is 2,99999....... = 3 ?
You seem to be satisfied if there is no real number between 2.999... and 3.If the 9s continue indefinitely, then it is equivalent. You can show this by showing that there is no real number between 2.999... and 3, which is the same as evaluating the infinite series represented by the decimal 2.999... .
Someone asking the question in the original post is probably not familiar with extending the real numbers or with alternative formulations of analysis. They have also either not understood or have never seen a proper definition of the set of real numbers, using which, they would immediately see the truth of the statement. The purpose is to lead them towards a motivation of the definition.You seem to be satisfied if there is no real number between 2.999... and 3.
What is a real number?
Perhaps an attempt to avoid inevitable locking of the thread?Any reason why you chose to use 2.99..=3 rather than the usual 0.999..=1?
I don't think they are that wrong in doing so. The Cauchy construction of the reals says that 2.999... is the equivalence class of the fundamental sequence (2.9, 2.99, 2.999, ...). But that very construction also says that 3 is the equivalence class of (2.9, 2.99, 2.999, ...). Thus 3=2.999...I have a feeling some think not about the number 2.999... but about sequence of numbers 2.9, 2.99, 2.999 ...