A quick google search brought up this page for doing arithmetic on repeating decimals. I’ve only read through the addition part so far but it seems interesting.

]]>The mathematical induction I was using only works for natural numbers so I can only say things are true for finite natural numbers.Wikipedia says there’s a form of transfinite mathematical induction using the ordinal numbers but I’m not very familiar with it.

]]>Let x be 0.999…(n)…9 with n 9’s after the decimal place. We’re not looking at the infinite string of 9’s yet, just at a finite, by variable length string of 9’s. It’s also worth nothing that if we look up the nth decimal place, there’s the last 9, then an infinite string of 0’s.

From the induction I did above, we can see that the general form of x² is 0.999…(n-1)…9800…(n-1)…01. Or, (n-1) many 9’s after the decimal place, then, starting from the nth decimal place, an 8, then a (n-1) string of 0’s and then, finally, 1.

If we make x an infinitely long string of 9’s, there’s no longer a point, n where we have the last 9. Which means, in x², there’s no longer a point, n, where we stick on the 8, then the (n-1) string of 0’s, then 1. Only more 9’s. If x = 0.999… infinitely repeating, then for any decimal point m, there’s a 9. Likewise, for x², for any decimal point m, there’s a 9. There’s nowhere to start sticking the 800…1 string.

I can prove it by contradiction if you like too, because that’s always a fun was to demonstrate things. Assuming that x=0.999…, where, for any decimal place n, there’s a 9. Again, this is the definition of 0.999… when mathematicians talk about it. Now let’s take x² and assume that, at some point m, there’s 8, plus an infinitely long string of 0’s and then a 1, because that makes it follow the same pattern as any other finitely long 0.99(lots more 9’s here)9², so like you’ve said, it makes sense that x² will follow the same pattern. As I showed above, the point m happens at the point where, in x, there’s a final 9, then an endless string of 0’s. Except this is a contradiction, because we’ve defined x such that that point does not exist in x. This is a contradiction, so something we’ve said is wrong. It can’t be our definition of x, so it must be false that there’s some point m, where there’s an 8 and then other stuff. In other words, for any decimal place n within x, there’s a 9, and the same winds up being true of x² (Or else the contradiction above). In other words, x² = x, which is only true for 1. Therefore, 0.999… = 1.

]]>“There are always zeroes to the end of any number, and there’s no reason that this won’t hold true for an infinite number.” Yes there is, a very good reason. Because there is no end to the number. If there is an end, it’s not an infinitely recurring decimal. Like I said, the definition of 0.999… is that, for any decimal place n, there’s a 9 in it. If there’s zeroes to the end, that’s a point, n, where there isn’t a 9 and that’s a contradiction. The general definition of any infinitely long decimal is that for any decimal point, n, there’s a non-zero n after it. You can’t say “Oh there’s infinitely many things before the 0, because that’s an absurd notion and that’s you trying to reify infinity. Like I said somewhere else, infinity isn’t “a really big number, bigger than you can possibly imagine”, it’s an endless number.

“To demonstrate this, we simply multiply a number like .999… by N*10, where N is the number of 9′s after the decimal point. You may not like multiplying by an infinite number and again this may distress your sensibilities, but if it can be done for literally any finite N, it should still be possible with an infinite one through induction. So what do you get my multiplying .999… by N*10? An infinite number of 9′s followed by a decimal point and then zeroes.”

Using induction, let’s do that. I’ve already demonstrated that 10x 0.999… is 9.999 with infinitely many 9’s after it, because if there aren’t infinitely many 9’s after the decimal place, it was never an infinitely recurring decimal. Likewise, if we multiply by 10 again, we get 99.999 with infinitely many 9’s after it, because if there aren’t infinitely many 9’s, it was never infinitely recurring. And so on. Each time we multiply by 10 again, we get infinitely many 9’s after the decimal place and when we go to 10^N (Which is, I assume, what you meant to say, 10 with N many 0’s after the 1), you wind up with infinitely many 9’s before and after the decimal place.

This shouldn’t surprise you either. You’re multiplying a number by an infinitely. You can get any infinitely large number you want doing that. This is why algebra on infinity isn’t allowed.

“I know that this doesn’t work like an infinity that you imagine, but the infinity you imagine is immune to math in the way that it typically works… that is, if it is an exception to the math I’m using, it ought to be an exception to the math that you use, as well. It is perfectly rational (given your standards) to look at a problem like 10*.999,.. and say that you don’t know what the product is.”

But it’s easy to calculate 10*0.999…. Like you’ve said, you just shift the decimal place over and you still wind up with infinitely many 9’s after the decimal place.

“It’s a double-standard to argue that this multiplication works but that multiplication like N*10*.999… is impossible.” That multiplication is possible because N is infinity, and standard algebra doesn’t allow multiplying by infinity. It’s as valid as multiplying by grapefruit.

Also, consider Hilbert’s paradox. Which, if you haven’t heard of it, you’ve got a hotel with infinitely many guests with infinitely many rooms and each room is occupied by a guest. A new guest shows up, but rather than saying your hotel is full, you move every guest to the next room (The guy in room 1 goes to room 2, the guy in room 2 goes to room 3 and so on infinitely), and now you have an empty room for your new guest and nobody winds up losing a room because you can always move the guy from room n to the room n+1. Likewise, if infinitely many new guests show up, you can get everyone to move from the room they’re in, to the room double what they’re in (ie, the guy in room 1 goes to room 2, the guy in room 2 goes to room 4, the guy in room n goes to room 2n) and now you’ve got infinitely many newly opened rooms for your new infinitely many guests. Both of these are very analogous to working with 0.999… except instead of guests in rooms, we’ve got 9’s in decimal places.

]]>Wani, you haven’t actually showed me how you get to .999… repeating by squaring .999… repeating except to say that it starts with an infinite string of 9’s and you still deny that anything can follow them. The only way I see you getting to that answer is using my induction method and then editing my result. I don’t get it… do you believe that I did the math right or not? This looks like the worst form of cherry-picking, taking only the parts of my argument that support your conclusion and ignoring the parts that don’t.

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