No.
Natural (or counting) numbers Integers Rationals Irrationals Transcendentals
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.
You can divide 65 by rationals and irrationals: Divided by a rational: 65 ÷ 13/2 = 10 Divided by an irrational: 65 ÷ √13 = 5√13
9x + 2x + 26 is an expression, not an equation (or inequality). It therefore cannot have a solution so that its solution set is the null set. There is nothing in the solution set - no integers, rationals, irrationals: nothing!
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
Yes, reals are rationals and irrationals.
Natural (or counting) numbers Integers Rationals Irrationals Transcendentals
Yes, there are countably infinite rationals but uncountably infinite irrationals.
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.
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.
By definition, the two sets do not overlap. This is because the irrationals are defined as the set of real numbers that are not members of the rationals.
Some would say that there is no intersection. However, if the set of irrational numbers is considered as a group then closure requires rationals to be a proper subset of the irrationals.
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.
The product of 2 rationals must be rational. The product of a rational and an irrational is irrational (unless the rational is 0) The product of two irrationals can be either rational or irrational.
Yes, every Cauchy sequence of real numbers is convergent. In other words, the real numbers contain all real limits and are therefore continuous, and yes the integers are discrete in that the set of integers only contains (very very few, with respect to the set of rationals) rational numbers, i.e. their values can always be accurately displayed unlike the set of reals which is dense with irrational numbers. It's so dense with irrationals in fact, that by comparison, the set of rationals can be called a null set, however that is a different topic.
0 belongs to the reals. It is a member of the irrationals, the rationals. It is also a member of the integers; It is a member (the identity) of the group of even integers, 3*integers, 4*integers etc with respect to addition.
You can divide 65 by rationals and irrationals: Divided by a rational: 65 ÷ 13/2 = 10 Divided by an irrational: 65 ÷ √13 = 5√13