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Basic Proof Writing Help continued.

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Given a series of mathematical statements, some of which are true and some of which are false.
    Prove ∀x: 3x2-5x-2=(3x+1)(x-2)


    2. Relevant equations
    x(px→qx)


    3. The attempt at a solution
    Statement: ∀x: 3x2-5x-2=(3x+1)(x-2)
    x{x∈ℝ I 3x2-5x-2=(3x+1)(x-2)}
    (1) Assume 3x2-5x-2 is true [Hypothesis]
    (2)
    3x2-5x-2 has two real roots. [Some Definition "?"]
    (3)
    (3x+1)(x-2) =3x2-5x-2
    =(3x+1)(x-2)
    Therefore 3x2-5x-2=(3x+1)(x-2) for all x by definition "?".

    What definition would I be using? Is this the right way to go about this proof? Should I be cataloging definitions and previously proved theorems?
     
  2. jcsd
  3. May 30, 2015 #2

    Mark44

    Staff: Mentor

    No. 3x2 - 5x - 2 is an expression, not a statement. Expressions have values, but "true" and "false" are not possible values. A statement, such as an equation or inequality or a sentence such as "I have three cats." can be true or false.

    The hypothesis here is "x is a real number."
    The conclusion is ##3x^2 - 5x - 2 = (3x + 1)(x - 2)##.
    The = symbol here is really ##\equiv##, since you are showing that the statement is identically true (for all real x), not just true for a small handful of value of x.
    Technically, no. The equation ##3x^2 - 5x - 2 = 0## has two real roots, but I think you're heading down the wrong track here.
    It seems to me that one could prove this by expanding (i.e., multiplying out) the right side.


    The above looks to me like you are proving that (3x + 1)(x - 2) is equal to itself, which is certainly true, but a trivial result.
     
  4. May 30, 2015 #3
    Hmmm... I can see this is a tautology involving two expressions. Does this really boil down to showing the LHS is equal to the RHS as you mentioned? Because I don't see any other way to do this since it is an equivalence between two expressions.
     
  5. May 30, 2015 #4
    ***update***
    x: 3x2-5x-2=(3x+1)(x-2)

    If x is a real number, then for all x, 3x2-5x-2=(3x+1)(x-2)
    (1) Assume U=ℝ
    (2) Let x∈ℝ
    (3) 3x2-5x-2=(3x+1)(x-2)
    =3x2-6x+x-2
    =3x2-5x-2
    Therefore 3x2-5x-2=(3x+1)(x-2) for all x, where x∈ℝ.
     
  6. May 30, 2015 #5
    In contrast to the following.
    The sum of the roots of: x2+5x+4=0 is equal to 5.
    (1) Let U=ℝ and x∈ℝ. Assume ∃x{x I x2+5x+4=0}
    (2) x2+5x+4=0 has real roots, a and b, for which a+b=5.
    (3) 0=x2+5x+4
    0=(a+4)(b+1)
    0=a+4 or 0=b+1
    a=-4 or b=-1
    a+b=5
    (-4)+(-1)=5
    -5≠5
    Therefore it is false that the sum of the roots of x2+5x+4=0 is 5.
     
  7. May 30, 2015 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would think that the simplest proof that [itex]x^2+ 5x+ 4= (x- 4)(x- 1)[/itex] would be to
    start from [itex](x- 4)(x- 1) [/itex] and show that it results in [itex]x^2+ 5x+ 4[/itex].
    Start with [itex](x- 4)(x- 1)= x(x- 1)- 4(x- 1)[/itex] (distributive law)
    and continue from there.
     
  8. May 30, 2015 #7

    Mark44

    Staff: Mentor

    No, the variable is x.
    Your conclusion is correct, but you are trying to make the work you show fit into a "one size fits all" pattern.

    The statement here is that the roots of the given quadratic equation add up to 5. All you need to do is to find the two roots, using factorization (which you did, sort of) or the quadratic formula. Unless your teacher is unusually pedantic, all of this "Let U=ℝ and x∈ℝ. Assume ∃x{x I x2+5x+4=0}" is completely unnecessary.

    Statement: The sum of the roots of: x2+5x+4=0 is equal to 5.

    Find roots:
    ##x^2 + 5x + 4 = 0##
    ##\Leftrightarrow (x + 4)(x + 1) = 0##
    ##\Leftrightarrow x = -4 \text{ or } x = -1##
    The two roots of the equation add to -5, so we conclude that the statement is false.
     
  9. May 30, 2015 #8
    wow, so much simpler. Thanks again for the help guys.
     
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