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Closure Property under Addition of Integers. If we add any two integers, the result obtained on adding the two integers, is always an integer. So we can say, that integers are closed under addition. Let us say ‘a’ and ‘b’ are two integers, either positive or negative. When we add the two integers, their result would always be an integer. So the property of closure for subtraction is not always true. 3) 12 + 0 = 12 Here, 12 and 0 both are whole numbers. The addition of them which is 12 again is also a whole number. So the property of closure is true. Closure property for multiplication: If a and b are whole numbers then their multiplication is.
The closure property means that a set is closed for some mathematical operation. That is, a set is closed with respect to that operation if the operation can always be completed with elements in the set. Thus, a set either has or lacks closure with respect to a given operation.
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For example, the set of even natural numbers, [2, 4, 6, 8, . . .], is closed with respect to addition because the sum of any two of them is another even natural number, which is also a member of the set. (Natural numbers are defined as the set: [1, 2, 3, 4, . . .].) It is not closed with respect to division because the quotients 6/2 and 4/8, for example, cannot be computed without using odd numbers (6/2 = 3) or fractions (4/8 = ½g;), which are not members of the set.
Knowing the operations for which a given set is closed helps one understand the nature of the set. Thus, one knows that the set of natural numbers is less versatile than the set of integers because the latter is closed with respect to subtraction, but the former is not. (Integers are defined as the set: [. . .-3, -2, -1, 0, 1, 2, 3, . . .].) Similarly one knows that the set of polynomials is much like the set of integers because both sets are closed under addition, multiplication, negation, and subtraction, but are not closed under division.
Particularly interesting examples of closure are the positive and negative numbers. In mathematical structure, these two sets are indistinguishable except for one property, closure with respect to multiplication. Once one decides that the product of two positive numbers is positive, the other rules for multiplying and dividing various combinations of positive and negative numbers follow. Then, for example, the product of two negative numbers must be positive, and so on.
The lack of closure is one reason for enlarging a set. For example, without augmenting the set of rational numbers with the irrationals, one cannot solve an equation such as x2 = 2, which can arise from the use of the pythagorean theorem. Without extending the set of real numbers to include imaginary numbers, one cannot solve an equation such as x2 + 1 = 0, contrary to the fundamental theorem of algebra.
Closure can be associated with operations on single numbers as well as operations between two numbers. When the Pythagoreans discovered that the square root of 2 was not rational, they had discovered that the rationals were not closed with respect to taking roots.
Although closure is usually thought of as a property of sets of ordinary numbers, the concept can be applied to other kinds of mathematical elements. It can be applied to sets of rigid motions in the plane, to vectors, to matrices, and to other things. For instance, one can say that the set of three-by-three matrices is closed with respect to addition.
Closure, or the lack of it, can be of practical concern, too. Inexpensive, four-function calculators rarely allow the user to use negative numbers as inputs. Nevertheless, if one subtracts a larger number from a smaller number, the calculator will complete the operation and display the negative number that results. On the other hand, if one divides 1 by 3, the calculator will display 0.333333, which is close, but not exact. If an operation takes a calculator beyond the numbers it can use, the answer it displays will be
Contents.Definitions This article will use the for the definitions of addition of the natural numbers, and the S(a). In particular:A1:a + 0 = aA2:a + S( b) = S( a + b)For the proof of commutativity, it is useful to define another natural number closely related to the successor function, namely '1'. We define 1 to be the successor of 0, in other words,1 = S(0).Note that for all natural numbers a,S( a)=S( a + 0)by A1=a + S(0)by A2=a + 1by Def. Of 1Proof of associativity We prove by first fixing natural numbers a and b and applying on the natural number c.For the base case c = 0,( a+ b)+0 = a+ b = a+( b+0)Each equation follows by definition A1; the first with a + b, the second with b.Now, for the induction. We assume the induction hypothesis, namely we assume that for some natural number c,( a+ b)+ c = a+( b+ c)Then it follows,( a + b) + S( c)=S(( a + b) + c)by A2=S( a + ( b + c))by the induction hypothesis=a + S( b + c)by A2=a + ( b + S( c))by A2In other words, the induction hypothesis holds for S( c).
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Therefore, the induction on c is complete.Proof of identity element Definition A1 states directly that 0 is a.We prove that 0 is a by induction on the natural number a.For the base case a = 0, 0 + 0 = 0 by definition A1.Now we assume the induction hypothesis, that 0 + a = a.Then0 + S( a)=S(0 + a)by A2=S( a)by the induction hypothesisThis completes the induction on a.Proof of commutativity We prove ( a + b = b + a) by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. We prove that 0 and 1 commute with everything).The base case b = 0 follows immediately from the identity element property (0 is an ), which has been proved above:a + 0 = a = 0 + a.Next we will prove the base case b = 1, that 1 commutes with everything, i.e. For all natural numbers a, we have a + 1 = 1 + a. We will prove this by induction on a (an induction proof within an induction proof). We have proved that 0 commutes with everything, so in particular, 0 commutes with 1: for a = 0, we have 0 + 1 = 1 + 0.
Now, suppose a + 1 = 1 + a. ThenS( a) + 1=S( a) + S(0)by Def. Of 1=S( S( a) + 0)by A2=S(( a + 1) + 0)as shown =S( a + 1)by A1=S(1 + a)by the induction hypothesis=1 + S( a)by A2This completes the induction on a, and so we have proved the base case b = 1. Now, suppose that for all natural numbers a, we have a + b = b + a. We must show that for all natural numbers a, we have a + S( b) = S( b) + a. We havea + S( b)=a + ( b + 1)as shown =( a + b) + 1by associativity=( b + a) + 1by the induction hypothesis=b + ( a + 1)by associativity=b + (1 + a)by the base case b = 1=( b + 1) + aby associativity=S( b) + aas shown This completes the induction on b.See also.References., Foundations of Analysis, Chelsea Pub Co.
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