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Proofs involving real numbers

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Let x [itex]\in[/itex] ℝ
    Prove that if 3x[itex]^{4}[/itex]+1≤x[itex]^{7}[/itex]+x[itex]^{3}[/itex], then x > 0

    2. Relevant equations

    None

    3. The attempt at a solution
    Assume
    3x[itex]^{4}[/itex]+1≤x[itex]^{7}[/itex]+x[itex]^{3}[/itex]
    then 0 ≤ -3x[itex]^{4}[/itex]-1≤x[itex]^{7}[/itex]+x[itex]^{3}[/itex]
    Then I assumed that each was greater than or equal to 0, which I thought gave the desired result. No examples in the book to really guide me...any help would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 16, 2012 #2
    That can't be right, no matter what value x has.



    You might want to think it like this instead: If x≤0, then x7+x3 is always smaller than 1+3x4. Why is this?
     
  4. Oct 16, 2012 #3
    Would it be because both powers of x are odd, which will mean it is always negative? So proving the contapositive would be an easier route than a direct proof?
     
  5. Oct 16, 2012 #4

    Mark44

    Staff: Mentor

    To be clearer, say what "it" refers to.
     
  6. Oct 16, 2012 #5
    Still not entirely sure if I did it correctly....this class is giving me such a headache. Not sure if its the book or just my lack of understanding but some of these topics covered are going straight over my head...



    Let x[itex]\in[/itex]ℝ. If 3x[itex]^{4}[/itex]+1 ≤ x[itex]^{7}[/itex] + x[itex]^{3}[/itex], then x > 0.
    Proof
    Contrapositive.
    If x≤ 0, then 3x[itex]^{4}[/itex]+1> x[itex]^{7}[/itex] + x[itex]^{3}[/itex]
    Assume x ≤ 0 for x [itex]\in[/itex] ℝ.
    Since for all values of x less than or equal to zero would produce a true statement.
    Since 3x[itex]^{4}[/itex]+1 < 0 for x ≤ 0 and 0 > x[itex]^{7}[/itex] + x[itex]^{3}[/itex] for x ≤ 0.


    Would this suffice as a proof? It seems like I don't know the tools to know how to get there, it always seems obvious when I look at the answer key, but have really no idea how to get there.
     
  7. Oct 16, 2012 #6
    Oh...and I would just like to point out how the book does this type of problem. Seems incredibly non-intuitive to me.

    Let x [itex]\in[/itex] ℝ. If x[itex]^{5}[/itex] -3x[itex]^{4}[/itex]+2x[itex]^{3}[/itex]-x[itex]^{2}[/itex] +4x - 1 ≥ 0, then x ≥ 0.
    Proof.
    Assume that x < 0. Then x[itex]^{5}[/itex] < 0, 2x[itex]^{3}[/itex] < 0, and 4x < 0. In addition, -3x[itex]^{4}[/itex] < 0 and -x[itex]^{2}[/itex] < 0.

    Thus x[itex]^{5}[/itex] -3x[itex]^{4}[/itex]+2x[itex]^{3}[/itex]-x[itex]^{2}[/itex] +4x - 1 < 0 - 1 < 0.
    as desired.

    This is what I have to go on for an example...there has got to be a better way to do these proofs. I'm just regurgitating the same problem, except with different numbers and yet I have no idea what I'm doing.
     
  8. Oct 16, 2012 #7

    Mark44

    Staff: Mentor

    You can omit the sentence above. Instead, show using inequalities why this is so.

    For any real x, 3x4 + 1 ≥ 1 > 0
    For any x ≤ 0, x7 ≤ 0 and x3 ≤ 0, hence x7 +x3 ≤ 0
    Then for any x ≤ 0, 3x4 + 1 > x7 +x3.
     
  9. Oct 17, 2012 #8
    Yes I think this is one of the most obvious cases where indirect proof is easier than direct proof. For a direct proof, you'd need to solve the equation 1+3x4-x3-x7=0.
     
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