Recurring decimal and GIF

Question

What is the difference between $2.\bar{9}$ and $3$? Are they really equal? Then why is $2.\bar{9} = 2$ and $[3]=3$? And in high school, why do we learn $2.\bar{9} = 3$? Is there any satisfying argument about this?

($[ \ ]$ denotes GIF)

Answer 1

Set $x=2.999\dots$. Then $10x=29.999\dots$. Now substract to obtain $10x-x=9x=29.999\dots-2.999\dots=27$. Thus $9x=27\Leftrightarrow x=3$. qed

Answer 2

Actually this is an interesting question: the discrepancy given by the GIF function you mention, or simply called floor function, on the number $2.9999\dots$ is due to the discountinuity of this function at any integer point.

For any number of the sequence $a_0=2$, $a_1=2.9$, $a_2=2.99$, $a_3=2.999$ and so forth the GIF function is indeed $2$ and hence $$\lim_{n}\lfloor a_n \rfloor=\lim_{n\to \infty}\lfloor 2.99\dots9 \rfloor=2$$ but if you exchange the limit and the function, then the limit $\lim_{n\to \infty}a_n=2.99\dots\equiv 3$ thus $$\lfloor\lim_{n\to\infty} a_n\rfloor=\lfloor 3\rfloor=3.$$

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Without sequences, this is saying that $$\lim_{x\nearrow 3^-}\lfloor x\rfloor=2\neq 3=\lim_{x\searrow 3^+}\lfloor x\rfloor.$$

Answer 3

It is not true that

$$\lfloor0.9999\cdots\rfloor=0.$$

You might argument that

$$\lim_{n\to\infty}{\lfloor0.\underbrace{999\cdots9}_n\rfloor}=\lim_{n\to\infty}0=0.$$

But this is not admissible as the $\text{floor}$ function is discontinuous (precisely at integer values).

The notation $$0.999\cdots$$ also $$0.\bar9$$ does not specify a finite number of nines, it is truly a second writing for $$1$$ and

$$\lfloor0.9999\cdots\rfloor=\lfloor1\rfloor\ne\lim_{n\to\infty}{\lfloor0.\underbrace{999\cdots9}_n\rfloor}.$$