phoneutria wrote:You're absolutely right of course. My apologies.
No need to apologize, but since you already did, let me accept it.
I gave what I thought was an answer to the question
Right. But if someone gives you something you didn't really ask for, even if its partly or entirely your fault, you have to note it, right?
The proof is only fallacious due to its implications, which are that anything equals anything, which obviously isn't true.
I am not sure I am following you. The proof merely shows that "2 = 1" not that "anything equals anything". And while we both agree that its conclusion is false, the fact that it is does not tell us where the flaw lies. (The flaw can't be in the conclusion.) And since this is a deductive argument involving definitional premises, the flaw must be in the logic since definitional premises cannot be false. So where exactly is the flaw?
The reason that this can occur is because you're using infinity as an operand.
But which part of the proof is wrong? Perhaps the very first line? Infinity plus infinity isn't equal to infinity?
And here is what I think:
The problem with the proof is that \(\frac{\infty}{\infty}\) does not equal to \(1\).
This is based on the premise that \(\infty + \infty = \infty\) stands for "When you take a number greater than every integer and add to it a number greater than every integer, you always get a number greater than every integer". And that is true even if "a number greater than every integer" is a contradiction in terms. But when you take a number greater than every integer and divide it by a number greater than every integer, you do not always get \(1\).
Yeah it doesn't tell you much, which is why I said "by itself this means nothing". It has to apply to something.
I think you misunderstood me. When I said "It doesn't tell me much", I meant "It doesn't tell me what I want to know" and what I want to know is how many arguments it accepts, what kind of arguments and what kind of value it outputs.
But that doesn't mean you can simply say "\(\infty 5\) and \(\infty(5)\) are equal to \(true\)", which indeed is why I earlier said that they're not valid expressions, and why I said in an earlier post that there are many invalid ways to use any operation. When you use it in a valid way, it does have an arity of one, where it accepts things like sums, products, integrals etc. - making it more like an operation on operations (since sums, products, integrals etc. are operations themselves). Trying to pass an integer as a value is just another way to invalidly use it in its specific capacity as an operator. For other operators you can pass integers - that's not a necessary requirement for an operation.
Right. So I made several assumptions:
1) You think infinity is a function. (Turns out to be correct.)
2) You think it is a function that can accept single argument. (Turns out to be correct.)
3) You think it's a function that can accept an integer. (Turns out to be incorrect.)
What I know now is that you think that infinity is a function that takes a single argument of type "function" making it a higher-order function.
What I don't know is:
1) What kind of function does it take? Any or some specific?
2) What kind of value does it output?
3) How does it map its input to its output?
4) Who else defines infinity this way?
"Let's keep the debate about poor people in the US specifically. It's the land of opportunity. So everyone has an opportunity. That means everyone can get money. So some people who don't have it just aren't using thier opportunities, and then out of those who are using them, then most squander what they gain through poor choices, which keeps them poor. It's no one else's fault. The end."
Mr. Reasonable
