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Solving $\frac{1}{x} < 4$ gives me $x > \frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \frac{1}{4}$.

My questions are:

Why does this inequality has two answers (preferably the intuition behind it)?

When using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?

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    $\begingroup$ $x$ can be negative...^^ $\endgroup$ – Probabilitytheoryapprentice Aug 27 '16 at 17:01
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    $\begingroup$ There is just one answer here, not two as you seem to think. It is not the case that $x>1/4$ is one answer and $X<0$ is another answer. Neither inequality by itself would be a correct answer. $\endgroup$ – Andreas Blass Aug 27 '16 at 17:14
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    $\begingroup$ "when using 1<4x it only gives me 1 answer. Aren't the two forms equivalent?" Nope. They are not. If $1/x < 1/4$ it's possible that $x < 0$. For $1 < 4x$ it is not possible. $\endgroup$ – fleablood Aug 28 '16 at 0:14
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    $\begingroup$ Draw a graph of the function $x \mapsto 1/x$ should give some intuition why there are two "zones" for the solution. $\endgroup$ – quid Aug 28 '16 at 0:26
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    $\begingroup$ Just try, for instance, $x=-1$. Is it true in this case that $1/x < 4$? Is it true that $1 < 4x$? $\endgroup$ – TonyK Aug 28 '16 at 12:04

13 Answers 13

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You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$.

Now suppose $x<0$. Then $$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is $(-\infty,0)\cup(1/4, \infty).$

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  • $\begingroup$ Wow!This answer got so many upvotes! ;-) $\endgroup$ – tatan Mar 24 '18 at 17:28
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Here is the solution $$\frac { 1 }{ x } <4$$$$ \frac { 1-4x }{ x } <0$$$$ \frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$ x\left( 1-4x \right) <0$$$$ x\left( 4x-1 \right) >0 $$

so $$x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 4 } ,+\infty \right) $$

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    $\begingroup$ nice approach to avoid the case division (+1) $\endgroup$ – b00n heT Aug 27 '16 at 17:06
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    $\begingroup$ As soon as I read that there were two answers, I started thinking "there has to be a quadratic hidden here." Thanks for revealing it! $\endgroup$ – Ross Presser Aug 28 '16 at 0:50
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    $\begingroup$ Well this is misleading. But the same means you could get solution of $x-1>0$ to be $(-\infty,0)\cup(1,+\infty)$. Your final presentation does need some justification (checking validity of each interval). $\endgroup$ – Ruslan Aug 28 '16 at 15:05
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    $\begingroup$ @Ruslan - Not quite. The solution multiplies by $x^2$ between lines 3 and 4, and we know $x^2 > 0$ for all real $x \neq 0$. So multiplying by $x^2$ is preserves the direction of the inequality for all $x \neq 0$. With $x - 1 > 0$ however, you'd multiply by $x$ which would require splitting the equality into two cases. However, I do concede that this maybe should have been emphasized in the answer. (Still, it's a very nice answer) $\endgroup$ – David E Aug 29 '16 at 18:39
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    $\begingroup$ @RossPresser The hidden quadratic can also be seen by inspection of the graph. 1/x is a hyperbola, therefore is a quadratic form. The obvious 2nd degree terms are just hidden by rotation :) $\endgroup$ – QuantumMechanic Aug 30 '16 at 17:12
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I am following the suggestion given by @quid in the comments because I like pictures:enter image description here

The orange/red line is the $x$-axis. The yellow line is the line $y=4$. The two blue curves are the graph of $y=1/x$. The solution to the inequality is the set of $x$ values for which the blue curve is below the yellow line. As @quid predicted, this picture gives some intuition for why there are two "zones" in the solution.

Note: originally I had $y=\frac{1}{4}$ which was incorrect, so I changed the answer to reflect the fact that it should be $y=4$ and re-plotted the graph.

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    $\begingroup$ I agree a picture helps, but feel this would be more helpful if you (a) indicated the solution set explicitly, (b) displayed a smaller $x$-range, (c) used the same scale on both axes and (d) made the markings legible $\endgroup$ – PJTraill Aug 31 '16 at 17:24
  • $\begingroup$ @PJTraill I don't feel that the first criticism is relevant to the actual question the OP asked, which was not for the exact value of the solution, but the intuition behind its two-part structure. The second and third criticism strike me as pedantic and unhelpful. The fourth criticism is just wrong -- click on the picture, and the full-size version opens in a new tab, where all of the markings are clearly legible. $\endgroup$ – Chill2Macht Aug 31 '16 at 18:46
  • $\begingroup$ It is (of course) up to you, though I thought of them as suggestions rather than criticisms; I hope you were not offended by them. I meant “readily legible as they appear in the answer”, as I prefer to see all relevant information at once. P.S. I do like the clean look of your diagram. $\endgroup$ – PJTraill Sep 1 '16 at 19:11
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The accepted answer is good, but I feel like you're really asking: why is there only one piece to the question, but two pieces to the answer?

This is actually a great question. Sometimes it happens that one piece turns into two (or more), like when you try to solve $x^2 = 9$ (which has "one piece") and get the two-piece solution $x = 3, -3$. Here, to figure out why one piece becomes two, you have to think about how the equation $y = x^2$ works.

So in our case we should think about how the equation $y = 1/x$ works. And when you think about it, you realize that you didn't really start with one piece. No matter what you plug in for $x$, the value of $1/x$ can never be zero. And that means when you write $1/x < 4$, really this gives you the TWO pieces

$$ 0 < 1/x < 4 $$

and

$$ 1/x < 0 $$

Basically, everything smaller than 4 but with zero removed. And that's why you end up with two pieces at the end -- because that's actually how many you started with!

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  • $\begingroup$ Yeah that was exactly what I was confused about! $\endgroup$ – Augusto Dias Noronha Aug 28 '16 at 11:53
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    $\begingroup$ Another way to look at it is that 1/x is a hyperbola and therefore is really a quadratic form, so you'd expect two "pieces" to the solution. $\endgroup$ – QuantumMechanic Aug 30 '16 at 17:10
  • $\begingroup$ Though you did well to nail what the questioner was confused about, I think you underplay the fact that turning one piece into one piece is only a reasonable expectation of a continuous function while $ x \mapsto 1/x $ is not continuous (or even defined) at $0$, though it is continuous on the pieces you start with and and bijective on its domain. $\endgroup$ – PJTraill Aug 31 '16 at 17:38
  • $\begingroup$ Just an idea: you are asked when an expression in x is smaller than 4, and this question has two answers: 1) the value of the expression is smaller than 4 and 2) the expression is negative, hence the two answers. $\endgroup$ – Dominique Sep 1 '16 at 11:06
  • $\begingroup$ $y = 5 - {(8x - 1)}^{2}$ is a continuous function but it still yields two solutions to $y < 4$. $\endgroup$ – joeytwiddle Sep 7 '16 at 6:12
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Just draw a graph of $1/x$ and you'll see 'why'.

Here is an image by WolframAlpha, with appropriate parts enhanced:

enter image description here

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Whenever you're writing $x \gt \frac{1}{4}$ you're assuming $x \gt 0$.

But for $x \lt 0$ you have $\frac{1}{x}\lt0\lt4$

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English doesn't have good words for this, so exactly what's going on can be a bit tricky to describe if you don't already understand the meaning.

It is true that every $x$ satisfying $x > \frac14$ does in fact satisfy $\frac 1x < 4$.

However, the converse fails: there are some $x$ that satisfy $\frac 1x < 4$ that do not satisfy $x > \frac 14$.

So, you have found a simple description of some of the $x$ that satisfy $\frac 1x < 4$ — but (presumably) you were being asked to describe all of the $x$ that satisfy $\frac 1x < 4$.

And there are indeed more of them: every $x$ satisfying $x < 0$ satisfies $\frac 1x < 4$.

Now, what is true is the following: if $\frac 1x < 4$, then it follows that at least one of the two statements "$x > \frac 14$" and "$x < 0$" is true. Thus, this gives a complete description of the solutions to $\frac 1x < 4$.

Put differently, for every $x$, the following bullet points are either both true or both false:

  • $x$ satisfies $\frac 1x < 4$
  • $x$ satisfies one of the statements "$x > \frac 14$", "$x < 0$".

Regarding your solution method, you forgot that multiplying by negative numbers reverses the sign of an inequality, and multiplying by zero turns any inequality into an equality. Here, we know that $x$ can't be zero, but it still could be either positive or negative, so you don't know the effect that multiplying by $x$ will have on the inequality.

The typical way to fix this problem is to break the problem into two parts: one part where you solve the case with the assumption $x<0$, and one part where you solve the case with the assumption $x>0$, and then you put the results together.

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we have $$\frac{1}{x}<4$$ is true if $$x<0$$ and if $x>0$ we get $$x>\frac{1}{4}$$

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  • $\begingroup$ What? $\frac{1}{x} < 4$ if $x < 0$? $\endgroup$ – nbro Aug 31 '16 at 17:19
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No, $$\frac 1x<4\text{ and }1<4x$$ are not equivalent.

You could think so just multiplying by $x$. But a rule says that an inequality is preserved when you multiply by a positive number and inverted with a negative one. So the right thing is

$$\begin{cases}x>0\to1<4x,\\x<0\to1>4x.\end{cases}$$

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we have $\frac{1}{x}>4$ then $ \frac{1}{x}-4>0$

that is , $$\frac{1-4x}{x}>0$$ but the domain of definition is $x\neq 0$

first of all you need to find the zeros and then study the its signs $$1-4x=0$$ then $$x=1/4 $$

$\begin{align} & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1/4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\ & \underline{\left. 1-4x\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \left. \frac{1-4x}{x}\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}$

as we observe from the above table that inequality is positive only when $0<x<\frac{1}{4}$

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I would like to harp on the meaning of your "two answers". When a question asking for a number satisfying some conditions leads to a unique number having that property we are fine. When there are two numbers having the required property then we can say two answers.

Here we have uncountably many real numbers $x$ such that $\frac1x < 4$. So even the region $x>4$ is infinitely many answers. One possible interpretation that could justify "two" could be number of connected components of the solution space of the question. This happens many times. The set $GL(n,\mathbf{R})$ of non-singular real $n\times n$ matrices has two connected components.

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$$\frac{1}{x} < 4,$$ $$\frac{1}{x} - 4 < 0,$$ $$\frac{1 - 4x}{x} < 0,$$ $1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.

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Here is an important aspect which should be always considered. If someone asks me:

Problem: Find the solution of \begin{align*} \frac{1}{x}<4 \end{align*} I would not answer the problem, but instead ask: What is the domain of $x$?

Please note the problem is not fully specified if the domain of $x$, the range of validity, is not given. This is crucial to determine the set of solutions.

Some examples:

Find the solution of

\begin{array}{lcl} \text{domain of }x\qquad&\qquad\text{inequality}\qquad&\qquad\text{solution}\\ \hline\\ \{x|x\in\mathbb{R}\setminus\{0\}\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad (-\infty,0)\cap(1/4,\infty)\\ \{x|x\in\mathbb{R}^{+}\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad (1/4,\infty)\\ \{\pi\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad \{\pi\}\\ \{x|x\in(0,1/4)\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad \emptyset \end{array}

Note: If a domain is not explicitly stated in the problem section of a book we should expect a corresponding statement somewhere else at the beginning of the chapter.

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