Check my solution of a simple inequality

Does anyone can solve the following equation

x^2 – x < 0

Here is the solutions of mine:

x^2 – x < 0
x(x – 1) < 0
x < 0, x < 1

Please advance

Re: Inequality

My dear friend!!!

Your problem seems to me rather simple, nevertheless you've made a mistake (perhaps, occasionally). However, the way of solving is appropriate and it comprises using interval's method:

x^2 - x < 0
x (x-1)<0

Finally, x>0 and x<1 simultaneously.

You should study basic theoretical aspects concerning inequalities solution

With respect,

Dmitrij

Re: Inequality

Dmitrij said:

My dear friend!!!

Your problem seems to me rather simple, nevertheless you've made a mistake (perhaps, occasionally). However, the way of solving is appropriate and it comprises using interval's method:

x^2 - x < 0
x (x-1)<0

Finally, x>0 and x<1 simultaneously.

You should study basic theoretical aspects concerning inequalities solution

With respect,

Dmitrij

Click to expand...

x > 0 ---> i dont agree with this soln

x <0 and x <1 ---> true soln

Later calculate the range for which inequality satisifies

x <0 and x <1
x <0 is also covered on x <1
hence
x <1 is complete solution !

Shiv

Re: Inequality

Please, check up my solution again very carefully and you'll realize that I'm absolutely right. The intervals method which was used for obtaining the solution is very well-known and leads the correct way for solving the inequality!!!

My solution is absolutely correct!!!!

With espect,

Dmitrij

Re: Inequality

Dmitrij said:

Please, check up my solution again very carefully and you'll realize that I'm absolutely right. The intervals method which was used for obtaining the solution is very well-known and leads the correct way for solving the inequality!!!

My solution is absolutely correct!!!!

With espect,

Dmitrij

Click to expand...

i agree with you
your solution is absolutely correct

0 < x < 1

Inequality

0 < x < 1 is correct Solution

assume x=-1 so x(x-1)=-1(-2)=+2 <0 !!
assume x=0 so x(x-1)=0(-1)=0 <0 !!
assume x=1 so x(x-1)=1(0)=0 <0 !!
assume x=0.5 so x(x-1)=0.5(-0.5)=-0.25 <0 true

let us think
firstly:
if we say the solution is only X<1 that is mean x belongs to interval ]-infinity.,1[
so all values which have -ve sign will produce
x(x-1)=+ve because our main condition that the multiplication od x and (x-1) must be less than zero
so at this moment we accept 0=<x<1
secondly:
if we try to to solve the equation of x(x-1) we will find that x(x-1)=0 incorrect
so x=0 refused
the solution is 0<X<1
the solution set (s.s)= ]0,1[

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Eng.