Inequality: x^5<x

[Guest]

Comment résoudre
[tex]x^5<x[/tex]

[Math Tutor]

[tex]x^5 - x < 0[/tex]
[tex]x(x^4 - 1) < 0[/tex]
[tex]x(x^2 - 1)(x^2 + 1) < 0[/tex]
[tex]x(x - 1)(x + 1)(x^2 + 1) < 0[/tex]

[tex](x^2 + 1) > 0[/tex] for every [tex]x[/tex]

x(x - 1)(x + 1) = 0 has solutions -1, 0, 1
let's choose a random number < -1 for example -2
-2(-2 - 1)(-2 + 1) < 0
then [tex]x\in (-\infty; -1)\cup (0, 1)[/tex]

[Guest]

x^5<x
x^5-x<0
x(x^4-1)<0
x<0 and x^4<1
x<0 and x<1
x lies in the interval (-infinity, 0)

[tazy]

|a+b|<=|a|+|b|
and when |a + b| = |a| + |b| ???

[Guest]

|a+ b|= |a|+ |b| if and only if
i) a and b are both positive
ii) a and b are both negative
iii) at least one of a or b is 0

[HallsofIvy]

x^5<x
x^5-x<0
x(x^4-1)<0
x<0 and x^4<1
x<0 and x<1
x lies in the interval (-infinity, 0)

No! No! No! Did you consider checking your answer? -1 is in that interval but (-1)^5- (-x)= 0, not less than 0. Or if x= -1/2, which is in that interval, (-1/2)^5- (-1/2)= -1/32+ 1/2= (16- 1)/32= 15/32> 0.

If x< 0 and x^4< 1 so that x^4- 1< 0 then you are multiplying to negative numbers which gives a positive number.

If the product of two numbers is negative then one must be positive and the other negative.
Case 1: x< 0, x^4- 1> 0. From x^4> 1, either x> 1 or x< -1. Since we also have x< 0, x< -1.
Case 2: x> 0, x^4- 1< 0. From x^4< 1, -1< x< 1. Since we also have x>0, we must have 0< x< 1.
x is in the set {x| x< -1} union {x|0< x< 1}.