1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?