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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.
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Which one is more irrational or rational?
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).
How do you write an irrational number in algebra?
There is no representation for irrational numbers: they are represented as real numbers that are not rational. The set of real numbers is R and set of rational numbers is Q so that the set of irrational numbers is the complement if Q in R.
What is the set of numbers that includes all rational and all irrational numbers?
the set of real numbers
What is the set of numbers including all irrational and rational numbers?
real numbers
Is an irrational number a real number?
An irrational number is any real number that cannot be expressed as a ratio of two integers.So yes, an irrational number IS a real number.There is also a set of numbers called transcendental numbers, which includes both real and complex/imaginary numbers. Of this set, all the real numbers are irrational numbers.
