.9 repeating equals 1
PostPosted:Thu Jan 20, 2011 5:59 pm
This is something that always bothered me. The most common explanation would be something either take .9(repeating) and multiply it by 10, subtract itself, and then divide by 9, or take .3(repeating) times it 3 which is equal to (1/3) * 3 = 1. Now I know since the ultimate result is true it is clearly safe to do this, but I believe Algebra is not supposed to deal with infinite numbers, so you're basically implying infinite quantites can be manipulated like a number. To me this is backwards. The only way an infinite string of 9s can be manipulated like a '1' is because by calculus we know that above number converges to 1 so we can manipulate that particular instance of infinite 9s. But then this doesn't mean you can always manipulate infinity as if it's just equal to X. Here's a case where you can't manipulate infinity.
1 + -1 + 1 + -1 + ... = 1 + -1 + 1 + -1 + ...
addition is associative, so ordering of parenthese doesn't matter
(1+-1) + (1+-1) + ... = 1 + (-1+1) + (-1+1) + ...
0 + 0 + ... = 1 + 0 + 0 + ...
0 = 1
The problem with the above proof is that the sum above is divergent and you can't just put paranetheses around that. I realize those two aren't exactly the same thing but they're both 'infinite' and are both representable as an infinite series. It seems to me the simple algebra explanation cheats in that it already knows the result converges so you can manipulate it like a number, but it's not a rigorous proof. Even if you use the geometric series, you'd still need calculus to argue that r^n = 0 if r < 1 limit n->infinity. I don't expect a rigorous background in calculus for anyone seeing this problem, but I think it should be pointed out at some point that it is because advanced mathematics exist so we know that some numbers converges to a finite number instead of divergent, which is why you can just manipulate it like a number instead of getting crazy results like the one I wrote above.
1 + -1 + 1 + -1 + ... = 1 + -1 + 1 + -1 + ...
addition is associative, so ordering of parenthese doesn't matter
(1+-1) + (1+-1) + ... = 1 + (-1+1) + (-1+1) + ...
0 + 0 + ... = 1 + 0 + 0 + ...
0 = 1
The problem with the above proof is that the sum above is divergent and you can't just put paranetheses around that. I realize those two aren't exactly the same thing but they're both 'infinite' and are both representable as an infinite series. It seems to me the simple algebra explanation cheats in that it already knows the result converges so you can manipulate it like a number, but it's not a rigorous proof. Even if you use the geometric series, you'd still need calculus to argue that r^n = 0 if r < 1 limit n->infinity. I don't expect a rigorous background in calculus for anyone seeing this problem, but I think it should be pointed out at some point that it is because advanced mathematics exist so we know that some numbers converges to a finite number instead of divergent, which is why you can just manipulate it like a number instead of getting crazy results like the one I wrote above.