Why can't people think for themselves?

[Olivér Perge]

...Not knowing the answer right off the bat is absolutely fine! You're not going to immediately know the answer to every question ever asked you throughout your entire life. The main thing is taking the time to try to figure out how to approach a topic you don't know how to do...

I just loved to read your post. Chris, you are a good teacher and a great person as well! I think people like you brings humanity to the next level again and again.

 

[Kenneth]

Yes yes, slopes but...

I belive people does not think because we are not supposed to. This world is ruled by mad people taking advantage of dumbness. If we all started to think and discussed the real problems of this world they would not last for long and they know that.

Up until about a 100 years ago all litterature in higer schools was written in latins and that with the argument "the average man shall not understand". It is better now but, medical educations are still there using latin terms all the time, most of the lingual terms are also latins and maths is mostly greek.

If the books in the schools was using the language people speaks they would grasp things much easier.

No wounder the student can't transform 500 metres into kilometres, he does not know the translation of the term kilo!

 

[mrCage]

I'm just wondering, but what's the deal with people sticking to dictionary definitions or things their teachers told them, even when they don't make sense?

There are actually people walking around who think 0 isn't a number (it is, and a very useful one at that), or the slope of a vertical line cannot be defined (it can, easily), or black means reflecting absolutely no light (not the way actual people use it), or the Christian God is violently anti-homosexual (he calls it an "abomination"... just like eating shellfish or pork).

Believe it or not, guys, authority figures can be wrong. Think for yourselves, do some basic research. It will make you smarter.

Goes also for cube solving. Why do 99.9% use fridrich? When 90% of these dont understand the algs ... And why is speed "everything" ??

Per

 

[cmhardw]

More importantly, do you see the point that Michael is trying to make?

The point that Michael is trying to make is that the definition of slope as we know it is completely arbitrary. There is no reason that our familiar \( \frac{y_2-y_1}{x_2-x_1} \) should be the definition for slope. A better way to phrase that is that this formula is a definition of slope. Vertical lines have an undefined slope not because of anything particularly special or magical about vertical lines. They have undefined slope because the accepted definition of slope as \( \frac{y_2-y_1}{x_2-x_1} \) happens to assign an undefined number to vertical lines. That is the entire reason.

In fact, I can provide a meaningful definition of slope such that it is still a scalar quantity, still lets us evaluate derivates and integrals and all that fancy awesome stuff, and yet a vertical line has slope -1

And now to put my money where my mouth is. In the rest of this post I will do all of what I claimed in this quote box. For those of you who typically do not read my very long posts then stop here. If you're a tl;dr forum browser than FOR YOUR OWN SAKE STOP READING NOW

--------------------------------------------------------------------------------------------

I will now define a new type of slope that is different from our normal definition of slope. I still want it to be related to slope in that as a line changes in steepness, the slope also changes. I also want to make sure that every line has a scalar slope, with the exception of the particular line that will be assigned the undefined slope based on my definition. There is nothing special about this particular line. It just gets the short straw in the drawing so to speak.

And now Ladies and Gentlemen I give you:

S-Slope

S-Slope will be defined by the letter S (where we use the letter m for standard slope). I will distinguish between my new S-slope by always calling it S-slope. I will refer to our more well known and loved standard slope by the name m-slope for the letter that usually represents it.


S-Slope is defined by the following formula (by me and because I say so - because slope definitions are completely arbitrary anyway!):

\( S = \frac{(x_2-x_1)+(y_2-y_1)}{(x_2-x_1)-(y_2-y_1)} \)

Now consider two points on a vertical line. Let's use \( (0,0) \) and \( (0,1) \)

\( S = \frac{(0-0)+(1-0)}{(0-0)-(1-0)} = \frac{1}{-1} = -1 \)

More generally, any two points \( (a,b) \) and \( (a,c) \) have slope:

\( S = \frac{(a-a)+(c-b)}{(a-a)-(c-b)} = \frac{c-b}{-(c-b)} = \frac{1}{-1} = -1 \)

Therefore vertical lines have slope -1 using this definition of slope.

----------

Now you may think that this definition of slope is artificially constructed such that it produces this nice result for vertical lines only. In fact, this definition of slope and the regular definition form a bijecton. Essentially this would mean that my definition of slope is mathematically equivalent, in some sense, to that of regular slope. I can take any slope:
\( m = \frac{b}{a} \) and map it into my definition of slope using the transformation:
\( S = \frac{a+b}{a-b} \)

You would need to use this transformation with the help of limits to map the undefined slope in m-slope to the corresponding slope in S-slope. Similarly with mapping the zero m-slope to the corresponding S-slope.

To prove the bijection you can do the following. It can be shown that S-slope between two points is the m-slope found after rotating those two points 45 degrees counterclockwise about the origin. This is in fact how I came to the definition of S-slope. A short proof sketch would be to show injection by showing that every S-slope maps to the m-slope of the same line rotated 45 degrees counterclockwise about any point on that line. Every m-slope maps to the S-slope of the line rotated 45 degrees clockwise about any point on that line. To prove surjection I can map onto any specified S-slope by first finding the m-slope of that line rotated 45 degrees counterclockwise about any point on that line. This m-slope will map to the specified S-slope.

Now as to my other claim that my definition of S-slope still let's you do Calculus.

Remember that in Calculus the definition of instantaneous slope at a point can be written more precisely as:
\( \lim_{c \to 0} \frac{f(x+c)-f(x)}{(x+c)-x} \)

You may recognize this as the m-slope between the points \( (x,f(x)) \) and \( (x+c,f(x+c)) \)

So now let's do the same thing, but use S-slope:
The instantaneous S-slope at a point is defined by:
\( \lim_{c \to 0} \frac{[(x+c)-x]+[f(x+c)-f(x)]}{[(x+c)-x]-[f(x+c)-f(x)]} \)
\( \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

So now let's look at something familiar, like the Power Rule. Using m-slope, the power rules states that:
\( \frac{d}{dx}(x^n)=n*x^{n-1} \)

We now wish to seek the S-slope Power Rule:
Using the above definition for instantaneous S-slope at a point, let's use \( f(x)=x^n \)

\( \lim_{c \to 0} \frac{c+(x+c)^n-x^n}{c-(x+c)^n+x^n} \)
Expanding this via the binomial theorem gives:

\( \lim_{c \to 0} \frac{c+x^n-x^n+\sum_{i=1}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i}}{c-x^n+x^n-\sum_{i=1}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i}} \)

\( \lim_{c \to 0} \frac{1+nx^{n-1}+\sum_{i=2}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i-1}}{1-nx^{n-1}-\sum_{i=2}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i-1}} \)

\( \frac{1+nx^{n-1}}{1-nx^{n-1}} \)

And so I give you the S-slope Power Rule for Derivatives
\( \left(\frac{d}{dx}\right)_S x^n = \frac{1+nx^{n-1}}{1-nx^{n-1}} \)

Other calculus concepts could be derived very similarly using the definition of S-slope and standard calculus techniques.

Chris

 

[Cubenovice]

And that's why we love you Chris!

Not that I understand too much of it...
I scored <17 out of a 100 possible points several times in math tests so I guess I amnot part of the target audience here.
LOL

 

[aronpm]

Chris, couldn't you also just use \( \frac{dx}{dy} = 0 \) as the slope? Your slope definition is very fascinating though.

 

[FMC]

nice post!now the slope of 45 degrees line is undefined

 

[irontwig]

Or you can just use degrees/radians, Chris.

 

[aronpm]

Chris, I was wondering, is this always true?

\( \left(\frac{d}{dx}\right)_S f(x) = \frac{1+f'(x)}{1-f'(x)} \)

 

[Kenneth]

Chris, I was wondering, is this always true?

\( \left(\frac{d}{dx}\right)_S f(x) = \frac{1+f'(x)}{1-f'(x)} \)

LOL, you was supposed to think yourself

 

[MichaelP.]

And what were our purposes again?

Is this really necessary?
"Our purposes", which I only used when justifying the word "needs", which Chris said was correct in the post below yours, are to show that vertical lines in math have a slope (of undefined).

The point that Michael is trying to make is that the definition of slope as we know it is completely arbitrary. There is no reason that our familiar \( \frac{y_2-y_1}{x_2-x_1} \) should be the definition for slope. A better way to phrase that is that this formula is a definition of slope. Vertical lines have an undefined slope not because of anything particularly special or magical about vertical lines. They have undefined slope because the accepted definition of slope as \( \frac{y_2-y_1}{x_2-x_1} \) happens to assign an undefined number to vertical lines. That is the entire reason.

Spoiler

And now to put my money where my mouth is. In the rest of this post I will do all of what I claimed in this quote box. For those of you who typically do not read my very long posts then stop here. If you're a tl;dr forum browser than FOR YOUR OWN SAKE STOP READING NOW

--------------------------------------------------------------------------------------------

I will now define a new type of slope that is different from our normal definition of slope. I still want it to be related to slope in that as a line changes in steepness, the slope also changes. I also want to make sure that every line has a scalar slope, with the exception of the particular line that will be assigned the undefined slope based on my definition. There is nothing special about this particular line. It just gets the short straw in the drawing so to speak.

And now Ladies and Gentlemen I give you:

S-Slope

S-Slope will be defined by the letter S (where we use the letter m for standard slope). I will distinguish between my new S-slope by always calling it S-slope. I will refer to our more well known and loved standard slope by the name m-slope for the letter that usually represents it.


S-Slope is defined by the following formula (by me and because I say so - because slope definitions are completely arbitrary anyway!):

\( S = \frac{(x_2-x_1)+(y_2-y_1)}{(x_2-x_1)-(y_2-y_1)} \)

Now consider two points on a vertical line. Let's use \( (0,0) \) and \( (0,1) \)

\( S = \frac{(0-0)+(1-0)}{(0-0)-(1-0)} = \frac{1}{-1} = -1 \)

More generally, any two points \( (a,b) \) and \( (a,c) \) have slope:

\( S = \frac{(a-a)+(c-b)}{(a-a)-(c-b)} = \frac{c-b}{-(c-b)} = \frac{1}{-1} = -1 \)

Therefore vertical lines have slope -1 using this definition of slope.

----------

Now you may think that this definition of slope is artificially constructed such that it produces this nice result for vertical lines only. In fact, this definition of slope and the regular definition form a bijecton. Essentially this would mean that my definition of slope is mathematically equivalent, in some sense, to that of regular slope. I can take any slope:
\( m = \frac{b}{a} \) and map it into my definition of slope using the transformation:
\( S = \frac{a+b}{a-b} \)

You would need to use this transformation with the help of limits to map the undefined slope in m-slope to the corresponding slope in S-slope. Similarly with mapping the zero m-slope to the corresponding S-slope.

To prove the bijection you can do the following. It can be shown that S-slope between two points is the m-slope found after rotating those two points 45 degrees counterclockwise about the origin. This is in fact how I came to the definition of S-slope. A short proof sketch would be to show injection by showing that every S-slope maps to the m-slope of the same line rotated 45 degrees counterclockwise about any point on that line. Every m-slope maps to the S-slope of the line rotated 45 degrees clockwise about any point on that line. To prove surjection I can map onto any specified S-slope by first finding the m-slope of that line rotated 45 degrees counterclockwise about any point on that line. This m-slope will map to the specified S-slope.

Now as to my other claim that my definition of S-slope still let's you do Calculus.

Remember that in Calculus the definition of instantaneous slope at a point can be written more precisely as:
\( \lim_{c \to 0} \frac{f(x+c)-f(x)}{(x+c)-x} \)

You may recognize this as the m-slope between the points \( (x,f(x)) \) and \( (x+c,f(x+c)) \)

So now let's do the same thing, but use S-slope:
The instantaneous S-slope at a point is defined by:
\( \lim_{c \to 0} \frac{[(x+c)-x]+[f(x+c)-f(x)]}{[(x+c)-x]-[f(x+c)-f(x)]} \)
\( \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

So now let's look at something familiar, like the Power Rule. Using m-slope, the power rules states that:
\( \frac{d}{dx}(x^n)=n*x^{n-1} \)

We now wish to seek the S-slope Power Rule:
Using the above definition for instantaneous S-slope at a point, let's use \( f(x)=x^n \)

\( \lim_{c \to 0} \frac{c+(x+c)^n-x^n}{c-(x+c)^n+x^n} \)
Expanding this via the binomial theorem gives:

\( \lim_{c \to 0} \frac{c+x^n-x^n+\sum_{i=1}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i}}{c-x^n+x^n-\sum_{i=1}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i}} \)

\( \lim_{c \to 0} \frac{1+nx^{n-1}+\sum_{i=2}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i-1}}{1-nx^{n-1}-\sum_{i=2}^n \left( \begin{array}{c}n \\i \\\end{array} \right)x^{n-i}c^{i-1}} \)

\( \frac{1+nx^{n-1}}{1-nx^{n-1}} \)

And so I give you the S-slope Power Rule for Derivatives
\( \left(\frac{d}{dx}\right)_S x^n = \frac{1+nx^{n-1}}{1-nx^{n-1}} \)

Other calculus concepts could be derived very similarly using the definition of S-slope and standard calculus techniques.

Chris

That's really impressive and interesting Chris. I still think it's pretty clear why we say the slope of a vertical line is undefined. As you mentioned earlier, your slope equation isn't useful in math except to define the slope of vertical lines, which in itself, seems like a relatively useless piece of information.

 

[cmhardw]

That's really impressive and interesting Chris. I still think it's pretty clear why we say the slope of a vertical line is undefined. As you mentioned earlier, your slope equation isn't useful in math except to define the slope of vertical lines, which in itself, seems like a relatively useless piece of information.

I don't think you read that part as clearly as it sounds. From my original crazy long post:

Now you may think that this definition of slope (S-slope) is artificially constructed such that it produces this nice result for vertical lines only. In fact, this definition of slope and the regular definition form a bijecton. Essentially this would mean that my definition of slope is mathematically equivalent, in some sense, to that of regular slope.

My S-slope is actually no less useful than the standard definition of slope. The only reason the whole world is not using my S-slope concept is that somebody came up with m-slope first, and decided that the whole world was going to use that. Stated another way, if my concept of S-slope is useless, then the standard definition of slope is equally as useless.

The decision of which slope formula to use was completely arbitrary. Now, the standard m-slope formula happens to be short, and makes things like the power rule for derivates rather short and nice. So that is probably the real reason why it was chosen over other slope formulas.

However, my S-slope is no less of a slope concept than what we currently consider the standard slope concept. In fact, the two concepts form a bijection: for all intensive purposes, they are the exact same concept, just appearing differently.

Chris

 

[Chrish]

LOL, you was supposed to think yourself

Thinking for yourself does not mean you are not allowed to ask questions. It may mean to THINK before you ask, because you might already know the answer, but there's nothing wrong with it.

 

[cmhardw]

Chris, I was wondering, is this always true?

\( \left(\frac{d}{dx}\right)_S f(x) = \frac{1+f'(x)}{1-f'(x)} \)

Wow, interesting observation. Yes it holds true always, but I will make your notation a little more precise:

\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)
\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(c+f(x+c)-f(x)\right) * \left(c-f(x+c)+f(x)\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(\frac{c+f(x+c)-f(x)}{c}\right) * \left(\frac{c-f(x+c)+f(x)}{c}\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \left[\left(1+\frac{f(x+c)-f(x)}{c}\right) * \left(1-\frac{f(x+c)-f(x)}{c}\right)^{-1}\right] \)
\( \left(\frac{d}{dx}\right)_S f(x) = \left[1+f'_m(x)\right] * \left[1-f'_m(x)\right]^{-1} \)
\( \left(\frac{d}{dx}\right)_S f(x) = \frac{1+f'_m(x)}{1-f'_m(x)} \)

Where \( f'_m(x) \) is defined as the m-slope (standard) derivative of f with respect to x.

Cool! S-Slope is looking to be easier and easier to deal with! I wonder why they came up with this silly m-slope concept in the first place :rolleyes:

Chris

 

[cmhardw]

Sorry for the double post, but my previous post is already dense with a long reply.

Not that I understand too much of it...
I scored <17 out of a 100 possible points several times in math tests so I guess I amnot part of the target audience here.
LOL

No I realize that not everyone will have studied this level of math, I tried to put comments in to explain the really big results such as the bijection (essentially a form of equivalence) between the two slope concepts. I wanted to put the math in as well, because I found it fascinating that things were working out so well the more I looked into it.

Chris, couldn't you also just use \( \frac{dx}{dy} = 0 \) as the slope? Your slope definition is very fascinating though.

Hi Aron, If I understand you correctly you mean why not look at the reciprocal of the m-slope formula as our new definition of slope? I wonder how this would affect things like the derivative? I would imagine that it would be like examining the derivative of the inverse function (assuming it exists) when using the comparable definition of derivative in that slope scheme. Don't quote me on that, I haven't looked at it yet, but that would be my initial guess.

nice post!now the slope of 45 degrees line is undefined

Yes, I wanted to construct a method such that the line y=x (a pretty standard function) was the line with undefined slope That is why I chose to rotate the two points 45 degree counterclockwise before finding the standard m-slope as the definition of S-slope. I wanted the undefined slope mapped to y=x and not the vertical line

Or you can just use degrees/radians, Chris.

Irontwig, I am interested in this. Do you mean to define slope of a line in terms of an angle with respect to some fixed line? For example the standard slope \( m=\frac{\Delta y}{\Delta x}=tan(\theta) \) if \( \theta \) is defined as the angle between our line and the horizontal.

so we could define a \( \theta \)-slope such that:
\( \theta \)-slope \( = \tan^{-1} \left(\frac{\Delta y}{\Delta x}\right) \)

I feel such a slope definition must have been looked into before. They even define slope as tangent of the angle theta on the mathworld page about slope. I imagine this would be much clearer than my, very arbitrary, definition of S-slope. The reason I chose S-slope to be such as it is was to make a point about how arbitrary slope definitions really are anyway. If I'm going to pick some arbitrary definition, why not go all out right? S-slope clearly already gives a fairly complicated derivative scheme, which I imagine will make integrals a fun chore to attempt However, S-slope still can be used to calculate derivates, which was the only point I was trying to make.

I think this theta-slope would be far more interesting and relevant than my S-slope to look into to be perfectly honest.

Chris

 

[Athefre]

There are actually people walking around who think 0 isn't a number (it is, and a very useful one at that), or the slope of a vertical line cannot be defined (it can, easily), or black means reflecting absolutely no light (not the way actual people use it), or the Christian God is violently anti-homosexual (he calls it an "abomination"... just like eating shellfish or pork).

Now that vertical slope has been discussed for five pages, I would like the rest of these explained. But, be careful with the last one.

I'm just wondering, but what's the deal with people sticking to dictionary definitions or things their teachers told them, even when they don't make sense?

...

Believe it or not, guys, authority figures can be wrong. Think for yourselves, do some basic research. It will make you smarter.

You should ask yourself why it matters so much to you if someone doesn't have the same knowledge, why it annoys you when someone says something incorrect. Depending on your current beliefs, you may or may not be surprised by the answer. You'll find it's the same answer to your question.

Just live your life and let others live theirs. If someone's actions are affecting someone else negatively, then you should care and try to change it.

 

[cmhardw]

If the books in the schools was using the language people speaks they would grasp things much easier.

No wounder the student can't transform 500 metres into kilometres, he does not know the translation of the term kilo!

I feel like this got lost after the cascade of slope posts. I think this is a very good point. If our terms were defined more clearly, or if students were taught the translations of the terms from a very young age, then I think more people would grasp these concepts much more quickly.

I've also seen our international students, who breeze through our metric system unit very easily, struggle with the Imperial units. I have to agree with them though that there really is not much rhyme or reason to these definitions at all

But it brings up the same point, if the terms were defined more clearly, or if the students were taught the meaning of those terms from a younger age, then there would be far less confusion in learning that topic.

Chris

 

[MichaelP.]

I don't think you read that part as clearly as it sounds. From my original crazy long post:



My S-slope is actually no less useful than the standard definition of slope. The only reason the whole world is not using my S-slope concept is that somebody came up with m-slope first, and decided that the whole world was going to use that. Stated another way, if my concept of S-slope is useless, then the standard definition of slope is equally as useless.

The decision of which slope formula to use was completely arbitrary. Now, the standard m-slope formula happens to be short, and makes things like the power rule for derivates rather short and nice. So that is probably the real reason why it was chosen over other slope formulas.

However, my S-slope is no less of a slope concept than what we currently consider the standard slope concept. In fact, the two concepts form a bijection: for all intensive purposes, they are the exact same concept, just appearing differently.

Chris

I was really referring to this. I interpreted it as leaving the slope of a vertical line undefined (and not redifing the slope formula) leads to cool things...

I disagree with this. A slope doesn't need to be a quantifiable number. It's just that cool things like Calculus and Modern Science happen when slope is a quantifiable number.

A line could be "really slopey" or "way totally not slopey" also. There is no need for a line to have a numerical slope, but it's arguably much better when it does have a quantifiable number slope.

Chris

 

[HaraldS]

Because everyone is lazy

 

[some1rational]

The definition of slope is as arbitrary as a choice of origin.

For example, I could call my '2' your '0' and all basic arithmetic would still be true. My '12' would be your '10', but relative to my own system, it's still 10 away from my 0 ('2'). Somewhat analogous to Chris's arbitrary choice of a definition of slope; in his case, the critical concept is the redefining of undefined slope to the y=x line, in this case, the critical concept is the redefining of '0' to '2'.

Now of course we use '0' as our origin for obvious reasons of simplicity, the point being the choice is arbitrary to begin with just as the choice of the line with 'undefined' slope was arbitrary. (which, incidentally, also brings along with it a choice of the line with '0' slope as the two must be orthogonal regardless of choice of a slope function)

Just giving another analogy. The most rewarding moments I recall in my mathematics lectures came when I bridged seemingly disconnected concepts, I hope this helped someone lol...

On a side note:
'0' and 'Infinity' (a la 'undefined', in the clase of slope) are significant abstract entities prevalent throughout mathematics.