How is $x \leq x^2$ false? There's an equation that says 
$$x \leq x^2$$
and $x \in \mathbb R$. 
What I can solve and clearly see is that this equation would be true for any value of '$x$' but then how come my maths teacher said that it can be false also? (he didn't explained why, probably because there were too many questions lined up).
So, over to you guys. ;-)
 A: Just try drawing a graph.

It is clear that $x>x^2\text{ whenever } 0\lt x\lt 1$
A: $$x\leq x^2\iff x^2-x\geq 0\iff x(x-1)\geq 0$$ $$\implies x\in (-\infty, 0]\cup[1, \infty)$$ Thus for all real values of $x\in (-\infty, 0]\cup[1, \infty)$ inequality holds 
while for all real values of $x\in(0, 1)$ the inequality is false 
A: If you've worked with inequalities, you find
$$ x^2 \ < \ x \ \ \Rightarrow \ \ x^2 \ - \ x \ < \ 0 \ \ \Rightarrow \ \ x \ (x-1) \ < \ 0 \ $$
has the solution $ \ 0 \ < \ x \ < \ 1 \ $  .  In fact, this is also the solution interval for $ \ x^n \ - \ x \ < 0 $ $ \Rightarrow \ \ x \ (x^{n-1}-1) \ < \ 0 \ $ for positive integers $ \ n \ \ge \ 2 \ $ , so $ \ x^n \ < \ x \ $ for those exponents on the same interval. (In fact, we can go further to say that the inequality holds true on that interval for all real exponents $ \ n \ > \ 1 \ $ .)
This is what causes the curves for the functions $ \ f(x) \ = \ x^n \ $ as $ \ n \ $ becomes larger and larger (with $ \ n \ > \ 1 \ $ ) to become flatter and flatter in the interval $ \ -1 \ < \ x \ < \ 1 \ $ . 
A: Perhaps a "clearer" more axiomatic way of seeing it is the following: 
$$ \forall x, y \in \mathbb{R} : x > 0 : y > 0  \implies xy > 0$$
Now consider $ 0 < x < 1$ we clearly have $1 - x > 0$ so necessarily
$$ x(1 - x) > 0 \iff x - x^2 > 0 \iff x > x^2$$
Similarily we may note that if $x > 1$ then $x - 1 > 0$ and again
$$ x(x - 1) > 0 \iff x^2 - x > 0 \iff x^2 > x$$
A: If $x\le0$, the inequation is verified as the LHS is negative and the RHS positive.
If $x>0$, you can simplify by $x$ and get $1\le x$.
So $x$'s such that $0<x<1$ make it false.
A: $$x \leq x^2\\x-x^2 \leq 0$$ now determine sign of $x-x^2\\=x(1-x)$

when $$0 <x<1 \rightarrow x-x^2 \leq 0 \\x \leq x^2 $$
A: Another way to look at it: we want to solve $$x\leq x^2 \Leftrightarrow0\leq x^2-x.$$ Let $$f:\mathbb R\rightarrow\mathbb R,~f(x)=x^2-x.$$ Then $f$ is an open up parabola. Therefore we have $f(x)<0$ if and only if $x$ lies between the two roots of $f$. This means: 


*

*$f(x)>0$ if $f$ has no roots at all

*$f(x)\geq 0$ if $f$ has exactly one root

*$f(x)\geq 0$ if $x\leq x_1$ or $x\geq x_2$ with $x_1<x_2$ being the two roots of $f$


Thus we only have to calculate the roots: $$f(x)=0 \Leftrightarrow x^2-x=0 \Leftrightarrow x(x-1)=0 \Leftrightarrow x=0\vee x=1.$$
We have two roots, therefore we can conclude: $x^2-x\geq 0$ if $x\leq 0$ or $x\geq 1$. For all $x$ between $0$ and $1$ we have $x^2-x<0$.
A: Note that $$x^2-x=\left(x-\frac 12\right)^2-\frac 14$$ and thus $$x^2-x\ge 0 \text{ iff }\left(x-\frac 12\right)^2\ge\frac 14$$ or$$\left| x-\frac 12\right|\ge \frac 12$$
But you can easily see that if $x=\frac 12$ then $x^2-x=-\frac 14$ and the inequality doesn't hold.
A: Take $\large x=\frac{1}{3}$. Is $\large \frac{1}{3} \le \left(\frac{1}{3} \right)^2 = \frac{1}{9}$?
