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Contrapositive proof

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    1st part , using induction to prvoe that if both x and y are positive then x<y implies x^n<y^n
    2nd part, prove the converse, that if both x and y are postive then x^n<y^n implies x<y

    2. Relevant equations
    my question is more on the second part. I understand that I have to use the indirect proof using contrapositive.


    3. The attempt at a solution
    I started with this
    If x and y are not positive then it does not imply the following x^n<y^n implies x<y
     
  2. jcsd
  3. Oct 5, 2008 #2

    HallsofIvy

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    First, what you have written here is irrelevant. The theorem (and therefore its contrapositive) talks only about positive numbers. "x and y both positive" is not part of the hypothesis, it is a "preliminary requirement" What is true if x and y are not positive does not matter. The theorem itself is simply "x^n< y^n implies x< y".

    Are you clear on what the contrapositive is here? The contrapositive requires that the hypothesis and conclusion be negated and reversed. "x and y both positive" is not part of the hypothesis, it is a "preliminary requirement" that stays the same.

    If x and y are positive numbers, then [itex]x^n\ge y^n[/itex] implies [itex]x\ge y[/itex].

    Also, while you can prove it using induction, you don't have to. [itex]x^n
    \ge y^n[/itex] implies that [itex]x^n- y^n\ge 0[/itex] and [itex]x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-2}[/itex]
     
    Last edited: Oct 5, 2008
  4. Oct 5, 2008 #3
    i do understand contrapositive
    but i'm new to it
    does it mean that i'm suppose to prove if x^n < y^n is false therefore it implies that x < y is false ?
    am i heading the correct direction ?
     
  5. Oct 5, 2008 #4
    i do not understand how did you arrive at this conclusion
     
  6. Oct 5, 2008 #5

    HallsofIvy

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    Yes, you are. Don't forget that you are assuming also that x and y are positive and that "x^n< y^n is false" is the same as "[itex]x^n\ge y^n[/itex]".

    What conclusion do you mean? If you are talking about the factoring (which is not a "conclusion"), what do you get if you multiply (x- y) and (xn-1+ yxn-2+ ...+ xyn-2+ yn-1? And what would it say if n= 2?
     
  7. Oct 6, 2008 #6
    What conclusion do you mean? If you are talking about the factoring (which is not a "conclusion"), what do you get if you multiply (x- y) and (xn-1+ yxn-2+ ...+ xyn-2+ yn-1? And what would it say if n= 2?[/QUOTE]


    sorry i meant like who did you factor that up ?
    is there a formula to help me with ?
     
  8. Oct 6, 2008 #7

    HallsofIvy

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    Did you kow that x2- y2= (x- y)(x+ y)? How about x3- y3= (x- y)(x2+ xy+ y2)?

    Yes, there is a formula: it is exactly what I gave: xn- yn= (x- y)(xn-1+ xn-2y+ ...+ xyn-2+ yn-1. And, again, you can prove that by multiply the right hand side.
     
  9. Oct 7, 2008 #8
    Thanks . Got it =)
     
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