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
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
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!!!
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!!!
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[