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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. ;-)

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closed as off-topic by colormegone, user91500, Claude Leibovici, Batominovski, Daniel W. Farlow Aug 3 '15 at 17:17

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    $\begingroup$ Consider any number $x$ in $(0,1)$. Then, the above inequality is totally false. $\endgroup$ – thanasissdr Jul 31 '15 at 7:31
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    $\begingroup$ @Anoneemus $(0,1)$ refers to all numbers between $0$ and $1$, exclusive. $\endgroup$ – Erick Wong Jul 31 '15 at 7:34
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    $\begingroup$ There are already some good answers, but I'll simply add some intuition. What happens the size of a (positive) number when it is multiplied? Well, it gets bigger if the multiplication factor is larger than 1, it stays the same when it is exactly 1, and it gets smaller when the multiplication factor is less than 1. So, remember that the square of a number is the number times itself. Therefore the square of a number will be smaller than the number itself if the given number is smaller than 1. $\endgroup$ – Eff Jul 31 '15 at 7:41
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    $\begingroup$ You might also find it helpful to look at the graph of $y = x$ overlaid on the graph of $y = x^2$. $\endgroup$ – William Hoza Jul 31 '15 at 9:57
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    $\begingroup$ Maybe it helps to look at the graphs of the two functions: wolframalpha.com/input/?i=plot+y%3Dx%2C+y%3Dx%5E2 $\endgroup$ – Martin Sleziak Jul 31 '15 at 12:08
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Take $\large x=\frac{1}{3}$. Is $\large \frac{1}{3} \le \left(\frac{1}{3} \right)^2 = \frac{1}{9}$?

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Just try drawing a graph.

enter image description here

It is clear that $x>x^2\text{ whenever } 0\lt x\lt 1$

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$$x\leq x^2\iff x^2-x\geq 0\iff x(x-1)\geq 0$$ $$\implies x\in (-\infty, 0]\cup[1, \infty)$$ 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

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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 \ $ .

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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.

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$$x \leq x^2\\x-x^2 \leq 0$$ now determine sign of $x-x^2\\=x(1-x)$
enter image description here

when $$0 <x<1 \rightarrow x-x^2 \leq 0 \\x \leq x^2 $$

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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 $x^2-x<0$.

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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 $$ x(x - 1) > 0 \iff x^2 - x > 0 \iff x^2 > x$$

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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.

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