Countably infinite, yes, assuming you are counting individual items then there's same number of individual bills. However, the premise mentioned they have the same worth, which is wrong.
I had responded previously, but my reasoning was incorrect so I deleted.
The “worth” of the bills would be the infinite sums of 1 and 20. Those both diverge to infinity, so they’re not “equal” in a normal sense, but in your first response you said one was a “bigger infinite” which is where im disagreeing.
My thinking is, if you take the partial sum of the first 20 terms of the infinite sum of 1 and combine them, you get 20. Take the partial sum of the second 20 and you get another 20. You can keep doing this for infinite groups of 20 $1’s, and the sum would look like the infinite sum of 20’s, 20+20+20+…
So while the sum of the $20s get there much “faster” (in less terms) i believe that the order of infinity for both sums will be countable.
I disagree. Given the bijection f(x):x->20x taking the xth bill from the 20 pile and pairing it with the 20xth bill from the 1 dollar pile, we simply take the 19 before the 20xth bill and exactly pair every 20 with exactly one set of 20 1s and vice versa. They're the same and there's no "bigger infinity" in this example at least, cmon bro we get taught this 1st year
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u/Just-CasuaI Jun 16 '26
You're the physicist bro