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It cannot be closed under the four basic operations (addition, subtraction, multiplication, division) because it is indeed possible to come up with two negative Irrational Numbers such that their sum/difference/product/quotient is a rational number, indicating that the set is not closed.
You will have to think of a different operation.
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Are a set of irrational numbers closed under exponents?
No. Sqrt (2) is irrational. Square it, or raise it to any even power, and it becomes rational. The set is not closed under exponentiation.
What answer choice shows that the set of irrational numbers is not closed under addition?
Hennd
Is the set of all negative numbers closed under the operation of multiplication Explain why or why not?
No. For a set to be closed with respect to an operation, the result of applying the operation to any elements of the set also must be in the set. The set of negative numbers is not closed under multiplication because, for example (-1)*(-2)=2. In that example, we multiplied two numbers that were in the set (negative numbers) and the product was not in the set (it is a positive number). On the other hand, the set of all negative numbers is closed under the operation of addition because the sum of any two negative numbers is a negatoive number.
What do rational and irrational numbers have in common?
Rational and irrational numbers are part of the set of real numbers. There are an infinite number of rational numbers and an infinite number of irrational numbers. But rational numbers are countable infinite, while irrational are uncountable. You can search for these terms for more information. Basically, countable means that you could arrange them in such a way as to count each and every one (though you'd never count them all since there is an infinite number of them). I guess another similarity is: the set of rational numbers is closed for addition and subtraction; the set of irrational numbers is closed for addition and subtraction.
Is the set of irrational numbers countably infinite?
No. The set of irrational numbers has the same cardinality as the set of real numbers, and so is uncountable.The set of rational numbers is countably infinite.
Is the set of irrational numbers closed under addition?
no it is not
Is the set of irrational numbers closed under mulriplication?
No. You can well multiply two irrational numbers and get a result that is not an irrational number.
Do irrational and rational form a negative number set?
No. Irrational and rational numbers can be non-negative.
Is the set of irrational numbers closed under subtraction?
No; here's a counterexample to show that the set of irrational numbers is NOT closed under subtraction: pi - pi = 0. pi is an irrational number. If you subtract it from itself, you get zero, which is a rational number. Closure would require that the difference(answer) be an irrational number as well, which it isn't. Therefore the set of irrational numbers is NOT closed under subtraction.
Is the square root of negative 6 irrational?
No, it is not irrational because it is a square root of a negative number - which falls into the set of Complex numbers. Irrational numbers can not have an imaginary component.
Are a set of irrational numbers closed under exponents?
No. Sqrt (2) is irrational. Square it, or raise it to any even power, and it becomes rational. The set is not closed under exponentiation.
What answer choice shows that the set of irrational numbers is not closed under addition?
Hennd
Which set of numbers is not closed under subraction?
The set of positive numbers, the set of negative numbers are two examples. Any subsets of these will also not be closed.
What are irrational numbers closed under?
Irrational numbers are not closed under any of the fundamental operations. You can always find cases where you add two irrational numbers (for example), and get a rational result. On the other hand, the set of real numbers (which includes both rational and irrational numbers) is closed under addition, subtraction, and multiplication - and if you exclude the zero, under division.
Is the set of all negative numbers closed under the operation of multiplication Explain why or why not?
No. For a set to be closed with respect to an operation, the result of applying the operation to any elements of the set also must be in the set. The set of negative numbers is not closed under multiplication because, for example (-1)*(-2)=2. In that example, we multiplied two numbers that were in the set (negative numbers) and the product was not in the set (it is a positive number). On the other hand, the set of all negative numbers is closed under the operation of addition because the sum of any two negative numbers is a negatoive number.
Are irrational numbers closed under multiplication?
No. To say a set is closed under multiplication means that if you multiply any two numbers in the set, the result is always a member of the set. If, say, the 2 numbers are radical 2 and radical 2 we have (1.4142...)(1.4142...) which by definition equals 2. The result is not an irrational number, so the set is not closed.
Is the set of irrational numbers closed under division?
No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.
