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What does the word paddock mean in math?
A paddock is a set that satisfies the 4 addition axioms, 4 multiplication axioms and the distributive law of multiplication and addition but instead of 0 not being equal to 1, 0 equals 1. Where 0 is the additive identity and 1 is the multiplicative identity. The only example that comes to mind is the set of just 0 (or 1, which in this case equals 0).
Do 0 and 3 have closure for addition?
A set of numbers is considered to be closed if and only if you take any 2 numbers and perform an operation on them, the answer will belong to the same set as the original numbers, than the set is closed under that operation. If you add any 2 real numbers, your answer will be a real number, so the real number set is closed under addition. If you divide any 2 whole numbers, your answer could be a repeating decimal, which is not a whole number, and is therefore not closed. As for 0 and 3, the most specific set they belong to is the whole numbers (0, 1, 2, 3...) If you add 0 and 3, your answer is 3, which is also a whole number. Therefore, yes 0 and 3 are closed under addition
Which of the following is an example of why irrational numbers are not closed under addition?
Don't know about the "following" but any irrational added to its additive inverse is 0, which is rational. Therefore, the set of irrationals is not closed under addition.
What sets are closed under division?
For example:* The set of real numbers, excluding zero * The set of rational numbers, excluding zero * The set of complex numbers, excluding zero You can also come up with other sets, for example: * The set {1} * The set of all powers of 2, with an integer exponent, so {... 1/8, 1/4, 1/2, 1, 2, 4, 8, 16, ...}
Is the set of irrational numbers closed under multiplication?
No. We go with the proof of a counter-example. pi is a well known irrational number. So is 1/pi. Then pi x (1/pi) = 1, a rational number. If you're not convinced that 1/pi is irrational as well, assume that 1/pi is rational, so that 1/pi = p/q, where p and q are integers and q is not 0 (implicitly, p is also not 0). Then pi = q/p, a contradiction to the fact that pi is not a rational number.
What operations is the set -1 0 1 closed to A Addition B Division C Multiplication D Subtraction?
Multiplication.
Why is the set of -1 0 and 1 closed under multiplication?
Because the product of any two elements is also an element of the set.
Is the set -1 closed with respect to multiplication?
No, it is not.
Is the set of all complex numbers x that have absolute value 1 closed under multiplication?
of course!
Is the set of whole numbers with 31 removed closed under the operation of multiplication?
No. Since -1 x -31 (= 31) would not be in the set.
What does the word paddock mean in math?
A paddock is a set that satisfies the 4 addition axioms, 4 multiplication axioms and the distributive law of multiplication and addition but instead of 0 not being equal to 1, 0 equals 1. Where 0 is the additive identity and 1 is the multiplicative identity. The only example that comes to mind is the set of just 0 (or 1, which in this case equals 0).
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.
Does the set of irrational numbers with the usual addition and multiplication form a field?
Strictly speaking, no, because the identity for addition 0, and the identity for multiplication, 1 are not irrationals.
Do matrices form an abelian group under multiplication?
More precisely, I think you're asking whether the set of n X n matrices forms an abelian group under multiplication. The answer is no (assuming n>1). For example(1 0)(0 1) = (0 1)(0 0)(0 0) (0 0),but(0 1)(1 0) = (0 0)(0 0)(0 0) (0 0). However, the set of n x n diagonalmatrices does form an Abelian set. This is true regardless of the direction of the diagonality, right-to-left or left-to-right. Note that the resulting matrix will also be diagonal, but always right-to-left.
Are odd numbered sets closed under addition and multiplication?
No. 1 + 3 = 4, which is not odd. In fact, no pair of odds sums to an odd. So the set is not closed under addition.
Is the set of all 2x2 invertible matrices a subspace of all 2x2 matrices?
I assume since you're asking if 2x2 invertible matrices are a "subspace" that you are considering the set of all 2x2 matrices as a vector space (which it certainly is). In order for the set of 2x2 invertible matrices to be a subspace of the set of all 2x2 matrices, it must be closed under addition and scalar multiplication. A 2x2 matrix is invertible if and only if its determinant is nonzero. When multiplied by a scalar (let's call it c), the determinant of a 2x2 matrix will be multiplied by c^2 since the determinant is linear in each row (two rows -> two factors of c). If the determinant was nonzero to begin with c^2 times the determinant will be nonzero, so an invertible matrix multiplied by a scalar will remain invertible. Therefore the set of all 2x2 invertible matrices is closed under scalar multiplication. However, this set is not closed under addition. Consider the matrices {[1 0], [0 1]} and {[-1 0], [0 -1]}. Both are invertible (in this case, they are both their own inverses). However, their sum is {[0 0], [0 0]}, which is not invertible because its determinant is 0. In conclusion, the set of invertible 2x2 matrices is not a subspace of the set of all 2x2 matrices because it is not closed under addition.
What are the limitation of the probability?
Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].Probability of an even must lie in the closed interval [0, 1].
