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https://math.answers.com/math-and-arithmetic/Is_the_set_of_integers_is_close_under_subtraction

Is the set of integers is close under subtraction? - Answers

Yes, because suppose that 'a' and 'b' are both arbitrary integers. Then (a-b) or (b-a) will then provide you with another integer. Suppose that the integer you are given from (a-b) is not unique. Then we have: (a-b)=c and (a-b)=c' Then, trivially, since (a-b)=(a-b), we have c=c'. Thus it is closed under subtraction.



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Is the set of integers is close under subtraction? - Answers

https://math.answers.com/math-and-arithmetic/Is_the_set_of_integers_is_close_under_subtraction

Yes, because suppose that 'a' and 'b' are both arbitrary integers. Then (a-b) or (b-a) will then provide you with another integer. Suppose that the integer you are given from (a-b) is not unique. Then we have: (a-b)=c and (a-b)=c' Then, trivially, since (a-b)=(a-b), we have c=c'. Thus it is closed under subtraction.



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https://math.answers.com/math-and-arithmetic/Is_the_set_of_integers_is_close_under_subtraction

Is the set of integers is close under subtraction? - Answers

Yes, because suppose that 'a' and 'b' are both arbitrary integers. Then (a-b) or (b-a) will then provide you with another integer. Suppose that the integer you are given from (a-b) is not unique. Then we have: (a-b)=c and (a-b)=c' Then, trivially, since (a-b)=(a-b), we have c=c'. Thus it is closed under subtraction.

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      Yes, because suppose that 'a' and 'b' are both arbitrary integers. Then (a-b) or (b-a) will then provide you with another integer. Suppose that the integer you are given from (a-b) is not unique. Then we have: (a-b)=c and (a-b)=c' Then, trivially, since (a-b)=(a-b), we have c=c'. Thus it is closed under subtraction.

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