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A bit more thinking

  1. Aug 5, 2008 #1
    "Define a * b, for a,b [tex]\in R[/tex] by a * b = ab+a+b. What we just have done is defined a new operation *, in terms of the well known ones multiplication and addition."

    Is there a real number which is the identity under the operation * ???
    Lets call this number e. A number for which a * e = e * a for all [tex]a \in R[/tex]?

    Furtheremore, does every [tex]a \in R[/tex] have an inverse under this operation? ...if so, what could it be? :rolleyes:
     
  2. jcsd
  3. Aug 5, 2008 #2

    HallsofIvy

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    Looks an awful lot like a homework problem to me!

    To answer the first question, you are looking for a number e such that a*e= ae+ a+ e= a. That means ae+ e= 0 so ae= e for all a. There is no real number for which that is true so this operation has no identity.

    (By the way "[itex]a*e= e*a[/itex]" is true but does NOT define the identity. for example that is true for any member of a commutative group.)
     
    Last edited: Aug 6, 2008
  4. Aug 5, 2008 #3

    Err... Doesn't zero have this property?
     
  5. Aug 5, 2008 #4
    Yep. a * 0 = a0 + a + 0 = a.
     
  6. Aug 6, 2008 #5

    HallsofIvy

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    Oh, bother!

    Okay, now that it has an identity, you need to find inverses for this operation. If b is a's inverse, then ab+ a+ b= 0. Solve for b.
     
  7. Aug 11, 2008 #6
    I think so too. But if e is the identity, we should have a*e = ae+a+e=a and e*a=ea+e+a=a
    => e(a+1)=0 and e = 0 :wink: And yes 0 is the only solutions that works for every [tex]a \in R[/tex]

    In order to find the inverse we want for every a, a real number b for which a*b=ab+a+b=e=0

    Then b(a+1)=0 => [tex]b= \frac{-a}{a+1}[/tex] and we DO have an inverse under this operation for every real number except -1, by letting the inverse of a be the number [tex]\frac{-a}{a+1}[/tex].

    :biggrin:
     
  8. Aug 11, 2008 #7

    gel

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    Take any invertible function f:R->R and you can define an operation a*b = f-1( f(a) f(b) ). The identity is f-1(1). The inverse of a is f-1(1/f(a)).
     
  9. Aug 11, 2008 #8
    Yep, the answer is 0.

    But just try different numbers to find an identity. Eventually you'll find it. Or you can do it algebraically:

    a * b = ab + a + b = a (true for identity, since a * b = a)

    ab + a + b = a

    ab + b = 0;

    (a+1)b = 0

    For a = -1, any b is an identity. For a [tex]\neq[/tex] -1, b = 0.
     
  10. Aug 11, 2008 #9
    Just trying numbers is not a good method, especially when there are uncountably many to choose from. And b=0 works even when a=-1.
     
  11. Aug 11, 2008 #10
    Yeah, solving it algebraically seems like a good approach.
     
  12. Aug 11, 2008 #11

    CRGreathouse

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    I think the definable numbers would suffice, and they're countable. So it's not too hard after all. o:)
     
  13. Aug 12, 2008 #12

    Mute

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    Note that since a and b are real numbers and multiplication and addition of real numbers are commutative, a*b = b*a for all real numbers a and b:

    a*b = ab + a + b = ba + b + a = b*a, where in the centre we use the fact that ab = ba for real numbers and a + b = b + a for real numbers.
     
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