Q: Are rational numbers closed under subtraction?

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Yes. In general, the set of rational numbers is closed under addition, subtraction, and multiplication; and the set of rational numbers without zero is closed under division.

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.

Rational numbers are closed under multiplication, because if you multiply any rational number you will get a pattern. Rational numbers also have a pattern or terminatge, which is good to keep in mind.

No.A set is closed under subtraction if when you subtract any two numbers in the set, the answer is always a member of the set.The natural numbers are 1,2,3,4, ... If you subtract 5 from 3 the answer is -2 which is not a natural number.

No. Zero is a rational number, but division by zero is not defined.

Related questions

Rational numbers are closed under addition, subtraction, multiplication. They are not closed under division, since you can't divide by zero. However, rational numbers excluding the zero are closed under division.

Yes, they are.

The set of rational numbers is closed under all 4 basic operations.

Yes. In general, the set of rational numbers is closed under addition, subtraction, and multiplication; and the set of rational numbers without zero is closed under division.

Subtraction.

To be closed under an operation, when that operation is applied to two member of a set then the result must also be a member of the set. Thus the sets ℂ (Complex numbers), ℝ (Real Numbers), ℚ (Rational Numbers) and ℤ (integers) are closed under subtraction. ℤ+ (the positive integers), ℤ- (the negative integers) and ℕ (the natural numbers) are not closed under subtraction as subtraction can lead to a result which is not a member of the set.

They are closed under all except that division by zero is not defined.

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.

yes, because an integer is a positive or negative, rational, whole number. when you subject integers, you still get a positive or negative, rational, whole number, which means that under the closure property of real numbers, the set of integers is closed under subtraction.

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.

A set of real numbers is closed under subtraction when you take two real numbers and subtract , the answer is always a real number .

No, they are not. An irrational number subtracted from itself will give 0, which is rational.