# Why can't people think for themselves?



## qqwref (Nov 16, 2010)

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.


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## joey (Nov 16, 2010)

You tell me.


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## cmhardw (Nov 16, 2010)

Thinking is too hard, I dunwanna


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## marthaurion (Nov 16, 2010)

venting much?


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## Chrish (Nov 16, 2010)

All things can only be interpreted one way, of course.


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## Johan444 (Nov 16, 2010)

Because they're not as smart and intelligent and awesome as you.

j/k


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## BigSams (Nov 16, 2010)

marthaurion said:


> venting much?


 
<3 lelouch dp. .. no homo lol.

I think people don't "think" because either they don't care, it's easier than learning the truth or they have too much faith in what you called authority figures. Or in fact they do think and act like this to see reactions, instigate so-called philosophical discussions (wayyy too many such threads on this forum) or just get attention.
My solution? Ignore mostly and reprimand/rage now and then


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## jiggy (Nov 16, 2010)

qqwref said:


> 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).


Naivety is not a crime. You simply cannot be an expert in absolutely everything in the world and to expect this of people is naive in itself...so I forgive you! 

What's there to get upset about?


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## Kirjava (Nov 16, 2010)

people are dumb. I ranted about this for a few hours last night on IRC.

someone should post logs for comic effect.


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## Toad (Nov 16, 2010)

But hey, by posting this qq is giving himself authority and therfore could be wrong? Ya get me?!

No.

Good. Cos he speaks truths.


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## Cubenovice (Nov 16, 2010)

Kirjava said:


> someone should post logs for comic effect.




At your service...


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## cmhardw (Nov 16, 2010)

Cubenovice said:


> At your service...
> 
> 
> (picture of logs)


 
I LOL'd


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## Chrish (Nov 16, 2010)

People never see themselves as dumb, they only see other people as dumb. Yet there are dumb people.

fixed error

only -- > other


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## Cubenovice (Nov 16, 2010)

Chrish said:


> People never see themselves as dumb, they only see only people as dumb. Yet there are dumb people.



And we all are better drivers than than all the other people on the roads too.

RE: thinking for yourself, This reminds me a little bit of "why can't people just use commons sense?" to which my answer always is "there is no such thing as common sense".

Which it proven every so many months by my wife when she, again..., uses an office chair *with wheels*, to stand on to grab something at height...


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## StachuK1992 (Nov 16, 2010)

cmhardw said:


> I LOL'd


 Same!
That was great.


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## cmhardw (Nov 16, 2010)

Michael,

I completely agree with the sentiment. I stand by my first post in this thread as an explanation of the prevailing attitude of _why_ people don't think for themselves.

I see this all the time in my math class. A perfect example is unit conversions. A student will call me over with an issue solving a problem. The problem reads: "Convert 500 meters into an equivalent measurement in kilometers."

A typical student reaction will be: "Well, I know there are 1000m in 1km, but does that mean I multiply the 500 and 1000, or divide?" My typical response to such a question is "You tell me." What I am implying here, and what I explain to them that I want them to do is to start brainstorming on how to figure out this question. I don't care whether or not you already know to multiply or divide on this problem, that's why you're in my class - to learn how to do many topics, one of which includes learning how to do unit conversions. However, in my class not knowing what to do is not an excuse to just throw your hands up in defeat.

I slowly prod them along to basically do something along the lines of "Try _something_, *anything*." I usually will have the students do *both* multiplication and division. Multiplying 500 * 1000 = 500,000 (implied km). Dividing gives 500/1000 = 0.5 (implied km).

I then ask them: "Which answer seems more _reasonable_?" More often than not this draws a blank stare from the student and an "I don't get it." I then tell them that they know one of these answers is very likely correct (one of them is), but they have to brainstorm and figure out which one makes more sense.

"How long is one kilometer?" I usually say. "Give me an example of something that is one kilometer away from you right now." They then guess some location on a different part of campus that is approx. 1km away.

"How long is 1 meter?" Same process, answer is usually "To the end of the table." (correct).

"So 500 meters, would be like 500 tables end to end right?" Student: "Right."

"So which answer (500,000km or 0.5km) seems to be the closest to 500 *tables* laid end to end?"

"Ooooooh, ok. 500,000km is *way* off."

"Exactly."

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

Again, I'm not implying that the students are "dumb" by any means. They don't understand the topic, and again that is fine. That's why they take this class, to learn that topic among many others.

What baffles me is the fact that so few students have a) the self confidence, and b) the drive, to brainstorm a situation to figure out how to do the problem. 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 don't understand the concept of "I'm stuck, so I give up." It makes more sense to say "I'm stuck, so let me try lots of different things until I hit on something that _seems_ like I'm on the right track."

That's the part that gets me. I'm sure I'm preaching to the choir here, but I had to throw in my two cents as well.

Chris


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## FatBoyXPC (Nov 16, 2010)

In fairness, having a definition from a set source gives us some sort of ability to say right or wrong. I think if we could freely define words as debates happen (as long as you have some sort of validity as to why it could be that way) then (some) things would never get settled.


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## Tyjet66 (Nov 16, 2010)

This inevitably leads to one thing, and one thing only:
Legalize cannabis!


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## Jason (Nov 16, 2010)

Come on, grow up, get over yourselves. Most people do actually think for themselves, but maybe simply don't give a toss about being able to define the slope of a vertical line. You're measuring people's ability to think for themselves using an extremely narrow spectrum of reasonning. Some people may simply have an appreciation of more aesthetic or non-quantifiable matters.


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## cmhardw (Nov 16, 2010)

Jason said:


> Come on, grow up, get over yourselves. Most people do actually think from themselves, but maybe simply don't give a toss about being able to define the slope of a vertical line. You're measuring people's ability to think for themselves using an extremely narrow spectrum of reasonning. Some people may simply have an appreciation of more aesthetic or non-quantifiable matters.


 
I teach math at an _Entertainment Arts_ school, so trust me in that I've heard this argument MANY times. Having an appreciation for the finer things in life is great, but what happens when you are presented with your first mortgage to sign? Or how are you going to manage your monthly bills such that you can pay your rent/food/other bills AND for your college education, and not go broke?

If the process of brainstorming through a topic in this class is too difficult (most problems are arguably between 2-5 steps), then how will you ever manage to balance your checking account with all of your bills taken into account? Learning to think for yourself is *not* a skill whose only use is in the classroom. It is an *essential* life skill that people need to hone and practice, *just like any other skill*.

It would be like saying "I want to get faster at my F2L, but I have an appreciation for the finer things in life so I don't want to drill my algs. I know that I'm capable of just thinking myself through to being faster." It just doesn't work that way.

Having an appreciation for the finer things in life is also not an excuse for not thinking for yourself. How would you get over an artistic *block* where your ideas stop flowing? You would need to try other side projects, work on your project from a different angle. Basically do _something_, *anything* that you feel could get you out of this block. How is this not the same kind of thinking for yourself that we ask people to do in a classroom?

I don't think it's any different.

Chris


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## Edward (Nov 16, 2010)

People don't think for themselves when the info is easily gained through other ways btw.


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## Owen (Nov 16, 2010)

Wait.. You're an authority figture.. Maybe you're wrong... but if you wrong that would make this statment..Wait... How...


Paradox.


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## waffle=ijm (Nov 16, 2010)

Edward said:


> People don't think for themselves when the info is easily gained through other ways btw.


 
same thing happened to me with some other pants.


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## Joël (Nov 17, 2010)

Thinking is not always the most important thing you need to do in order to survive and reproduce.


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## MichaelP. (Nov 17, 2010)

qqwref said:


> 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.




Nuh-uh. HOYVUX.


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## ~Phoenix Death~ (Nov 17, 2010)

cmhardw said:


> Thinking is too hard, I dunwanna


 
I think you just summed it up <3


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## The Puzzler (Nov 17, 2010)

I totally agree some people just follow exactly what they here and those people will never do anything super amazing in their lives.
BTW why did waffle get banned?


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## Edward (Nov 17, 2010)

The Puzzler said:


> I totally agree some people just follow exactly what they here and those people will never do anything super amazing in their lives.
> BTW why did waffle get banned?



"too many pants"


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## StachuK1992 (Nov 17, 2010)

Edward said:


> "too many pants"


 
same thing happened to me with some *other* pants.


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## The Puzzler (Nov 17, 2010)

Ok.... Nice Waffle. 

On topic: I really hope those people learn there is more to the world then their little clouded perception.


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## aronpm (Nov 17, 2010)

cmhardw said:


> I teach math at an _Entertainment Arts_ school, so trust me in that I've heard this argument MANY times. Having an appreciation for the finer things in life is great, but what happens when you are presented with your first mortgage to sign? Or how are you going to manage your monthly bills such that you can pay your rent/food/other bills AND for your college education, and not go broke?


 
Mathematics _is_ one of the finer things in life! 



MichaelP. said:


> Nuh-uh. HOYVUX.


Calculus.


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## EVH (Nov 17, 2010)

I have tried researching how to define the slope of a vertical line but have never been able to find out how. Could someone explain?

OT: I definitely agree with you, but for some it just seems easier to retain information that they are told, then research it themselves.


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## Ashmnafa (Nov 17, 2010)

About the finding the slope of a vertical line, a lot of people (including myself) still believe that because they haven't been taught the necessary information to find the slope of a vertical line.

As far as I know, slope is rise/run. And if the run is zero, then wouldn't the slope be undefined? I am curious.


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## Edward (Nov 17, 2010)

Ashmnafa said:


> About the finding the slope of a vertical line, a lot of people (including myself) still believe that because they haven't been taught the necessary information to find the slope of a vertical line.
> 
> As far as I know, slope is rise/run. And if the run is zero, then wouldn't the slope be undefined*?* I am curious.


A question mark :O
BLASPHEMY!
Think for yourself!


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## Ashmnafa (Nov 17, 2010)

Edward said:


> A question mark :O
> BLASPHEMY!
> Think for yourself!



I have tried, and I have failed, unless you care to explain, Edward.


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## number1failure (Nov 17, 2010)

qqwref said:


> 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.


 
THANK YOU!!!!


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## Edward (Nov 17, 2010)

Ashmnafa said:


> I have tried, and I have failed, unless you care to explain, Edward.


 
D: This isn't about me


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## Dene (Nov 17, 2010)

Ashmnafa said:


> About the finding the slope of a vertical line, a lot of people (including myself) still believe that because they haven't been taught the necessary information to find the slope of a vertical line.
> 
> As far as I know, slope is rise/run. And if the run is zero, then wouldn't the slope be undefined? I am curious.


 
I'd suggest something like this: The slope of a vertical line is vertical. Makes sense to me as long as you know what "vertical" means. Alternatively, you would say it doesn't have a slope, because that would require some sort of tilting. A vertical line is not tilting to either side therefore it has no slope.


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## Ashmnafa (Nov 17, 2010)

Dene said:


> I'd suggest something like this: The slope of a vertical line is vertical. Makes sense to me as long as you know what "vertical" means. Alternatively, you would say it doesn't have a slope, because that would require some sort of tilting. A vertical line is not tilting to either side therefore it has no slope.



A horizontal line has no slope, but I don't see why a vertical line couldn't either I guess.

But when you have a zero as the denominator, that is undefined, so that should mean that the slope of a vertical line is undefined, so I guess the above explanation wouldn't work.


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## oprah62 (Nov 17, 2010)

Ashmnafa said:


> A horizontal line has no slope


 
Wrong.


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## Ashmnafa (Nov 17, 2010)

oprah62 said:


> Wrong.


 
Sorry, allow me to correct myself. A horizontal line has a slope of zero.


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## MichaelP. (Nov 17, 2010)

Dene said:


> I'd suggest something like this: The slope of a vertical line is vertical. Makes sense to me as long as you know what "vertical" means. Alternatively, you would say it doesn't have a slope, because that would require some sort of tilting. A vertical line is not tilting to either side therefore it has no slope.


 
A slope needs to be a quantifiable number. You should be able to plug it into equations, and because a vertical line's slope is (Y2-Y1)/0 , and you cannot divide by 0, the slope is undefined.


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## oprah62 (Nov 17, 2010)

Ashmnafa said:


> Sorry, allow me to correct myself. A horizontal line has a slope of zero.


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## Ashmnafa (Nov 17, 2010)

MichaelP. said:


> A slope needs to be a quantifiable number. You should be able to plug it into equations, and because a vertical line's slope is (Y2-Y1)/0 , and you cannot divide by 0, the slope is undefined.


 
That sounds like a better explanation of what I just said. Thank you Mr. Perkins.


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## aronpm (Nov 17, 2010)

oprah62 said:


>


 


> 15:06 < oprah> A horizontal line has 0 slope right?
> 15:06 < oprah> guy on forums is wrong
> 15:07 <+aronpm> oprah: if you have 0 bananas you have no bananas



Yeah.


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## BigGreen (Nov 17, 2010)

But what about people who are incapable of thinking for themselves like the mentally challenged


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## oprah62 (Nov 17, 2010)

aronpm said:


> Yeah.


 
Posted that before I asked lol. 
I had this weird way of remembering it. On flat ground (horizontal) you're happy because you're safe, but if you're falling down (vertical) you're screaming NOOOOOOO.


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## MichaelP. (Nov 17, 2010)

oprah62 said:


> Posted that before I asked lol.
> I had this weird way of remembering it. On flat ground (horizontal) you're happy because you're safe, but if you're falling down (vertical) you're screaming NOOOOOOO.



HOYVUX FTW


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## cmhardw (Nov 17, 2010)

MichaelP. said:


> *A slope needs to be a quantifiable number.* You should be able to plug it into equations, and because a vertical line's slope is (Y2-Y1)/0 , and you cannot divide by 0, the slope is undefined.


 
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


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## MichaelP. (Nov 17, 2010)

cmhardw said:


> 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


 
In that case we don't need any numbers at all. We can just say "super big" or "eency weency". In my opinion this isn't really relevant to the discussion. I used "needs" because for our purposes, the slope needed to be quantifiable.


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## Dene (Nov 17, 2010)

And what were our purposes again?


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## cmhardw (Nov 17, 2010)

MichaelP. said:


> In that case we don't need any numbers at all. We can just say "super big" or "eency weency". In my opinion this isn't really relevant to the discussion. I used "needs" because for our purposes, the slope needed to be quantifiable.


 
Ok point taken. I will agree that it is much better to have a quantifiable number for the slope, as this clearly leads to lots of fancy shmancy maths 

However, and this hits on Michael point about an undefined slope:



> A slope needs to be a quantifiable number. You should be able to plug it into equations, and because *a vertical line's slope is (Y2-Y1)/0* , and you cannot divide by 0, the slope is undefined.



Why is a vertical line slope (Y2-Y1)/0?

In fact, why is slope even defined as \( \frac{y_2-y_1}{x_2-x_1} \) at all?

Hint: I'm not being facetious here, there actually is an answer to this question, and it is very closely related to Michael's point that a vertical line absolutely *can* have a defined slope.

Chris


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## Cubenovice (Nov 17, 2010)

Dene said:


> I'd suggest something like this: The slope of a vertical line is *magi*cal.


*fixed*


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## cmhardw (Nov 17, 2010)

Cubenovice said:


> *fixed*


 
Ralph, again I LOL'd  Well played sir... well played


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## BeautifullyDecayed. (Nov 17, 2010)

Cubenovice said:


> Which it proven every so many months by my wife when she, again..., uses an office chair *with wheels*, to stand on to grab something at height...


 
I do this all the time. I don't see the problem with it, often office chairs are adjustable so can get higher and especially if you're on carpet it isn't that hard to balance and usually it is the only chair around.


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## blade740 (Nov 17, 2010)

If you must put a numerical value to it, I would say that a vertical line has a slope of infinity. Common sense tells me that zero goes into any number infinity times, and so n/0 should be infinity. 

Or, to look at it another way, as a line tilts from 0 degrees (horizontal) to 90 degrees (vertical) the slope approaches infinity.


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## cmhardw (Nov 17, 2010)

blade740 said:


> If you must put a numerical value to it, I would say that a vertical line has a slope of infinity. Common sense tells me that zero goes into any number infinity times, and so n/0 should be infinity.
> 
> Or, to look at it another way, as a line tilts from 0 degrees (horizontal) to 90 degrees (vertical) the slope approaches infinity.


 
*Directed not at anyone in particular, but the thread in general*

The point I am trying to make is the same point as Michael was trying to make in the OP. I'll prove to you what I mean. A vertical line's slope is undefined. Like Michael, I disagree that this *must always* be the case. A vertical line's slope *can* be defined. 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

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

Chris


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## Hadley4000 (Nov 17, 2010)

I don't think people are incapable of thinking for themselves, I think they're just too lazy.


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## Zubon (Nov 17, 2010)

It seems that recently (especially because of the internet) that there are more and more people "questioning authority". Unfortunately it is going to extremes and irrational conspiracy theories are spreading like a disease. It is one thing to be skeptical and think for yourself after considering all the evidence, but abandoning rationality and blindly following some conspiracy theory that is labeled as "thinking outside the box" is very different.

It is good to think for yourself, but if you don't have all of the unbiased facts, all you will end up doing is regurgitating crap that other people say.

There are some thing that you learn in school that are either plain wrong or over simplifications. It is good to question authority as long as you have a valid reason to do so. I am pretty sure that there are more free thinkers now than there have ever been.


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## Cubenovice (Nov 17, 2010)

Interesting discussion going on here.
Since I'm not that smart or educated in the subject of math I'll try to summarize my learnings of this thread

Horizontal line: slope is zero / something = zero
Vertical line:
The general Idea: slope is something / zero = undefined so we're not even sure if we can use the term "slope" for this situation.
However it seems that scalable quantity actually can a be assigned. (Chris)
Right?


Let's complicate things for myself and compare parallel lines:
To my understanding two straight lines are considered to run parallel when they have identical slope.

Take two horizontal lines, both have an identical slope (zero): they are parallel
Now we take two vertical lines:
General idea: both have an "undefined" slope. So my question is: can we even claim that they are parallel? (Math people?)
Scalable quantity idea: can we use this to show that two vertical lines have identical "scalable quantity" and thus? are parallel?


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## Olivér Perge (Nov 17, 2010)

cmhardw said:


> ...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.


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## Kenneth (Nov 17, 2010)

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!


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## mrCage (Nov 17, 2010)

qqwref said:


> 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


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## cmhardw (Nov 17, 2010)

> 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.



cmhardw said:


> 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


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## Cubenovice (Nov 17, 2010)

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


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## aronpm (Nov 17, 2010)

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


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## FMC (Nov 17, 2010)

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


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## irontwig (Nov 17, 2010)

Or you can just use degrees/radians, Chris.


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## aronpm (Nov 17, 2010)

Chris, I was wondering, is this always true?

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


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## Kenneth (Nov 17, 2010)

aronpm said:


> 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


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## MichaelP. (Nov 17, 2010)

Dene said:


> 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). 



cmhardw said:


> 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.
> 
> 
> 
> ...


 
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.


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## cmhardw (Nov 17, 2010)

> 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


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## Chrish (Nov 17, 2010)

Kenneth said:


> 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.


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## cmhardw (Nov 17, 2010)

aronpm said:


> 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  

Chris


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## cmhardw (Nov 17, 2010)

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



Cubenovice said:


> 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.



aronpm said:


> 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.



FMC said:


> 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 



irontwig said:


> 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


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## Athefre (Nov 17, 2010)

qqwref said:


> 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.



qqwref said:


> 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.


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## cmhardw (Nov 17, 2010)

Kenneth said:


> 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


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## MichaelP. (Nov 17, 2010)

cmhardw said:


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


 

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...



cmhardw said:


> 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


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## HaraldS (Nov 17, 2010)

Because everyone is lazy


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## some1rational (Nov 17, 2010)

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.


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## mrCage (Nov 18, 2010)

This is going way off topic. But it reminds me of Euclidean geometry. It says that through a point outside a straight line we have exactly ONE straight parallell to that line. There are 2 other geometries possible. We have no such line or we have infinitely many such lines ... Not in ANY way cube/ing related!!

Per


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## riffz (Nov 19, 2010)

LMAO at Chris's post on S-slope. Seriously, you put so much detail and effort into everything you do.


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## Stefan (Nov 19, 2010)

riffz said:


> LMAO at Chris's post on S-slope. Seriously, you put so much detail and effort into everything you do.


 
I just wish he put a little more effort into making it short enough for me to read...


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## cmhardw (Nov 19, 2010)

Stefan said:


> I just wish he put a little more effort into making it short enough for me to read...


 
Stefan I knew it would probably go past your threshold  That's, in part, why I added the tl;dr warning.

I wanted to prove every claim I had made to a fairly high degree of formality, or at least give a proof sketch. I gave a proof sketch for my claim of the bijection between S-slope and m-slope, as making the claim without the proof is meaningless since it's a core part of my argument that m-slope and S-slope are basically no different from each other. I also wanted to show a walkthrough proof of how S-slope can still be used to calculate S-slope derivatives in a meaningful way, since that was a mighty big claim for me to make without providing a proof of it. On the next page Aron's question leads to a general proof of how to calculate S-slope derivatives for any function, but at the time I had only done this for the S-slope power rule.

How else could I have shortened it, other than removing the commentary which I added to let people follow along who maybe are not as familiar or comfortable with these sorts of math topics? My audience is made up of people who both understand all the math I am showing, and those who understand only some of the math I am showing. I was trying to accommodate as many people as possible.

The flow beyond the tl;dr warning was basically the following. Define S-slope. Show an example for two points on a vertical line, then any vertical line. Define the mapping between S-slope and m-slope. Provide a proof sketch of the bijection. Walk through a proof of the S-slope power rule for derivatives. The extra wording, I hope, is only the commentary to explain that to the lay person who is only casually following the argument.


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## Rpotts (Nov 19, 2010)

cmhardw said:


> In fact, the two concepts form a bijection: *for all intensive purposes*, they are the exact same concept, just appearing differently


 
HAHAHAHAA I had to, your posts were _too_ perfect, and no one pointed it out yet. lolz 



Spoiler



I thought that it was intensive purposes for the majority of my life, it wasn't til I saw it written out that I had my epiphany


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## Stefan (Nov 19, 2010)

cmhardw said:


> How else could I have shortened it


 
I guess I'd have to read it to answer that 

Well, also has to do with the topic, not just the length. I'm just not interested enough in an alternative slope definition.


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## cmhardw (Nov 19, 2010)

For those who do not want to read the longer version of my post, here are the Cliff notes.

Definitions
Let m-slope be defined by:
\( m = \frac{y_2-y_1}{x_2-x_1} \)

Let S-Slope be defined by:
\( S = \frac{(x_2-x_1)+(y_2-y_1)}{(x_2-x_1)-(y_2-y_1)} \)

Any m-slope \( m = \frac{b}{a} \) maps into S-slope via the mapping \( S = \frac{a+b}{a-b} \)
This mapping can be shown to be a bijection between m-slope and S-slope.

Calculus with S-slope
Let f(x) be a continuous function in a neighborhood around point x. The instantaneous S-slope at the point \( \left(x,f(x)\right) \) is defined by:

\( \lim_{c \to 0} \frac{[(x+c)-x]+[f(x+c)-f(x)]}{[(x+c)-x]-[f(x+c)-f(x)]} \)
\( \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

Derivatives in S-slope
Let \( \left(\frac{d}{dx}\right)_S f(x) = \lim_{c \to 0} \frac{c+f(x+c)-f(x)}{c-f(x+c)+f(x)} \)

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

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


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## mrCage (Nov 23, 2010)

Just for those who do not what a bijection is!!

Per


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