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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.
iaskdumbstuff
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    note that in the second part we have $x<0$ as first condition on $x$ – user Apr 15 '18 at 11:27
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    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. – Bill O'Haran Apr 15 '18 at 11:28

4 Answers4

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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.

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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 $x \le 1\implies x<0$
user
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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 $x<0$.

Eff
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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 $$x\in [1,\infty)$$

Mohammad Riazi-Kermani
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