Bob
]]>Must read
]]>Edit- I have no idea btw...
]]>Let r = 0.99999 recurring.
r ∈ [0.9,1) That is r lies in the half open region bounded by 0.9 and 1 but is strictly less than 1.
Similarly, r ∈ [0.99,1) and r ∈ [0.999,1) etc.
So r is in the intersection of all these regions.
But it is a property of the real numbers that, for any two reals, a and b, you can always find a value between them.
So the half open regions cannot have a non empty intersection.
Therefore, r cannot exist.
This is actually a proof by contradiction that r ≥ 1. In other words the assumption that r < 1 leads to the absurdity that r does not exist.
By the exact same method, you can prove by contradiction that r > 1 is also impossible.
Suppose r > 1. Then r ∈ (1, 1.1], r ∈ (1, 1.01], r ∈ (1, 1.001], and by the same argument it will be found that r does not exist.
Conclusion: If r exists, then r = 1.
]]>I did say, in another thread, that I would try to give the Shilov proof that 0.9999recurring doesn't exist.
It has been a struggle for me and what follows may not be correct, so I'll apologise now.
Firstly, I couldn't read the small mathematical text on my Kindle. But the nice people at Kindle HQ suggested an app. that means I can read it on my laptop. That helped a lot.
But the relevant section is well on through the first chapter and you only want a small bit. So, could I condense, omit and edit in order to get to where you want? Here's my attempt:
Let r = 0.99999 recurring.
r ∈ [0.9,1) That is r lies in the half open region bounded by 0.9 and 1 but is strictly less than 1.
Similarly, r ∈ [0.99,1) and r ∈ [0.999,1) etc.
So r is in the intersection of all these regions.
But it is a property of the real numbers that, for any two reals, a and b, you can always find a value between them.
So the half open regions cannot have a non empty intersection.
Therefore, r cannot exist.
You may not like this, but it seems to be a requirement for a consistent real number theory. And it is not without precedent. He has already, earlier, shown that 0 has no multiplicative inverse, which is why you mustn't divide by zero.
I think that's it. I don't claim that's how he puts it nor that I like it but there you are. You could always read the book yourself.
Meanwhile, if he is right, then it follows that any discussion of r is meaningless.
Bob
]]>Fortunately, the great majority are good.
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