1) The mathematical system has real numbers as elements and defines the operation @ as a@b=(a-b)+a/b. 2) The operation @ is not closed, commutative, or associative for real numbers since division by zero is undefined and order of operations affects results. 3) There is no identity element for @ since no single element i satisfies a@i=i@a=a for all real numbers a.

1. Examples of Mathematical Systems(Blitzer, Thinking Mathematically, 4e, Section 13.1) By Gideon Weinstein

2. Example 1 Mathematical System = Elements and Operation(s) Elements = “Things” are the integers Operation # = “Action on two” is defined by a#b=ab+(a+b) Now we check each of the properties and see which ones the integers have under # Closure, Commutative, Associative, Identity, Inverse

3. a#b=ab+(a+b) Closure: is a#b always an integer? YES, because ab and (a+b) are integers because we know integers are closed under addition and multiplication

4. a#b=ab+(a+b) Commutative: is a#b = b#a all the time? Better check by working it out…. Compute a#b and get ab+(a+b). Compute b#a and get (ba)+(a+b). Is ab+(a+b) = ba+(b+a) always true? YES, because ba=ab and b+a=a+b because the integers are commutative for addition and multiplication

5. a#b=ab+(a+b) Associative: is (a#b)#c=a#(b#c)? YES, left as an exercise for the listener. HINT: use the same trick as before to reduce it to a question about whether the integers are associative under multiplication and addition.

6. a#b=ab+(a+b) Identity: Is there a special element i so that a#i=a? Try to figure it out. We know a#i=ai+(a+i) by the definition of #. And we are trying to solve a#i = a. So we need to solve a=ai+(a+i) for i a=ai+(a+i) a=ai+a+i by simplifying parentheses 0=ai+i by subtracting a from both sides 0=i(a+1) by factoring out the i 0/(a+1)=i by dividing by (a+1) 0=i because zero divided by anything is zero So this algebra tells us that 0 acts as the identity Let’s double-check: a#0=a0+(a+0)=0+a=a YES, 0 is the identity for #

7. a#b=ab+(a+b) Inverse: Since 0 is the identity, this becomes the question that given an a, is it possible to find an integer x so that a#x=0? As before, we try to figure it out by solving the equation a#x=0 for x. Since a#x=ax+(a+x), you need to try to solve ax+(a+x)=0 for x ax+(a+x)=0 ax+a+x =0 by simplifying parentheses ax+x+a =0 because addition of integers is commutative x(a+1)+a =0 by factoring x(a+1)=-a by subtracting a from both sides x=-a/(a+1) by dividing by a+1 on both sides. It is NOT always an integer, so the inverse does NOT always exist. NO, # does not have the Inverse Property

8. Example 2 Mathematical System = Elements and Operation(s) Elements = “Things” are the real numbers Operation @ = “Action on two” is defined by a@b=(a-b)+a/b Now we check each of the properties and see which ones the real numbers have under @ Closure, Commutative , Associative , Identity, Inverse

9. a@b=(a-b)+a/b Closure: is a@b always a real number? NO. It is true that (a-b) is always a real number because real numbers are closed under subtraction. But a/b is not always a real number because a/0 is undefined and therefore not a real number. we know integers are closed under addition and multiplication.

10. a@b=(a-b)+a/b Commutative: is a@b = b@a all the time? Check by working it out…. Compute a@b and get (a-b)+a/b. Compute b@a and get (b-a)+b/a. Is (a-b)+a/b= (b-a)+b/a always true? It really doesn’t feel like it could be true because division and subtraction give different results when done in the opposite order. Let’s try a couple of numbers like 5 and 10. 5@10=(5-10)+5/10=-5+0.5=-4.5 10@5=(10-5)+10/5=5+2=7 -4.5 and 7 and NOT equal, so NO, @ is not commutative.

11. a@b=(a-b)+a/b Associative: is (a@b)@c=a@(b@c)? NO, left as an exercise for the listener. HINT: finding a counterexample.

12. a@b=(a-b)+a/b Identity: TWO questions, because @ is not commutative Is there a special element i so that a@i=a AND i@a=a for any value of a? It turns out the algebra is quite horrific, so it would not be a textbook or exam question. The answer is NO, by the way. It might be legitimate to ask something like “Could the number 2 be the identity element for the operation @?” You would check by calculating if a@2=a=2@a, and you’d very quickly see it is not true.

13. a@b=(a-b)+a/b Inverse There can’t be an inverse for every element, because the identity doesn’t exist. The same horrific algebra done before does show that 1@1=1 (and you might stumble on it just by exploring @ for some numbers), so in some sense 1 is it’s own inverse and its own identity, but that is not the standard way to use those terms