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Supplement 13.1 examples

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Two examples of mathematical systems

Two examples of mathematical systems

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  • 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