3
$\begingroup$

I know it's probably a stupid question, but I'm confused. I have a set {$x\in\mathbb R, \frac{1}{x} \le 1$} that I want to represent as interval/s.

Thinking about it logically, I know that the set is $x\in]-\infty, 0[$U$[1, +\infty[$.

However, when trying to solve the inequality, I can't seem to get the answer. What am I doing wrong?

I take $\frac{1}x \le 1$, and I split it into 2 cases:

  1. if $x > 0$, then $x \ge 1$,
  2. if $x < 0$, then $x \le 1$, which is every element of $\mathbb R$. Where am I going wrong? Thanks.
$\endgroup$
  • 1
    $\begingroup$ note that in the second part we have $x<0$ as first condition on $x$ $\endgroup$ – user Apr 15 '18 at 11:27
  • 2
    $\begingroup$ Your answer is right. You just forgot that, in case 2, since $x<0$ and $x\leqslant 1$, you must ne restricted to the most constraining one. $\endgroup$ – Bill O'Haran Apr 15 '18 at 11:28
6
$\begingroup$

In your second analysis you must intersect the conditions within each case.

In 1. you got $x>0$ and $x\geq 1$. The conjuction of these two is $x\geq 1$.

In 2. you got $x<0$ and $x\leq 1$. The conjunction of these two is $x<0$. The idea is that the solution $x\leq 1$ must be taken into account together with the assumptions that were made to reach it, $x<0$.


It is more common to forget the assumptions when applying nonequivalent transformations, like multiplying by $x$ in this inequality. Applying equivalent transformations the need to intersecting stays with you until the end.

$$\frac{1}{x}\leq1\Leftrightarrow 0\leq 1-\frac{1}{x}=\frac{x-1}{x}$$

You see that $\frac{x-1}{x}$ is non-negative, when either both factors are non-negative, or both are non-positive.

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

It is correct indeed the solutions are $x<0$ and $x\ge 1$ since

  1. for $x > 0$ we have $x \ge 1\implies x\ge1$
  2. for $x < 0$ we have $x \le 1\implies x<0$
| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Why does $x\le 1 \implies x <0$? Thanks. $\endgroup$ – iaskdumbstuff Apr 15 '18 at 11:28
  • 1
    $\begingroup$ @user494405 we have two conditions ($x<0$ and $x\le 1) \implies x<0$ $\endgroup$ – user Apr 15 '18 at 11:29
3
$\begingroup$

Consider the two cases $x>0$ and $x<0$ separately.

If $x>0$ then the direction of the inequality remains if you multiply by $x$, hence $$\frac1x \geq 1 \implies \frac1x\times x\geq 1\times x\implies 1\geq x. $$

If $x<0$, then the equality changes direction if you multiply by $x$, hence $$\frac1x \geq 1\implies \frac 1x\times x\leq 1\times x \implies 1\leq x, $$ which is true for all $x<0$, hence we have either $1\geq x$ or $x<0$.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

For positive values of $x$, $$1/x \le 1 \iff x\ge 1$$

For negative values of $x$, $1/x$ is negative so it is less than $1.$

For $x=0$, $1/x$ is undefined

Thus the answer is $$x\in [1,\infty)$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.