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[palsch]

Anyway, I find the "that method won't work every time" argument to be pretty much silly. It would be like someone telling you that it's a bad a idea to drive to the grocery store without looking it up up google maps. Because "pull out your driveway, turn left, turn right, turn left" isn't a method that will work to get you to every place. And that's true. But if those directions get you to the grocery store, and you're going to the grocery store, and you know this...there's nothing wrong with doing that rather than using the method that "works in every case" of printing out a map and directions.

In day to day life, you are right. But if you are taking a course on how to navigate using google maps, just because you happen to know one of the routes the teachers are using as examples doesn't mean you shouldn't use them to learn the skill you are being taught.

With many methods you do have to put in the time and effort to know them well enough for future application. If you are using shortcuts to avoid doing that sort of drill then I understand the teachers being annoyed.

[Vector]

Eh. . . a function f is a pair of values from the product X x Y (consisting of all ordered pairs (x, y) such that x is in X and y is in Y), with every (x, y) and (x, y') in f having the property that y = y'.  Sometimes these functions can be determined by an equation, but most of the functions out there cannot.

...see...this is why people have a difficult time understanding math.

And it's because I was taught math in a way that apparently works well for most people that I had a difficult time understanding math.  This is what works for me.

[Pnx]

Eh. . . a function f is a pair of values from the product X x Y (consisting of all ordered pairs (x, y) such that x is in X and y is in Y), with every (x, y) and (x, y') in f having the property that y = y'.  Sometimes these functions can be determined by an equation, but most of the functions out there cannot.

I... I had reread that a few times before my brain clicked things together and said, "Ohhh that makes sense... err... I think".

Just to be clear, you are using "X x Y" to mean X * Y (X multiplied by Y), right?

[MagmaMcFry]

Eh. . . a function f is a pair of values from the product X x Y (consisting of all ordered pairs (x, y) such that x is in X and y is in Y), with every (x, y) and (x, y') in f having the property that y = y'.  Sometimes these functions can be determined by an equation, but most of the functions out there cannot.

I... I had reread that a few times before my brain clicked things together and said, "Ohhh that makes sense... err... I think".

Just to be clear, you are using "X x Y" to mean X * Y (X multiplied by Y), right?

X is the set of values that you can place into the function, and Y is the set of values that all return values are from. X x Y is not multiplication of integers, but the cartesian product of those two sets (the set of all ordered pairs with the first element from X and the second element from Y).

Example: {1, 3, 6} x {2, 6} = {(1, 2), (1, 6), (3, 2), (3, 6), (6, 2), (6, 6)}

[Pnx]

That makes more sense, I guess. And it would better describe functions that have broken up domains.

[Vector]

And better suggest that most functions aren't continuous, differentiable, or any other special condition you could care to place on them.  There is a long, long list of benefits to the formal definition that doesn't depend on any defining relation.

[Putnam]

I'm guessing most here means "an infinitely large set, but with a higher cardinality than the set of functions that do have these limitations"?

[Vector]

If I remember correctly, the set of differentiable functions has measure zero.  That basically means that, proportionally speaking, they might as well not exist.  Less than one grain of sand on an entire beach.

[Putnam]

Ah, so... like the set of integers vs. the set of complex numbers or something.

[Pnx]

So basically if I've gotten the gist of it, you're saying that the relationship between X₁ and Y₁ in the ordered pair (X₁,Y₁) can be completely different from the relationship between X₂ and Y₂ in (X₂,Y₂), despite them both being part of the same function... If I'm doing my math speak right.

[Vector]

In a certain sense, yes.

So basically if I've gotten the gist of it, you're saying that the relationship between X₁ and Y₁ in the ordered pair (X₁,Y₁) can be completely different from the relationship between X₂ and Y₂ in (X₂,Y₂), despite them both being part of the same function... If I'm doing my math speak right.

Right, like the function {(1, 3), (2, 4), (3, 4), (4, 1), (5, 2)}.  We know what each thing maps to, so it's a function!  It's just not a very pretty one.

[MagmaMcFry]

I'm guessing most here means "an infinitely large set, but with a higher cardinality than the set of functions that do have these limitations"?

Well, maybe Vector's just referring to a straightforward measure over all functions where the set of continuous functions has measure zero, but the set of real functions has cardinality 2^c and the set of continuous real functions have cardinality c, so your definition of "most" is also correct.

4xNinja fake-edit:

If I remember correctly, the set of differentiable functions has measure zero.  That basically means that, proportionally speaking, they might as well not exist.  Less than one grain of sand on an entire beach.

What measure are you referring to?

[Vector]

Yeah, the usual metric over all functions inherited from the Euclidean norm.

[MagmaMcFry]

Yeah, the usual metric over all functions inherited from the Euclidean norm.

Derp, confused metric with measure (being German and stuff).

[Vector]

Hmm. . . I can't quite remember the specific measure--it's been a while.  I think it's the adaptation of the Lebesgue measure to the function space.