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Multiplicative identity proof

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that there is at most one real number b with the property that br=r for all real numbers r. (Such a number is called a multiplicative identity)

    Note: to show there is a unique object with a certain property, show that (1) there is an object with the property and (2) if objects A and B have the property, then A=B.

    2. Relevant equations
    It looks like the statement is false.

    3. The attempt at a solution
    Let r=0, then b(0)=(0).
    b can then equal anything because anything times 0 is 0, so when r=0, there is more than one real number b with the property that br=r. The statement is false.

    Am I right, or is this problem really a lot harder than that?
  2. jcsd
  3. Sep 19, 2007 #2


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    1 is an object with that property. If br=r for all r and ar=r for all r then put r=a in the first equation and r=b in the second. If you are talking about reals, then multiplication is commutative. Conclusion? I can't make sense of your other argument.
    Last edited: Sep 19, 2007
  4. Sep 20, 2007 #3
    Well, what I'm saying is that the statement is not true because if r=0 (because r can be any real number and 0 is a real number) then b can be equal to anything, not just one fixed number. b=1, b=2, b=5^.5, whatever, you know? So the statement must be false. That's what I mean with my argument.

    Now I have a question on your's... how does putting r=a and r=b solve my problem? How does it prove that for all real numbers b, b can be at most one real number?
  5. Sep 20, 2007 #4


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    It says for ALL r, not just r=0. If a*r=r for all r then a*b=b, if b*r=r for all r then b*a=a. If the product is commutative then a*b=b*a. What does that tell you about a and b?
  6. Sep 20, 2007 #5

    oh, oh, oh! That note thing in the instructions... I've been avoiding it the whole time. Just follow it and it's solved... okay, so you said 1 is an object of that property, meaning that step one is finished, step two is to prove that a=b, and then step 2 is finished, meaning the proof is done... I getcha.

    I'm still confused on the whole r is any real number thing. It does say for all r, but you can't just exclude 0 though, right? Shouldn't the statement then say something like "for all r, except 0"? This is really bugging me...
  7. Sep 20, 2007 #6
    These are not equivalent:

    there exists a b, such that for all r, br = r

    for all r, there exists a b, such that br = r
  8. Sep 20, 2007 #7


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    0 works fine. We want a*r=r for all r, and if r=0 then a*0=0. No need to make a special case out of r=0.
  9. Sep 20, 2007 #8
    Aaaahhhh, okay. I get it now. Thanks. lol. Sometimes it takes me a while. Sorry about that.
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