Can we discuss this?

[Leonhard]

Can we prove this? Can we discuss the reasons for this being the case?

Yes! But I'd love to see your response to my posts first.

[Leonhard]

I'm heading to bed though as its getting late over here. Goodnight.

[Teallaura]

Night, Leo!



Wait, how are we defining 'night'?

[pancreasman]

Can we prove this? Can we discuss the reasons for this being the case?

You can't actually 'prove' a definition. Definitions are just ... Definitions. It's all part of setting up a mathematical system.

It's like chess: the rules are that bishops can only move diagonally. If you move your bishop horizontally you're not playing chess anymore.

[Paprika]

1/0 is undefined.

I'm not arguing for the result of 1/0 to be defined, but that the operation of division be extended to having 0 as divisor.

[Leonhard]

I'm not arguing for the result of 1/0 to be defined, but that the operation of division be extended to having 0 as divisor.

I'm not sure what you're talking about then. If 1/0 can't have any value, what meaning does it say to extend division to include 0 as a divisor?

[Paprika]

I'm not sure what you're talking about then. If 1/0 can't have any value, what meaning does it say to extend division to include 0 as a divisor?

I distinguish between whether the operation is defined over the values used, and whether the result of the operation is defined:

Using the basic definition of infinite series, the operation of infinite series is defined for 1-2+3-4+..., ie. a_i=i*(-1)^(i+1); the operation 'makes sense' even though the result is not defined. Are you with me at this point?

[Leonhard]

I distinguish between whether the operation is defined over the values used, and whether the result of the operation is defined:

Using the basic definition of infinite series, the operation of infinite series is defined for 1-2+3-4+..., ie. a_i=i*(-1)^(i+1); the operation 'makes sense' even though the result is not defined. Are you with me at this point?

I think so, but if you want to be completely pedantic: there's no such thing as 'operation of infinite series'. If you mean that there's a partial sum for infinite series consisting of



and that this terms partial sum is divergent, and therefore that result its result is undefined... Yes we agree. Though there are alternative summation methods that can in fact produce a convergent value from the partial sum.

Did you follow that?

[Zymologist]

There's an extra "e" in the title. Vowels cost money!

Ye olde discussione on mathematicse.

[Paprika]

I think so, but if you want to be completely pedantic: there's no such thing as 'operation of infinite series'. If you mean that there's a partial sum for infinite series consisting of



and that this terms partial sum is divergent, and therefore that result its result is undefined... Yes we agree. Though there are alternative summation methods that can in fact produce a convergent value from the partial sum.

Did you follow that?

That was pretty elementary.

Therefore we can define the operation of division for the divisor 0 as lim (x->0) a/x even though the result is not defined, hence we get rid of that inelegant hole.

[Leonhard]

Therefore we can define the operation of division for the divisor 0 as lim (x->0) a/x even though the result is not defined, hence we get rid of that inelegant hole.

And what value would you plug into that "inelegant" hole?

[Paprika]

And what value would you plug into that "inelegant" hole?

Why should I need to?

[Leonhard]

Why should I need to?

How else are you going to get rid of a hole? The function f(x) = 1/x, remains defined for all x, except, x = 0.

[Paprika]

How else are you going to get rid of a hole?

The hole I was referring to was that the division operator was not defined for all real divisors. That hole is plugged, but not every hole can be plugged.

[Boxing Pythagoras]

The hole I was referring to was that the division operator was not defined for all real divisors. That hole is plugged, but not every hole can be plugged.

You haven't actually defined it for all Real divisors, though. You've simply kicked the can down the road, a bit.

The is undefined. So, if you say that the division operator over a divisor of 0 is equivalent to this limit, you still have not defined it.