You need to read Cantor's diagonal argument (Google it and go for the Wikipedia page). It does not require a great deal of prior knowledge to appreciate the proof.
Proof By Contradiction:Claim: R\Q = Set of irrationals is countable.Then R = Q union (R\Q)Since Q is countable, and R\Q is countable (by claim), R is countable because the union of countable sets is countable.But this is a contradiction since R is uncountable (Cantor's Diagonal Argument).Thus, R\Q is uncountable.
You can choose an irrational number to be either greater or smaller than any given rational number. On the other hand, if you mean which set is greater: the set of irrational numbers is greater. The set of rational numbers is countable infinite (beth-0); the set of irrational numbers is uncountable infinite (more specifically, beth-1 - there are larger uncountable numbers as well).
real numbers
the set of real numbers
Yes. The set of real numbers is closed under addition, subtraction, multiplication. The set of real numbers without zero is closed under division.
No, it is uncountable. The set of real numbers is uncountable and the set of rational numbers is countable, since the set of real numbers is simply the union of both, it follows that the set of irrational numbers must also be uncountable. (The union of two countable sets is countable.)
In mathematics, when a set is uncountable, it means that it has a cardinality greater than that of the set of natural numbers. For example, the set of real numbers is uncountable because there is no bijection between it and the set of natural numbers. It implies that the set is infinite and dense in some sense.
It is uncountable, because it contains infinite amount of numbers
This would be a literally impossible task, because the set of all real numbers is uncountable and every real number except 0 is a factor of 38. Probably the questioner meant "integral factors", for which the answer is 1, 2, and 19.
They are not. They are countably infinite. That is, there is a one-to-one mapping between the set of rational numbers and the set of counting numbers.
Proof By Contradiction:Claim: R\Q = Set of irrationals is countable.Then R = Q union (R\Q)Since Q is countable, and R\Q is countable (by claim), R is countable because the union of countable sets is countable.But this is a contradiction since R is uncountable (Cantor's Diagonal Argument).Thus, R\Q is uncountable.
Uncountable
uncountable
uncountable
You can choose an irrational number to be either greater or smaller than any given rational number. On the other hand, if you mean which set is greater: the set of irrational numbers is greater. The set of rational numbers is countable infinite (beth-0); the set of irrational numbers is uncountable infinite (more specifically, beth-1 - there are larger uncountable numbers as well).
Transport is both countable and uncountable as a noun.
The noun 'daytime' is an uncountable noun.