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Homework Help: Inequalities 2

  1. Sep 30, 2008 #1
    i was wondering if i'm using the right approach for this
    the question reads
    is the following statement true for all x and y : 'If x<y then x^2<y^2'
    then it follows by asking about 'if x^2<y^2'

    i am currently using case analysis to do this
    by considering whether x and y are positive or negative
    is there any other more efficient way to do this ?

    what axiom should i use in this case ? because squaring both sides instead of multiplying by a common factor

    thanks
     
  2. jcsd
  3. Sep 30, 2008 #2

    mathman

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    Probably not. I assume x and y are supposed to be real numbers.
     
  4. Sep 30, 2008 #3
    so what would you suggest for me to prove these statements ?
    induction ?
     
  5. Sep 30, 2008 #4
    eg x=-2 y= -1
    then this statement will not hold
    but how should i put in down
     
  6. Sep 30, 2008 #5

    HallsofIvy

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    For goodness sake, yes, do a "case analysis by considering whether x and y are positive or negative"!

    In particular, you should consider the case x= -4 and y= -3!

    Now, what do you by "if x^2< y^2"? If x^2< y^2, then what?
     
  7. Oct 1, 2008 #6
    If x and y are both positive, we have given x<y, so x^2<xy <y^2.
     
  8. Oct 1, 2008 #7
    you have the right answer.

    you disproved the statement by giving a counter example.

    "eg x=-2 y= -1
    then this statement will not hold
    but how should i put in down"

    first assume that the statement is true
    if x < y for all x,y implies that x^2 < y^2, then we have that (-2 < -1 implies that) 4 < 1
    but 4<1 is nonsense.

    QED
     
  9. Oct 2, 2008 #8
    this is what i came up with for the first part

    Assume that x<y is true , therefore x^n < y^n for n >0 is also true
    therefore x^2 < y^2 is true
    but the statement does not hold
    eg x = -4 y =-3

    should i put it down that way ? is that sufficient ?
     
  10. Oct 2, 2008 #9
    the second part says ' if x^2 > y^2 then x>y 0 ' . questions is whether it is true for all x and y

    again ... didn't i just prove that the statement for part does not hold
    so isn't it just copying the first part as my second part ?
     
  11. Oct 2, 2008 #10
    Um, then you can't say x^2 < y^y is true because it ISN'T. I don't think you should say a statement is true and then give a counterexample as to why it is not.

    You should consider different cases, such as when n is odd or even.
     
  12. Oct 2, 2008 #11
    how would then put in down ?
    because i assume that the statement is true
    then i recheck with the base statement

    isn't that's hwo we do it ?
     
  13. Oct 2, 2008 #12
    Hmm ok I see how the original question was posed. Well, can you be more general? I mean you've deduced that x = -4 and y = -3 would render the statement false. What about just considering -x and -y? How does this affect the inequality. There shouldn't be that much casework.
     
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