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This explanation is collected from an old textbook by Stoughton Bell & Judah Rosenblatt, Title INTRODUCTORY CALCULUS, 1966, p.5. SECTION 2.10

Let X denote the number defined by 1.999... where every digit after the decimal point is 9. Where is the number X? Solution given by the author is as follow. We know that: X is between 1.9 and 2.0 X is between 1.99 and 2.00 X is between 1.999 and 2.000 ..... so on. (A) To assume that X lies to the left of 2 would contradict one of the statements in the above table. Therefore X cannot lie to the left of 2. (B) X does not lie to the right of 2, based on the initial value given. From (A) and (B), we conclude that X equals 2.

I easily get the (B) but (A) is kind of confusing. Isn't 1.999.... is always to the left of 2 no matter hiw far we go? Where is the contradiction the author indicates here? Thank you.

Kafi Kfishna
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  • Pick any number $a$ to the left of 2. In other words, $a<2$. Then $X>a$ for any choice of such an $a$. – Shean Jul 01 '23 at 15:29
  • In the text, note that if X is to the left of 2 it is a nonzero distance away, too. So there is some amount of $9$s that $X<1.99999999\cdots9999$ which is obviously false – FShrike Jul 01 '23 at 15:29
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    In fact, either $X=2$ or there is a number bigger than all of the numbers $1.9, 1.99, 1.999, 1.9999, 1.99999,\ldots$ but smaller than $2$. What would be that number, and what would be $2-X$? In other words, existence of $X$ implies that there exist infinitesimally small real numbers. Do they, and how do you perform calculations on them? (Have a look at Nonstandard analysis.) –  Jul 01 '23 at 15:39

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