inequalities

**psolaki** · September 21st 2011, 06:37 PM

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

**Soroban** · September 21st 2011, 09:15 PM

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: . $0 \:<\:x\:<\:1$

**psolaki** · September 22nd 2011, 01:59 AM

Originally Posted by Soroban

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??

**HallsofIvy** · September 22nd 2011, 06:46 AM

Originally Posted by psolaki

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.

**psolaki** · September 22nd 2011, 05:41 PM

Originally Posted by HallsofIvy

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

**skeeter** · September 22nd 2011, 06:01 PM

Originally Posted by psolaki

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 $x < 0$

**psolaki** · September 22nd 2011, 06:32 PM

Originally Posted by skeeter

$\frac{x-1}{x}>x+1$

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

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

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

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

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

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

**HallsofIvy** · September 23rd 2011, 06:54 AM

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?

Thread: inequalities

LinkBack

Thread Tools

Display

inequalities

Re: inequalities

Re: inequalities

Re: inequalities

Re: inequalities

Re: inequalities

Re: inequalities

Re: inequalities

Similar Math Help Forum Discussions

inequalities

inequalities

Inequalities

Inequalities

inequalities

Search Tags