Greatest integer value of $[2.999...]$

Question

What happens to the greatest integer value in case of such non-terminating decimals ?What is $[2.999...]$ ? Is it $2$ or $3$ ? Does it have connection with limit ?

Answer 1

$$\begin{array}{rcl} x &=& 2.999\cdots \\ 10x &=& 29.99\cdots \\ 10x-x &=& 27 \\ 9x &=& 27 \\ x &=& 3 \end{array}$$

Therefore, $2.999 \cdots = 3$.

Therefore, $\lfloor 2.999 \cdots \rfloor = \lfloor 3 \rfloor = 3$.

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What might be confusing to you is that $\lfloor 2.\underbrace{999\cdots9}_n \rfloor = 2$ for any finite $n$. That means that the limit as $n \to \infty$ is indeed $2$, which proves that $\lfloor \cdot \rfloor$ is not continuous at $x=3$.

Answer 2

Does it have connection to limit?

Yes, any non-terminating decimal $D$ can be considered as a limit of numbers $D_n$ that are written as $D$'s finite decimal substrings of length $n$. The sequence $\{D_n\}$ is monotonic non-decreasing and bounded from above, so the limit of $D_n$ is guaranteed to exist as $n\to\infty$.

(This answers part of your question re:limits. Use in conjunction with the other answer.)

Answer 3

Another way to look at this, is to consider the following simple proof:

Let $a,b\in \mathbb{R}$ such that $a\geq b$, then:

$$2a=a+a\geq a+b\geq b+b=2b$$

Dividing by 2 yields:

$$a\geq\frac {a+b}{2}\geq b$$

That is, there is always a real number in between any two arbitrarily selected real numbers. We say the real numbers are "complete".

So indeed $2.\overline {999}=3$

So, as stated above, $\lfloor 2.\overline {999}\rfloor=\lfloor 3\rfloor=3$

Answer 4

Yes, its value derives from the limit of the sum of geometric progression:

$$[2.999...]=[2+0.9+0.09+...]=[2+\lim_\limits{n\to\infty} \frac{0.9\cdot \left(1-0.1^n\right)}{1-0.1}]=[2+\frac{0.9}{1-0.1}]=[3]=3.$$