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Problem with implication and bi-implication.

  1. Sep 17, 2008 #1

    ehj

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    I have a problem which is best explained with an example:

    sqrt(A)=B => A=B^2 <=> B=-sqrt(A) or B=sqrt(A)

    Since A=B^2 <=> B=-sqrt(A) or B=sqrt(A) I should be able to reverse the order of this into

    B=-sqrt(A) or B=sqrt(A) <=> A=B^2

    But if this is done in the first equation you obviously run into a problem:

    sqrt(A)=B => B=-sqrt(A) or B=sqrt(A)

    What is it that I'm doing wrong, what wasn't allowed in above calculation?
     
  2. jcsd
  3. Sep 17, 2008 #2
    "sqrt(A)=B => B=-sqrt(A) or B=sqrt(A)". Notice the "or" there, if sqrt(A)=B then obviously the statement "B=-sqrt(A) or B=sqrt(A)" holds, since only one of B=-sqrt(A) or B=sqrt(A) needs to be true, and B=sqrt(A) is true...
     
  4. Sep 17, 2008 #3

    HallsofIvy

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    I presume you mean "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" and "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))". I see nothing wrong with either of those.
     
  5. Sep 17, 2008 #4

    ehj

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    okay great
     
  6. Sep 17, 2008 #5
    HalsofIve the sqroot of 9 is 3 and not 3 or -3 but how do you guarantee that in the definition of sqroot??

    If you write sqroot(A)=B <===> B^2=A ,only............. then sqroot(9)=B <===> B^2=9 <==> (B=3v B =-3) AND hence sqroot(9)= 3 or -3

    So what is the proper definition of the sqroot??
     
  7. Sep 20, 2008 #6
  8. Sep 20, 2008 #7

    HallsofIvy

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    The proper definition of square root is, of course, "sqrt(a) (for a a non-negative real number) is the positive real number b such that b2= a".

    Now, back to the implications. Using that definition,
    "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" is correct.

    The statement "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))" is also a correct statement but has nothing to do with the definition of sqrt(A). It is true simply because "A=> (A or B)" is a tautology.

    In particular, I did NOT write, nor would I, "sqrt(A)=B <===> B^2=A". That is false.
     
  9. Sep 20, 2008 #8

    ehj

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    So if I have an expression, for instance a speed squared equals something, but I'm only interested in writing the positive solution, it would only be mathematically correct to write:

    v^2 = b <= v=sqrt(b)

    Right?
     
  10. Sep 20, 2008 #9

    HallsofIvy

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    If you are only interested in the positive solution, you don't need to write "[itex]\pm[/itex]". Although I think it would be better to write v^2= b => v= sqrt(b) since the implication is going that way.
     
  11. Sep 20, 2008 #10

    ehj

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    But you said earlier: "(B=-sqrt(A) or B=sqrt(A)) <=> A=B^2" is correct."
    The statement "sqrt(A)=B => (B=-sqrt(A) or B=sqrt(A))" is also a correct statement"
    Which gives you that sqrt(A)=B => A=B^2
    If that is rearranged and B replaced with v and A with b you get:
    v=sqrt(b) => v^2=b
    which is the same as
    v^2=b <= v=sqrt(b)
    Which was what I asked about the correctness of earlier.
    If what you said is also true, that v^2=b => v=sqrt(b) you have implications both ways and actually a biimplication: v^2=b <=> v=sqrt(b) which is not correct! So it has to be either.. But you seem to be giving two different answers HallsofIvy :P
     
  12. Sep 21, 2008 #11

    HallsofIvy

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    No, it is the difference between "or" and "and".

    If "A is true" then "either A is true or B is true" is a tautology.

    "sqrt(A)= B=> (B= -sqrt(A) and B= sqrt(A))" is a false statement: "sqrt(A)= B" does NOT give "B= -sqrt(A)".

    But "sqrt(A)= B=> (B= -sqrt(A) or B= sqrt(A)" is a true statement because "sqrt(A)= B" does give "B= sqrt(A)" and so the "or" statement is true whether "B= -sqrt(A)" is true or false.

    I did not and would not say "v2= b <=> v= sqrt(b)".

    I would say "v2= b <=> (v= sqrt(b) or v= -sqrt(b))"
     
  13. Sep 21, 2008 #12

    ehj

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    "I did not and would not say "v2= b <=> v= sqrt(b)"." I know, I didn't say you did. I completely agree with your latest post, but I don't see it's relevance? I never mentioned "and" in any of the math?
    The only problem I have now is that you said that v^2=b => v=sqrt(b)
    It gives rise to a problem.
    Do you agree that v=sqrt(b) => v^2=b ?
    This has to be correct since I'm simply squarring both sides, applying the _function_ f(x) = x^2 on both sides. By the definition of a function, it relates only one element in the codomain for each element in the domain, which makes it impossible for the second statement v^2=b to be false. So v=sqrt(b) => v^2=b has to be correct. This is the same as v^2=b <= v=sqrt(b)
    You wrote in an earlier post that v^2=b => v=sqrt(b). Clearly one of these has to be wrong since we would otherwise have a biimplication between the two statements (which is wrong, like you said yourself). You agree that A => B and B => A can be put together to A <=> B right? This post seems to be very much like the last i posted :/
     
  14. Sep 21, 2008 #13
    I think in post #9, Halls meant to switch the order of the statements, so that we don't have this awkward <= in the middle, but forgot to actually do it.
     
  15. Sep 21, 2008 #14

    HallsofIvy

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    Please read ALL of post 9:
    I have added the emphasis.

    This was in response to post 8:
    where again I have added the emphasis.
     
  16. Sep 21, 2008 #15
    Maybe i'm reading you wrong, but assuming all is real numbers, it is always true that

    v = sqrt(b) => v^2 = b

    and

    v = -sqrt(b) => v^2 = b

    but

    v^2 = b => v = +/- sqrt(b) (sqrt meaning the principle square root)

    if you know that v is positive also then you of cause have

    v^2 = b => v = sqrt(b).

    But for it to mathematical correct it would be better to write

    (v^2 = b and v > 0) => v = sqrt(b).
     
  17. Sep 21, 2008 #16

    ehj

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    So you say that saying "v is positive" is like adding a condition like

    v^2=b and v>0 => v=sqrt(b)
     
  18. Sep 21, 2008 #17
    yeah or else i would write

    v^2 = b => v = +/- sqrt(b)

    and then as a physicist, write something like: Because of the context (of some given exercise fx.), only the positive solution makes sence so the negative is rejected.
     
  19. Sep 21, 2008 #18

    ehj

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    Maybe you can even say v^2=b and v>=0 <=> v=sqrt(b)
    where >= means bigger than or equal to
    ?
     
  20. Sep 22, 2008 #19
    The plus-or-minus sign is an evil shorthand used to confuse students of algebra.

    This is actually a case of abuse of the equals sign. When you see a teacher write that x^2 = 4 implies x = +/- 2, he or she means something that cannot be expressed using standard algebraic notation. He or she *means* to say that the *set of solutions* to that equation is {-2, 2}. He or she *means* to say that x = -2 OR x = 2. But it can't be both, and you cannot correctly express it using a single equals sign.

    One thing that algebra teachers often neglect to teach their students is that the sqrt function is in fact NOT the inverse of squaring! The equation sqrt(x^2) = x is true sometimes, but not all the time. It's only true when x >= 0! Squaring is an operation that has no inverse. Once you square something, you can't undo it safely.
     
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