Marrow wrote: wisnij wrote:
Marrow wrote:Well if you say it doesn't work that way then you never actually can finish calculating 10x because there is always another 9 to multiply by 10 so you can never end up with a number to subtract the .9 from to get an answer. You can't say that one continues forever but another one can terminate at the same number of significant digits +1. By that it means the answer to 10x-.9999=Hold on I am not done with the last one yet.
There's no need to enumerate all the infinite digits in 10x
, so that's not a problem. Simply knowing that there are a (countable) infinity of digits is enough to be able to use certain properties of infinities.
Remember the Grand Hotel puzzle where everyone had to shift down one room to make room for a new guest? Now picture the opposite operation: the guy in room 1 leaving and everyone else shifting one room in the other direction. Multiplying by 10 ・i.e. moving the decimal place one point to the right ・is almost exactly analogous to the latter case. For every digit right of the decimal place in 0.999... there is a corresponding digit to the right of the decimal place in 9.999..., and it all matches up. You can subtract the one from the other to come out with the integer 9, nice and even.
I just don't understand why you believe that you don't have to do one thing to both parts and still come up with an equation(why does that word look like its spelled wrong, well its better than equasion like I put first also thinking it to be wrong?). You can't do one thing to one of the x's and not do it to the other, even if they are on the same side of the equasion x must still equal =. Very SIMPLE rule from grade school is that ANYTHING multiplied by 10 ends in a 0. This applies to .999 repeating as well.
Grade school math also teaches us that pi = 22/7. Later on in life we learn which things we learned were only approximations used to gloss over a more complicated truth.
Let's phrase it this way instead:
Code: Select all
(1) x = 0.999...
(2) 10x = 9.999...
(3) x + 9 = 0.999... + 9 [from eq. 1]
(4) 10x = x + 9 [from eqs. 2 & 3]
(5) 10x - x = 9
(6) 9x = 9
(7) x = 1
(8) 0.999... = 1 [from eqs. 1 & 7]
Look, if you don't like the algebraic proof, take your pick of the others
. They all come to the same conclusion: within the standard real numbers, 0.999... = 1. The proof using Dedekind cuts is particularly nice.
Marrow wrote:The hotel is pointless to bring up but since you did I will argue it anyway. With the hotel you were trying to make x (the number of rooms) = y (the number of guests). Personally I would have just stuck the new guy at the end and left everyone else in the hotel alone because its his fault for getting there so late. If x=y in the hotel, why does x not have to be the same as x in this problem.
Tell my why you don't have to enumerate all the infinite digits in 10, not just that you don't and the last nine just drops off the unmultiplied x.
no end! What part of "infinite" don't you understand? The rooms in the hotel continue forever, just as do the digits in 0.999.... There's no "last room" to stick the new guy in, and there's no last digit that can be dropped or appended to.
Marrow wrote:You can't have a countable infinity, that defies the concept of infinity, and anything will work out if you only use certain properties of something while ignoring others. I can turn lead into gold if I ignore certain properties of it having the wrong number of molecules and such if I just paint it yellow. Unfortunately this only works with psychology.
You have a lot of learning to do. An infinity is countable
if it is equal to the cardinality of the set of integers. That is, every element of an infinite set of this type could be uniquely numbered ("counted") with an integer. The set of digits in a real number, for example, is countably infinite set. On the other hand, the set of all real numbers is uncountable, as Cantor proved.