Undefined Values: Is 1/z = 0 When R = 0?

[Gear300]

Let us say that 1/x + 1/y = 1/z; z is a function of R, so that z = 1/R
For R = 0, z does not exist in set C, in which C is the most general set for this case.
However, is it possible to say that 1/z = 1/(1/0) = 0/1 = 0 for R = 0?...is it valid?

 

[HallsofIvy]

Let us say that 1/x + 1/y = 1/z; z is a function of R, so that z = 1/R
For R = 0, z does not exist in set C, in which C is the most general set for this case.
However, is it possible to say that 1/z = 1/(1/0) = 0/1 = 0 for R = 0?...is it valid?

No, it would not be. Since z is not defined when R= 0, neither is 1/z. It would be valid to say that the limit of 1/z, as R goes to 0, is 0.

 

[Gear300]

Does that imply that there is some sort of discrepancy when we look at it as 1/z = 1/(1/R) = R?

 

[Gear300]

...o_o...

 

[Office_Shredder]

In a manner of speaking you can think of it that way. This is basically a question of defining a function carefully.

A function is defined as a map from a set A to a set B (notation: f:A -> B) where each element a in A is assigned a unique element in B (written f(a)). Note not all elements in B must be mapped to, and elements in B can have more than one element in A mapping to them. A is called the domain, B is called the co-domain

If you have f:A->B a function and g:B->C a function (g could also map D->C for D a subset of B) , then g(f(x)) is a function g(f):A->C (as any value in A is carried to a value in C). So if we take f(x)=1/x, the domain and codomain of this function is going to be R-{0} (R being the set of real numbers... alternatively it could be the set of complex numbers) Then f(f(x)) is a function that maps R-{0} to R-{0} where every element x is mapped to itself. But the function isn't defined at 0 itself, as 0 isn't in the domain (so f(f(0)) is as meaningful as f(f(apple)) or f(f(red)).

It's possible to add additional values (such as an infinity symbol) to the codomain in order to allow 0 to be in the domain, and in such cases you'll find f(f(0)) ends up being 0, but these are special cases

 

[Gear300]

It's possible to add additional values (such as an infinity symbol) to the codomain in order to allow 0 to be in the domain, and in such cases you'll find f(f(0)) ends up being 0, but these are special cases

What is infinity (I've come up with a vague definition, but I don't think it suffices in more modern script)?

 

[slider142]

There are many different types of infinities used in mathematics. The infinity that Office_Shredder is referring to is the infinity of the Projectively Extended Real Numbers, which is defined in the link. The extended real number sets unfortunately do not have a field algebra, so you will not see them used much in everyday arithmetic.

 

[HallsofIvy]

Does that imply that there is some sort of discrepancy when we look at it as 1/z = 1/(1/R) = R?

Yes, there is a discrepancy.
If z= 1/R, then 1/z= R for all R except R= 0.

A variation often seen in beginning Calculus is this: $$\frac{x^2- 4}{x- 2}= x+2$$ for all x except x= 2. Any good textbook will make that point.

 

[Gear300]

Yes, there is a discrepancy.
If z= 1/R, then 1/z= R for all R except R= 0.

A variation often seen in beginning Calculus is this: $$\frac{x^2- 4}{x- 2}= x+2$$ for all x except x= 2. Any good textbook will make that point.

There are many different types of infinities used in mathematics. The infinity that Office_Shredder is referring to is the infinity of the Projectively Extended Real Numbers, which is defined in the link. The extended real number sets unfortunately do not have a field algebra, so you will not see them used much in everyday arithmetic.

Oh...I see...

 

[eliotargy]

Yes, there is a discrepancy.
If z= 1/R, then 1/z= R for all R except R= 0.

A variation often seen in beginning Calculus is this: $$\frac{x^2- 4}{x- 2}= x+2$$ for all x except x= 2. Any good textbook will make that point.

I thought that was just called a removable discontinuity?

 

[HallsofIvy]

Yes it is a "removable discontinuity". What is your point?

 

[eliotargy]

I thought that was just called a removable discontinuity?

So why is there a discrepency? And why the original question from the other guy?