# Rational and irrational numbers

1. Dec 23, 2014

Are numbers which tend to an integer for example 1, that is 0.9999... And 1.00....01 irrational?

[where I got the doubt: There was a question in my textbook--->
$$f(x) = x^2+ax+1 \text{ if x is rational}$$
$$f(x) = ax^2+bx+1 \text{ if x is irrational}$$
If f(x) is continuous at x=1 and 2 find a and b.
Why I had the doubt: Since if numbers that tend to 1 and 2 are irrational then 1 and 2 are the roots of the equation $x^2+ax+1=ax^2+bx+1$]

2. Dec 23, 2014

### HallsofIvy

Staff Emeritus
Your question makes no sense. 0.999... by which I assume you mean "the limit of the sequence 0.9, 0.99, 0.999, ..." is 1. There is NO such number as $0.000...1$ because you cannot have a number with an "infinite number of 0s ended with a 1"- and infinite sequence has NO end. If you mean, as you should, "the limit of the sequence 0.01, 0.001, 0.0001, ...", then that number is 0. No, 1 and 0 are NOT irrational.

But I have no idea what you mean by "if numbers that tend to 1 and 2 are irrational". Given any number, whether rational or irrational, there exist sequences of rational number and sequences or irrational number that converge to it.

3. Dec 23, 2014

### lavinia

Every number is the limit of both rationals and irrationals.

What then does continuity at 1and 2 tell you then about the two expressions for f at 1and 2?

Last edited: Dec 23, 2014
4. Dec 23, 2014

Il rather post the question in the homework forum. Maybe I'm totally wrong. You can close the thread.

5. Dec 23, 2014