Math Is Fun Forum

I assume you mean 0.999... 0.9 = 0.1
And yes, 0.999... + 0.1 = 1.1, because as I said above, 1.0999... = 1.1, by the same principles that we used to describe 0.999... = 1

Yes, using much the same strategy.

Math doesn't lie, but it can be wrong if the fomulae used don't make sense.  If 1 + "infinite difference!" = 1.1, then 1.1 - 1 = "infinite difference!", therefore "infinite difference!" = 0.1
That means that by your idea 0.999... + 0.1 = 1? but we know that 0.999... + 0.1 = 1.0999... which is definitely not 1. So there's obviously a problem with your formulae.

I'm not sure I understand your logic. You're basically saying that if I use some impossible means to get to a place that doesn't exist, and then can't get back without making use of the same impossible means, that means that the place that doesn't exist, must exist?

I'm not sure I understand your argument here. There are two key differences between 0.000...1 and 0.111...
1/ In order to define the "last digit" is equal to something, there has to be a "last digit" which means it has to be a finite number of digits. It's impossible for a finite number of digits to contain an infinite number of zeros.
2/ 0.111... is a repeating digit, which can repeat for an infinite number of digits without changing. 0.000...01 implies an infinite number of digits that repeat for a while, and then suddenly change. What could possibly produce that change? 1/9 = 0.111... is reached because when you divide 1 / 9 you see that 9 goes into 10 once, leaving a remainder of 1, and then 9 goes into 10 once, leaving a remainder of 1, and then ... etc etc. The same result each time propogating the same result means that the same digit will keep appearing without end.

In order for a "last digit" to be different, that means that at some point, you have to conclude that 9 goes into 10 twice, with no remainder, which would then give you the 0.111...2 result, with a "last digit" that is different from the rest. Of course, math doesn't work like that, the same simple algebra can't spontaneous yield a different result after reaching one result an infinite number of times, nor can ANYTHING happen "after" something else has happened an infinite number of times, because the first thing is still happening.

So, 0.000...1 or 0.111...2 are impossible numbers, but 0.111... is quite possible.

Now ksa-boy, I hope that when you submit your homework you make sure to put luca-deltodesco's name on it. ;-)

Why does it have to start from somewhere? It's a number, it's not starting anywhere and it's not going anywhere... it just is. I think this concept of numbers "starting" somewhere (stage one?) is one of the confusing ones, because it makes no sense. A number is a number. 5 is 5. 15 is 15, it doesn't "start" as 1 and later "become" 15 just because you write the numbers from left to right. Similarily, 0.999... doesn't "start" as 0.9, just because you wrote that number first before you started adding extra 9s to the end. Otherwise you just proved that 1.9999.... = 1, because that's what it "starts" as, which makes no sense either.

By using arithmatic, (A/B) x B = A, as has been shown to you many times before already. The fact that you insist that D <> 0, because C < A seems to be based entirely on inspection, that you THINK 0.999... looks smaller than 1, because the first digit of the number is smaller. And it's true that intuitively, ignoring all the math behind it (basically what you're doing) that does appear to be the case. And that's exactly why this is such a controversal proof, because it going completely against intuition. But the math backs up the proof and shows that your intuition is wrong. Just like if you'd asked somebody a hundred years ago if they thought that a 400 tonne hunk of metal would fly for any distance, they'd tell you that it's impossible because it's completely against their intuition that anything that heavy could get off the ground. Turns out their intuition is wrong, because now we see 747s and the new Airbus A380 do it with relative ease, because the math supports it. Just like the Wright brothers, you'll have to learn to get past your intuition and see that the math doesn't lie.

What is the mathmatical basis for this assertion? This is only true if D <> 0, and you can't assume that D <> 0 when you're trying, in essance, to prove that D <> 0 (and the rest of us are proving that D = 0 through various methods). You need a proof for D <> 0 before you can use it as fact to determine the rest of your proofs.

The problem is that if they're the same length, and it's an infinite length, that means that there's an infinite number of 0s. And 0.000... (with an infinite number of 0s) = 0, so D = 0, so the above assertions are false. You cannot possibly have an infinitely long string of 0s with a 1 on the end, it's not possible within the realms of mathmatics. If you continue to assert that it is possible within your definition of math, I'll have to assume that you're inventing your own version of math where 1 > 1, and then I can't argue against you because I don't understand your personal "math" system.

What? So 0.999... = 0.9? Does that mean that 0.5555 = 0.5? And 0.0004 = 0.4? And 0.9991 = 0.1 (because you can remove the 9s, since they're the same). I'm confused how you came up with that one!

aaaahhh... I feel so old now!

I've been "grown up" for years now, so it's kinda hard for me to say what I wanna be. ;-)

I agree that C = 0.999..., the way that you wrote it. However, You also have to agree that (A/B) x B = A. Ignoring the values that you've assigned to A and B, even with the brackets there, it's basic math. If I ask you what (A/F) x F is, without knowing the value of F, the answer is still A. By definition, division and multiplication are the inverses of each other, and order of operations is irrelevant (so (A/B) x B = (AxB)/B = A either way).

The other problem with the above is that you're starting with an assumption, and then trying to prove it using your initial assumption. You're starting with the assumption that C < A, and then using that proof (f2, f3) to prove (f4) that A cannot equal C. But A can equal C for cases where D = 0, and since you're starting with A and B and solving for C and D, you can't automatically assume that C < A and D > 0, you have to assume nothing about them, and THEN prove that it's true. So ignoring your initial assumptions, we can see that A/B x B = A, therefor C = A, therefor since A = C + D, D = 0, and all the rest of the formulae fit into place.