No
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Of course not.Number if irrational numbers is larger than number of rational numbers.To be more exact: There is no one-to-one mapping of set of rational numbersto the set of irrational numbers. If there would be such a mapping, their cardinality(see Cardinality ) would be same.In reality, rational numbers are countable (cardinality alef0)real numbers, as well as irrational numbers are not countable (cardinality alef1).These are topics inwikipedia.org/wiki/Transfinite_numbertheory
The set of irrational numbers is larger than the set of rational numbers, as proved by Cantor: The set of rational numbers is "countable", meaning there is a one-to-one correspondence between the natural numbers and the rational numbers. You can put them in a sequence, in such a way that every rational number will eventually appear in the sequence. The set of irrational numbers is uncountable, this means that no such sequence is possible. All rational and irrationals (ie real numbers) are a subset of complex numbers. Complex numbers, in turn, are part of a larger group, and so on.
an irrational number is any real number that cannot be expressed as a ratio a/b, where a and bare integers, with b nonzero, and is therefore not a rational number.Informally, this means that an irrational number cannot be represented as a simple fraction. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals. As a consequence of Cantor's proof that the real numbers are uncountable (and the rationals countable) it follows that almost all real numbers are irrational.[1]When the ratio of lengths of two line segments is irrational, the line segments are also described as being incommensurable By Paul Philip S. Panis
Yes, finite numbers are always countable.
yes * * * * * No. Rational and irrational numbers are two DISJOINT subsets of the real numbers. That is, no rational number is irrational and no irrational is rational.