Let's say you have a computer program that counts natural numbers from 1 to infinity. Between printing the number [math]k[/math] and [math]k + 1[/math], it waits exactly [math]k[/math] seconds.
It's clear that it would take [math]1 + 2 + 3 + ...[/math] seconds for the program to finish the counting. Any sane person would now argue that this would be infinite amount of time, i.e. the program never finishes its task. A mathematician would claim that the program only needs [math]-\frac{1}{12}[/math] seconds, i.e. we would immediately get the results out.
It's obvious that the mathematician is wrong with this one -- Anyone can test this out using a simple script.
Jason Gray
Actually, -1/12 means you got the result in the past
Liam Campbell
Sorry op, I actually ran the program and got it all printing a fraction of a second before I started it.
Carson Myers
clearly this is incorrect, because there is no computer that can count up to infinity, so there is no computer that can execute such program.
but could imagine a settings where gods solve problem that require infinite time to calculate with this trick of making the computer wait k seconds each operation, it would be pretty rad actually
Benjamin Gutierrez
Since numbers are stored in memory spaces of n bytes, your program would eventually only print the number 111111... (n times) after 11111... seconds
Ryan Rivera
You [math]\textbf{can}[/math] make a computer to count to infinity, and you only need a constant amount of memory for that, too. Just make it print some symbol after another, and interpret it as a natural number in unary.
Jeremiah Kelly
Ok but then if would print - 1/12, because the series of printed values 1+1+1+1..... is less than the series of second spent waiting 1+2+3+...., but the series 1+1+1+1 is greater than - 1/12, therefore
-1/12
Jason Evans
>because the series of printed values 1+1+1+1..... is less than the series of second spent waiting 1+2+3+.... You misunderstood. If you have [math]k[/math] symbols on the screen, you wait for [math]k[/math] seconds. Then you put another symbol after the old ones, so that you have [math]k + 1[/math] symbols, and you wait for [math]k + 1[/math] seconds.
>It does not matter how many times you call your print function, or what is the argument, your program will always have - 1/12 as output Nope. You can't have [math]-\frac{1}{12}[/math] symbols on the screen. Nor is the program even designed to erase existing symbols -- Only to add new ones.
But I just realized that the program still wouldn't work, because the next wait time would still need to be stored in memory.
Ryder Lee
This. Mathematicians are 1/12 of a second faster than OP starting the program.
Vedä vittus
Jace Sanders
i was about to run the program but the numbers appeared on my screen before i compiled it
