There are an infinite number of subsets: All rationals other than 1 All rationals other than 2, etc All rationals other than 1.1 All rationals other than 2.1, etc, etc. All integers
No. Fractions do not include irrational numbers. And although there are an infinite number of both rationals and irrationals, there are far more irrational numbers than rationals.
There are an infinite number of rationals between 1 and 6.
It is not possible to answer this question sensibly, since rational numbers form a continuum. So for any pair of rational numbers surrounding sqrt(85), it is possible to give another pair of rationals that surrounds it but such that the rationals are closer together. And this sequence is infinite.
yes
Yes, there are countably infinite rationals but uncountably infinite irrationals.
There are an infinite number of subsets: All rationals other than 1 All rationals other than 2, etc All rationals other than 1.1 All rationals other than 2.1, etc, etc. All integers
No. Fractions do not include irrational numbers. And although there are an infinite number of both rationals and irrationals, there are far more irrational numbers than rationals.
There are an infinite number of rationals between 1 and 6.
There are more irrational numbers than rational numbers. The rationals are countably infinite; the irrationals are uncountably infinite. Uncountably infinite means that the set of irrational numbers has a cardinality known as the "cardinality of the continuum," which is strictly greater than the cardinality of the set of natural numbers which is countably infinite. The set of rational numbers has the same cardinality as the set of natural numbers, so there are more irrationals than rationals.
No, nowhere near. Georg Cantor proved that the number of rational numbers is countably infinite whereas the irrationals are uncountably infinite. If you take the number of rationals to be N then the number of irrationals is of the order 2^N.
It is not possible to answer this question sensibly, since rational numbers form a continuum. So for any pair of rational numbers surrounding sqrt(85), it is possible to give another pair of rationals that surrounds it but such that the rationals are closer together. And this sequence is infinite.
Yes. There is an injective function from rational numbers to positive rational numbers*. Every positive rational number can be written in lowest terms as a/b, so there is an injective function from positive rationals to pairs of positive integers. The function f(a,b) = a^2 + 2ab + b^2 + a + 3b maps maps every pair of positive integers (a,b) to a unique integer. So there is an injective function from rationals to integers. Since every integer is rational, the identity function is an injective function from integers to rationals. Then By the Cantor-Schroder-Bernstein theorem, there is a bijective function from rationals to integers, so the rationals are countably infinite. *This is left as an exercise for the reader.
yes
We use Q for Rationals... which is repreentative of Quocentia (Quotient), since rationals are RATIOs or fractions.
Yep.
It could be the set denoted by Q- (the non-positive rationals) or Q+ (the non-negative rationals).