Supplement 13.1 examples

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Two examples of mathematical systems

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Supplement 13.1 examples

  1. 1. Examples of Mathematical Systems(Blitzer, Thinking Mathematically, 4e, Section 13.1)<br />By Gideon Weinstein<br />
  2. 2. Example 1<br />Mathematical System = Elements and Operation(s)<br /> Elements = “Things” are the integers<br /> Operation # = “Action on two” is defined by a#b=ab+(a+b)<br />Now we check each of the properties and see which ones the integers have under #<br />Closure, Commutative, Associative, Identity, Inverse <br />
  3. 3. a#b=ab+(a+b)<br />Closure: is a#b always an integer? <br />YES, because ab and (a+b) are integers because we know integers are closed under addition and multiplication<br />
  4. 4. a#b=ab+(a+b)<br />Commutative: is a#b = b#a all the time? <br />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<br />
  5. 5. a#b=ab+(a+b)<br />Associative: is (a#b)#c=a#(b#c)? <br />YES, left as an exercise for the listener. <br />HINT: use the same trick as before to reduce it to a question about whether the integers are associative under multiplication and addition. <br />
  6. 6. a#b=ab+(a+b)<br />Identity: Is there a special element i so that a#i=a? <br />Try to figure it out. We know a#i=ai+(a+i) by the definition of #. <br />And we are trying to solve a#i = a. So we need to solve a=ai+(a+i) for i<br />a=ai+(a+i)<br />a=ai+a+i by simplifying parentheses<br />0=ai+i by subtracting a from both sides<br />0=i(a+1) by factoring out the i<br />0/(a+1)=i by dividing by (a+1)<br />0=i because zero divided by anything is zero<br />So this algebra tells us that 0 acts as the identity<br />Let’s double-check: a#0=a0+(a+0)=0+a=a<br />YES, 0 is the identity for #<br />
  7. 7. a#b=ab+(a+b)<br />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<br />ax+(a+x)=0<br />ax+a+x =0 by simplifying parentheses<br />ax+x+a =0 because addition of integers is commutative<br />x(a+1)+a =0 by factoring<br />x(a+1)=-a by subtracting a from both sides<br />x=-a/(a+1) by dividing by a+1 on both sides. <br />It is NOT always an integer, so the inverse does NOT always exist. <br />NO, # does not have the Inverse Property<br />
  8. 8. Example 2<br />Mathematical System = Elements and Operation(s)<br /> Elements = “Things” are the real numbers<br /> Operation @ = “Action on two” is defined by a@b=(a-b)+a/b<br />Now we check each of the properties and see which ones the real numbers have under @<br />Closure, Commutative , Associative , Identity, Inverse <br />
  9. 9. a@b=(a-b)+a/b<br />Closure: is a@b always a real number? <br />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. <br />
  10. 10. a@b=(a-b)+a/b<br />Commutative: is a@b = b@a all the time? <br />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.<br />5@10=(5-10)+5/10=-5+0.5=-4.5<br />10@5=(10-5)+10/5=5+2=7<br />-4.5 and 7 and NOT equal, so NO, @ is not commutative. <br />
  11. 11. a@b=(a-b)+a/b<br />Associative: is (a@b)@c=a@(b@c)? <br />NO, left as an exercise for the listener. <br />HINT: finding a counterexample. <br />
  12. 12. a@b=(a-b)+a/b<br />Identity: <br />TWO questions, because @ is not commutative<br />Is there a special element i so that a@i=a AND i@a=a for any value of a? <br />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.<br />
  13. 13. a@b=(a-b)+a/b<br />Inverse<br />There can’t be an inverse for every element, because the identity doesn’t exist. <br />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<br />

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