How to solve inequality $1-\frac{1}{x}\geq 0$ correctly?

Question

When I try to solve this inequality I get the wrong results: $$1-\frac{1}{x}\geq 0$$ Assuming $x\neq 0$

It's really embarrassing that I can't see what I'm doing wrong. So please help me, telling me where the maths goes wrong :D

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$$1-\frac{1}{x}\geq 0 \Leftrightarrow 1\geq\frac{1}{x} \Leftrightarrow x\geq1$$

But this is wrong, since I can easily see $1-\frac{1}{x}\geq 0$ when x<0. So what am I doing wong?

Answer 1

As an alternative, we have that

$$1-\frac{1}{x}\geq 0 \iff \frac1x\le 1$$

and

• for $x>0\quad x\cdot \frac1x\le x\cdot1 \implies x\ge 1$
• for $x<0\quad x\cdot \frac1x\ge x\cdot1 \implies x\le 1$ that is $x<0$

Also a graphical check for this simple inequality may help

Answer 2

It's $$\frac{x-1}{x}\geq0,$$ which gives $$[1,+\infty)\cup(-\infty,0).$$

We used the intervals method.

In the expression $(x-a)^k$ the natural number $k$ is named a degree of the point $a$.

We need to solve $$\frac{(x-1)^1}{(x-0)^1}\geq0.$$ The degree of $1$ is $1$ and the degree of $0$ is $1$.

If we go through a point $1$ so the sign of $x-1$ is changed because the degree of this point is odd.

If we go through a point $0$ so the sign of $x-0$ is changed because the degree of this point is odd.

For $x>1$ the expression $\frac{x-1}{x}$ has a sign $+$.

Thus, we obtain the following signs: $++++0----1++++$.

It's better to draw an $x$-axis and to put there points $0$ and $1$, which gives the following sequence of sings: $+$$-$$+$ and we obtain the answer: $[1,+\infty)\cup(-\infty,0).$

Another example.

Let we need to solve $\frac{(x-2)^2}{(x-1)(x-3)}\geq0.$

The degree of $2$ is even (the sign is not changed) and degrees of $1$ and $3$ are odds (the sign is changed).

Also, the sign of the expression $\frac{(x-2)^2}{(x-1)(x-3)}$ for $x>3$ is $+$.

Thus, we have the following sequence of sings: $$+--+,$$ which gives the answer: $$(-\infty,1)\cup(3,+\infty)\cup\{2\}$$

Answer 3

Set apart forbidden value $x=0$, you are with a quotient:

$$\dfrac{x-1}{x}>0$$

Here is a very convenient "recipe": the sign of the quotient $N/D$ is the same as the sign of the product $N*D=x(x-1)$.

As this product is a quadratic beginning with a positive coefficient for $x^2$, with real roots $0,1$, we can conclude that this product is $>0$ outside the interval $[0,1]$ delimited by these roots.

Answer 4

What you did wrong was the last double-implication.

In general, you can turn

$$ A \le C $$ into $$ AB \le CB $$ only if you know that $B \ge 0$. When $B < 0$, $$ A \le C $$ is equivalent to $$ AB \ge CB $$ instead!

For example, we all agree that $$ 1 \le 4, $$ right? Now multiply both sides by $-1$, and you get $$ -1 ??? -4 $$ What should replace the "???" Answer: a $\ge$ sign, not a $\le$ sign!

Back to your equation. You had $$ 1 \ge \frac{1}{x} $$ and you wanted to multiply by $x$ ... but depending on whether $x$ is positive or negative, the result will be different.

So split into two cases: You say

"In the case where $x > 0$, this is equivalent to $$ x \ge 1 $$ so we get, as solutions, all positive numbers greater than or equal to $1$.

In the case where we look for solutions $x$ with $x < 0$, our equation is equivalent (after multiplying both sides by $x$, a negative number!) to $$ x \le 1. $$ Of course, not all solutions of this equation satisfy our assumptions for this part of the work: numbers like $\frac{1}{2}$ are less than $1$, so they satisfy the equation, but they're not negative, so they don't satisfy the assumptions. Which negative numbers are less than or equal to $1$? All of them.

So the solutions overall are

1. All negative numbers, and
2. All positive numbers greater than or equal to $1$.