Controversial concepts in mathematics

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Re: Controversial concepts in mathematics

Postby obsrvr524 » Wed Jan 20, 2021 3:54 pm

I had to read back to see where this discussion of \(\infty \) being an operator came from. I think I can see that it is just another language/communication problem where something has been shortened to the point of being confusing.

In this notation -
    \(\sum_{i=1}^{5}a_i\)
What is meant and I think originally stated is -
    \(\sum_{i=1}^{i=5}a_i\)
Meaning to iterate the summation operation until i=5.

And in this notation -
    \(\sum_{i=1}^{\infty}a_i\)
What is meant is -
    \(\sum_{i=1}^{i=\infty}a_i\)
Meaning to iterate the summation operation until i= {the highest possible number} -even though it doesn't exist.
Or -
    \(\sum_{i=1}^{i=Highest Possible Number}a_i\)

And from something I recently read in order to make that more logically rational the notation became -
    \(\sum_{i=1}^{i=>\infty}a_i\)
Implying "toward" the highest possible number so as to avoid the issue of i becoming a non-existent number.

When those details were left out, the idea of what was meant by -
    \(\sum_{i=1}^{\infty}a_i\)
became ambiguous.

I just had this very same conversation with Certainly real concerning other word usage shortcuts that cause ambiguity and confusion.

So in maths, \(\infty\) is actually referring to "Highest Possible Number".
And in that summation notation \(\sum_{}^{}a_i\) it is referring to a "direction toward the highest possible number" - as the DESTINATION of the Summation Operation.

At no time is \(\infty\) meant as an operator - always a destination of "highest possible number" even though it is known there is no highest to be reached.
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Thu Jan 21, 2021 5:20 am

It is not clear whether it means "the highest possible number" or "non-specific number greater than every integer". I am inclined to believe it refers to the latter.
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sat Jan 23, 2021 9:14 am

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?
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Re: Controversial concepts in mathematics

Postby MagsJ » Sun Jan 24, 2021 3:30 pm

_
Sock-puppets.. there are numerous and many.
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