Two arithmetic variables have a linear relationship if they maintain a constant ratio. If you are moving at constant speed then the distance moved is directly proportional to the time for which you have been travelling. If an item in a shop costs $5 then, usually, two items will cost $10, but there may be a discount for the second item, resulting in a nonlinear relationship.doglover9754 said:So, my pre-algebra class is learning about linear relationships, but can anyone help explain it to me a little more? I tried google, but honestly, I want to hear other people before jumping to conclusions. So, my question is, what is a linear relationship?
A set of ordered pair can form a line when graphed.doglover9754 said:So, my pre-algebra class is learning about linear relationships, but can anyone help explain it to me a little more? I tried google, but honestly, I want to hear other people before jumping to conclusions. So, my question is, what is a linear relationship?
Keep in mind doglover9754 is a pre-algebra student.WWGD said:In a slightly different way, a relationship is linear if multiplying the input value by k will give rise to an output k times higher, i.e., L(kx)=kL(x) * and evaluating at a sum of variables will give the sum of the evaluation at each value, i.e., L(x+a)=L(x)+L(a) . Think of relations that are not that way. Does , say, doubling input usually give rise to doubling output? Does evaluating at a sum always produce sum of each evaluation?
* This arguably follows from the second: L(kx)=L(x+x+...+x)=L(x)+L(x)+...L(x) (k times). So second one is arguably the fundamental one
More specifically, if the set of ordered pairs fall on a line then the relationship is "affine".symbolipoint said:A set of ordered pair can form a line when graphed.
Amen. For example, discussing why for a linear function L, L(kx) = kL(x) is very likely well beyond any prealgebra level.symbolipoint said:Keep in mind doglover9754 is a pre-algebra student.
I always thought that latter description (through the Origin) was joint or direct variation. (Still, linear).jbriggs444 said:More specifically, if the set of ordered pairs fall on a line then the relationship is "affine".
If the set of ordered pairs fall on a line and the line goes through the origin then the relationship is "linear".
That is " a fine" distinction to make . I agree, I get pissed when people use linear when they really mean "affine" , it creates confusion over the long run.jbriggs444 said:More specifically, if the set of ordered pairs fall on a line then the relationship is "affine".
If the set of ordered pairs fall on a line and the line goes through the origin then the relationship is "linear".
I am only familiar with descriptions which fit y=mx+b and Ax+By=C being "linear"; and that joint and direct variation are as y=kx or x=ky. Also being "linear". This much should be adequate for the asking member, doglover9754.WWGD said:That is " a fine" distinction to make . I agree, I get pissed when people use linear when they really mean "affine" , it creates confusion over the long run.
The issue is that linear relationships are satisfied only by objects going through the origin. If you agree that a linear map L should satisfy ## L(v+w)=L(v)+L(w) ## , then ##y=mx+b ## does not satisfy that: ##y(v+w)=m(v+w)+b =mv+mw+b \neq y(v)+y(w)=mv+b)+m(w+b)=mv+mw+2b## unless b=0 . The map y=mx+b , strictly speaking is the composition of the linear map x with a translation ( by b units), which is not linear . EDIT: What will ultimately create confusion beyond if one goes beyond a very basic level and notices that most of the concepts : kernel of a linear map, basis, etc. , do not make sense for affine maps. But, yes, I guess it depends on the choice of pedagogy, which I cannot make for others.symbolipoint said:I am only familiar with descriptions which fit y=mx+b and Ax+By=C being "linear"; and that joint and direct variation are as y=kx or x=ky. Also being "linear". This much should be adequate for the asking member, doglover9754.
Hey, too much fancy talk (even if academically good). I can only report the extent that my education took me. We did not deal with affine-anything in College Algebra, so the finer distinction you are giving, I still do not understand. Original poster most likely has not gone beyond Intermediate Algebra yet, so he would tend to recognize something fitting with y=mx+b or Ax+By=C, as a linear relationship.WWGD said:The issue is that linear relationships are satisfied only by objects going through the origin. If you agree that a linear map L should satisfy ## L(v+w)=L(v)+L(w) ## , then ##y=mx+b ## does not satisfy that: ##y(v+w)=m(v+w)+b =mv+mw+b \neq y(v)+y(w)=mv+b)+m(w+b)=mv+mw+2b## unless b=0 . The map y=mx+b , strictly speaking is the composition of the linear map x with a translation ( by b units), which is not linear . EDIT: What will ultimately create confusion beyond if one goes beyond a very basic level and notices that most of the concepts : kernel of a linear map, basis, etc. , do not make sense for affine maps. But, yes, I guess it depends on the choice of pedagogy, which I cannot make for others.
It's a tricky thing/thin line: do you (over)simplify at the beginning to make concepts easier to understand and leave the student unprepared forsymbolipoint said:Hey, too much fancy talk (even if academically good). I can only report the extent that my education took me. We did not deal with affine-anything in College Algebra, so the finer distinction you are giving, I still do not understand. Original poster most likely has not gone beyond Intermediate Algebra yet, so he would tend to recognize something fitting with y=mx+b or Ax+By=C, as a linear relationship.
Note that "College Algebra" was much like Intermediate level of Algebra - still mostly dealing with generalized arithmetic using variables, graphs of functions, transcendental functions, a brief algebraic introduction to Limits not yet to the level of Calculus.
Maybe we were being taught somewhat wrong just for keeping simplicity of instruction.
Because the OP is in a pre-Algebra class, much of the discussion so far in this thread is akin to drinking from a firehose. In my view, it's best to give simple descriptions that don't cover every possibility, but qualify them by saying that these are simplified explanations that will be made more precise in later courses.WWGD said:It's a tricky thing/thin line: do you (over)simplify at the beginning to make concepts easier to understand and leave the student unprepared for
more complex material later on or do you introduce complexities from the get go so student becomes accustomed to it from the get-go; harder
at the beginning , maybe easier over the long run? Of course this is a false dichotomy, but it is hard to know which way to go.
Still, it is a choice. No reason why one can't just have an informal view of the more advanced material without the pressure to master it. It may allow for easier time with the material if she is interested in going beyond this level. EDIT I for one, wish I had had an understanding earlier that not all mathematical objects I would eventually encounter were going to be as nice/simple/full of structure as Euclidean space, particularly without the pressure of having to understanding the material. I'm giving the OP access to the best of both worlds.Mark44 said:Because the OP is in a pre-Algebra class, much of the discussion so far in this thread is akin to drinking from a firehose. In my view, it's best to give simple descriptions that don't cover every possibility, but qualify them by saying that these are simplified explanations that will be made more precise in later courses.
Something is worth saying: One learns to solve simple (or easy) examples before learning to solve more complicated examples. Maybe how to teach topics follows this same idea.WWGD said:Still, it is a choice. No reason why one can't just have an informal view of the more advanced material without the pressure to master it. It may allow for easier time with the material if she is interested in going beyond this level. EDIT I for one, wish I had had an understanding earlier that not all mathematical objects I would eventually encounter were going to be as nice/simple/full of structure as Euclidean space, particularly without the pressure of having to understanding the material. I'm giving the OP access to the best of both worlds.
But this is true as well: the earlier you start breaking down the harder stuff, the earlier you are likely to absorb it. At least I want to provide a glimpse, not intended to be a lesson but just a quick look into more advanced work. I see all time in other sites undergrads approaching advanced-level topics they know they will not fully ( or even partially) understand on the first ( maybe second, third, etc.) review, but the advanced material remains there, marinating at a subconscious level, and, upon seeing it later on , it starts making sense. This is not likely to happen when you wait till the last minute to study the material. EDIT: I don't think there is a clearly correct answer here; I don't see it as a black/white right/wrong issue.symbolipoint said:Something is worth saying: One learns to solve simple (or easy) examples before learning to solve more complicated examples. Maybe how to teach topics follows this same idea.
WWGD said:Still, it is a choice. No reason why one can't just have an informal view of the more advanced material...
WWGD said:That is " a fine" distinction to make
StoneTemplePython said:0\- - - -
Ugh... I can't resist: your reference earlier to affine spaces:reminds me that you "con vex" people with too much precision too early.
And you really think that any of this would be helpful to someone who is not even in algebra yet, and for whom the concepts of "linear map," "composition," and "translation" have not been presented yet?WWGD said:The issue is that linear relationships are satisfied only by objects going through the origin. If you agree that a linear map L should satisfy ## L(v+w)=L(v)+L(w) ## , then ##y=mx+b ## does not satisfy that: ##y(v+w)=m(v+w)+b =mv+mw+b \neq y(v)+y(w)=mv+b)+m(w+b)=mv+mw+2b## unless b=0 . The map y=mx+b , strictly speaking is the composition of the linear map x with a translation ( by b units), which is not linear .
Right. See posting numbers #5 through #19, approximately.Mark44 said:And you really think that any of this would be helpful to someone who is not even in algebra yet, and for whom the concepts of "linear map," "composition," and "translation" have not been presented yet?
We have a system of tags for threads: B (roughly high school and below), I (undergrad), and A (postgrad), respectively. The expectation is that a response to a B post, such as this one, will be geared to that level, and not wander off into college-level or beyond topics.