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Math Help - inequalities

  1. #1
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    inequalities

    in trying to solve the following inequality : \frac{1}{x(1-x)}>0 the following proof went through my mind ,but i am not sure if it is correct:

    Proof:

    Assume 1-x<0 ,then the inequality becomes ,if we multiply across : 1/x <0 ,which implies that x<0.

    BUT since 1<x ( 1-x<0) we have 1<0 , a contradiction,hence \neg(1-x)<0.

    BUT since [(1-x)<0 or (1-x)>0] ,and since \neg(1-x)<0 we have (1-x)>0

    Hence the inequality is satisfied for all ,x<1
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  2. #2
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    Re: inequalities

    Hello, psolaki!

    \text{Solve the inequality: }\:\frac{1}{x(1-x)}\:>\:0

    There are two "critical values": . x \:=\:0,\,1

    The two values divide the number line into three intervals.

    . . \begin{array}{ccccc}----\,- & * & -- & * & ----\,- \\ & 0 && 1 \end{array}


    Test a value of x in each interval.

    \begin{array}{ccccccccc}\text{On }(\text{-}\infty,0)\!: & x &=& \text{-}1  & \Longrightarrow & \dfrac{1}{(\text{-}1)(2)}&=& \text{negative} \\ \\ \text{On }(0,1)\!: & x &=& \tfrac{1}{2} & \Longrightarrow & \dfrac{1}{\left(\frac{1}{2}\right)\left(\frac{1}{2  }\right)} &=& \text{positive} \\ \\ \text{On }(1,\infty)\!: & x &=& 2 & \Longrightarrow & \dfrac{1}{(2)(\text{-}1)} &=& \text{negative} \end{array}


    The inequality is true on the interval (0,\,1)

    Answer: . 0 \:<\:x\:<\:1

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  3. #3
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    Re: inequalities

    Quote Originally Posted by Soroban View Post
    Hello, psolaki!


    There are two "critical values": . x \:=\:0,\,1

    The two values divide the number line into three intervals.

    . . \begin{array}{ccccc}----\,- & * & -- & * & ----\,- \\ & 0 && 1 \end{array}


    Test a value of x in each interval.


    \begin{array}{ccccccccc}\text{On }(\text{-}\infty,0)\!: & x &=& \text{-}1  & \Longrightarrow & \dfrac{1}{(\text{-}1)(2)}&=& \text{negative} \\ \\ \text{On }(0,1)\!: & x &=& \tfrac{1}{2} & \Longrightarrow & \dfrac{1}{\left(\frac{1}{2}\right)\left(\frac{1}{2  }\right)} &=& \text{positive} \\ \\ \text{On }(1,\infty)\!: & x &=& 2 & \Longrightarrow & \dfrac{1}{(2)(\text{-}1)} &=& \text{negative} \end{array}


    The inequality is true on the interval (0,\,1)

    Answer: . 0 \:<\:x\:<\:1


    Thank you Soroban,but can you help me ,and tell me where my solution is wrong??

    Because i have the following dilemma.

    Suppose we wanted to solve out the inequality: \frac{x-1}{x}>x+1

    And suppose that we followed the same argument that i used for solving the inequality in my opening post:

    proof:

    Assume x>o,then the inequality becomes,if we multiply across: x-1>x^2+x,which implies that 1+x^2<0 ,a contradiction ,hence \neg(x>0).

    BUT since [x<0 or x>0] ,and since \neg(x>0) we have x<o

    Hence the inequality is true for all x<0.

    So how can the same TYPE of argument in one case give the right solution and the wrong solution in another case??
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  4. #4
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    Re: inequalities

    Quote Originally Posted by psolaki View Post
    Thank you Soroban,but can you help me ,and tell me where my solution is wrong??

    Because i have the following dilemma.

    Suppose we wanted to solve out the inequality: \frac{x-1}{x}>x+1

    And suppose that we followed the same argument that i used for solving the inequality in my opening post:

    proof:

    Assume x>o,then the inequality becomes,if we multiply across: x-1>x^2+x,which implies that 1+x^2<0 ,a contradiction ,hence \neg(x>0).

    BUT since [x<0 or x>0] ,and since \neg(x>0) we have x<o
    Your logic is wrong. Yes, because x can't be 0 on the left, we must have x> 0 or x< 0. Since x> 0 does not satisfy the inequality, we must have x< 0.

    Hence the inequality is true for all x<0.
    NOT necessarily all x< 0.

    So how can the same TYPE of argument in one case give the right solution and the wrong solution in another case??
    You haven't used the full argument.
    \frac{x+1}{x}> x+1
    if x< 0, multiplying by x reverses the inequality:]
    x+ 1> x(x+ 1)= x^2+ x
    Subtracting x+ 1 from both sides
    0> x^2- 1= (x- 1)(x+ 1)
    In order that the product of two numbers be negative the two numbers must have opposite signs. Either
    x-1> 0 and x+ 1< 0 or x-1< 0 and x+ 1> 0.

    x- 1> 0 gives x>1 but remember that we are assuming x< 0 so this is not a valid solution.

    x-1< 0 gives x< 1 but since x< 0, that gives nothing new. x+1> 0 gives x> -1. -1< x< 0 satisfies both conditions.

    The solution set to the inequality is -1< x< 0.
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    Re: inequalities

    Quote Originally Posted by HallsofIvy View Post
    Your logic is wrong. Yes, because x can't be 0 on the left, we must have x> 0 or x< 0. Since x> 0 does not satisfy the inequality, we must have x< 0.


    NOT necessarily all x< 0.


    You haven't used the full argument.
    \frac{x+1}{x}> x+1
    if x< 0, multiplying by x reverses the inequality:]
    x+ 1> x(x+ 1)= x^2+ x
    Subtracting x+ 1 from both sides
    0> x^2- 1= (x- 1)(x+ 1)
    In order that the product of two numbers be negative the two numbers must have opposite signs. Either
    x-1> 0 and x+ 1< 0 or x-1< 0 and x+ 1> 0.

    x- 1> 0 gives x>1 but remember that we are assuming x< 0 so this is not a valid solution.

    x-1< 0 gives x< 1 but since x< 0, that gives nothing new. x+1> 0 gives x> -1. -1< x< 0 satisfies both conditions.

    The solution set to the inequality is -1< x< 0.
    The inequality is not \frac{x+1}{x}> x+1, but \frac{x-1}{x}>x+1

    Your argument does not work for the inequality in concern
    Last edited by psolaki; September 22nd 2011 at 06:03 PM.
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    Re: inequalities

    Quote Originally Posted by psolaki View Post
    The inequality is not \frac{x+1}{x}> x+1, but \frac{x-1}{x}>x+1
    \frac{x-1}{x}>x+1

    \frac{x-1}{x} - (x+1) > 0

    \frac{x-1}{x} - \frac{x(x+1)}{x} > 0

    \frac{x-1}{x} - \frac{x^2 + x}{x} > 0

    \frac{-(x^2+1)}{x} > 0

    \frac{x^2+1}{x} < 0

    since x^2+1 > 0 for all x , the solution set is x < 0
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    Re: inequalities

    Quote Originally Posted by skeeter View Post
    \frac{x-1}{x}>x+1

    \frac{x-1}{x} - (x+1) > 0

    \frac{x-1}{x} - \frac{x(x+1)}{x} > 0



    \frac{x-1}{x} - \frac{x^2 + x}{x} > 0

    \frac{-(x^2+1)}{x} > 0

    \frac{x^2+1}{x} < 0

    since x^2+1 > 0 for all x , the solution set is x < 0
    Thank you Skeeter ,but if you read my opening post again, you will notice ,that I did not ask for a solution of the inequality,but if my own solution was correct
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  8. #8
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    Re: inequalities

    Yes, and you have been told repeatedly that the your solution is NOT correct and you have been told why. What more do you want?
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