago, Conjurer said:

Khan academy is now an accredited university.

...no, it isn't. 

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3 minutes ago, Conjurer said:

It is a necessary step to finding the instantaneous value of anything in Newtonian Physics.  This is like asking if Newton knew how to do Newtonian Physics.

If it is really necessary, it should be so much easier for you to present a single example of such a calculation, even without having to go all the way back to the 17'th century. Can you do this? Remember that I am asking about dividing by zero. Dividing by an infinitesimal is no problem, each of us who have been following this thread already knows how to do that. 

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3 minutes ago, taeto said:

If it is really necessary, it should be so much easier for you to present a single example of such a calculation, even without having to go all the way back to the 17'th century. Can you do this? Remember that I am asking about dividing by zero. Dividing by an infinitesimal is no problem, each of us who have been following this thread already knows how to do that. 

The derivative is the equation of the line that intersect a curve of a function at exactly one point.  In order to accomplish this, the change in x and y has to become zero, or it will intersect at two points on the curve.  All you have to do is cancel out the change in x on the bottom with the change in x on the top in the equation of the derivative.  It doesn't matter if the change in x is zero.  You just have to write the word limit x->infinity in front of it.  Then writing that makes it correct by some sort of ancient alien magic.

 

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7 minutes ago, Conjurer said:

All you have to do is cancel out the change in x on the bottom with the change in x on the top in the equation of the derivative.  It doesn't matter if the change in x is zero.

...yes, it does. The definition of the limit of a function specifically excludes the value of the function of the limiting value from being relevant.

"For all epsilon greater than 0, there exists a delta greater than 0 such that for any 0 < |x - a| < delta, |f(x) - L| < epsilon"

 

Edited by uncool

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7 minutes ago, Conjurer said:

The derivative is the equation of the line that intersect a curve of a function at exactly one point. 

If that were the case, then \(y = 1\) would be a derivative of the function that maps \(x\) to \(\sqrt{x}\) for \(x \geq 0. \) Since the line with equation \(y=1\) intersects the graph of this function in exactly one point \((1,1)\).

Edited by taeto

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2 minutes ago, taeto said:

If that were the case, then \(y = 1\) would be a derivative of the function that maps \(x\) to \(sqrt{x}\) for \(x \geq 0. \) Since the line with equation \(y=1\) intersects the graph of this function in exactly one point \(1,1)\).

The equation y = 1 or y = constant is excluded from this working.  It is a vertical line, so it fails the vertical line test, and it is not a function.  It has to be a function.

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1 minute ago, Conjurer said:

The equation y = 1 or y = constant is excluded from this working.  It is a vertical line, so it fails the vertical line test, and it is not a function.  It has to be a function.

...y = 1 is a horizontal line, not a vertical one; f(x) = 1 is a function. 

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1 minute ago, uncool said:

...y = 1 is a horizontal line, not a vertical one; f(x) = 1 is a function. 

My bad, then the answer to the derivative of that would be zero.

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1 minute ago, Conjurer said:

The equation y = 1 or y = constant is excluded from this working.  It is a vertical line, so it fails the vertical line test, and it is not a function.  It has to be a function.

It is still a line though. If you draw it in the xy-plane it even looks suspiciously like a horizontal line, not a vertical line. 

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1 minute ago, Conjurer said:

My bad, then the answer to the derivative of that would be zero.

you're missing the point taeto is making. 

Let me put it quite simply: no, the process of taking the derivative never involves a division by 0. Period. 

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1 minute ago, taeto said:

It is still a line though. If you draw it in the xy-plane it even looks suspiciously like a horizontal line, not a vertical line. 

The derivative of a horizontal line is zero, and the derivative of a vertical line is infinite.  In those cases, you cannot resolve infinity or zero.

1 minute ago, uncool said:

you're missing the point taeto is making. 

Let me put it quite simply: no, the process of taking the derivative never involves a division by 0. Period. 

The derivative of a line intersects a point on the curve, because the height or distance between the two points in question is zero.  If h is not zero, then it would intersect at two points.

You are missing the logic behind finding the limit.  When you take a limit you are discovering what the value would be when it actually is zero by looking where it approaches on a graph.  Then you don't have to graph it to see this every time.  

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7 minutes ago, uncool said:

Let me put it quite simply: no, the process of taking the derivative never involves a division by 0. Period. 

He seems to sense that a vertical line does not have a finite slope. Give him credit for that.

10 minutes ago, Conjurer said:

The derivative of a line intersects a point on the curve, because the height or distance between the two points in question is zero. 

You have an education degree right? What is your answer if a student asks "which two points in question"?

Edited by taeto

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11 minutes ago, Conjurer said:

You are missing the logic behind finding the limit.  When you take a limit you are discovering what the value would be when it actually is zero by looking where it approaches on a graph.  Then you don't have to graph it to see this every time.  

You are missing the logic behind the idea of taking limits, which is to avoid actually dividing by zero. The fact that you seem to think that the process is "magical" is a strong indicator of why. 

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!

Moderator Note

Ok. I think we have had enough of conjurer demonstrating their incompetence. (Again)

I think the original question about hyperreals has been answered. If not, please start a new thread (in which conjurer will not be allowed to post).

 

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