Solve ##\dfrac{1}{x}<4##

[RChristenk]

Homework Statement

Solve $\dfrac{1}{x}<4$

Relevant Equations

Inequalities

$\dfrac{1}{x}<4$

For $x>0$:

$1<4x \Rightarrow x>\dfrac{1}{4}$

For $x<0$:

$1<4x \Rightarrow \dfrac{1}{4}<x \Rightarrow x<-\dfrac{1}{4}$

But the problem is $x<0$ works in the original expression instead of just $x<-\dfrac{1}{4}$, so from calculations alone I missed $[-\dfrac{1}{4}, 0)$. I feel conceptually I'm not understanding inequalities or my calculations are wrong.

 

[Office_Shredder]

When you multiply both sides by a negative valued x, the inequality flips direction.

$4x<1$.

You also didn't handle what comes after that very well in your attempt- if you did believe $1/4<x$ then there would be no solutions since x has to be negative. Think through carefully what the next step needs to be here.

Sometimes it helps to say if we're assuming x is negative, replace it with -y and assume y is positive.

 

[RChristenk]

I'm getting confused from assuming $x<0$, because doing the calculation results in $x<\dfrac{1}{4}$, which by itself means $x$ is positive in $[0,\dfrac{1}{4})$. But I think I'm starting to understand I must keep in mind the starting assumption $x$ is negative, so $x<0$ is a fixed condition, despite the calculation ending up showing $x<\dfrac{1}{4}$.

 

[docnet]

I'm getting confused from assuming $x<0$, because doing the calculation results in $x<\dfrac{1}{4}$, which by itself means $x$ is positive in $[0,\dfrac{1}{4})$. But I think I'm starting to understand I must keep in mind the starting assumption $x$ is negative, so $x<0$ is a fixed condition, despite the calculation ending up showing $x<\dfrac{1}{4}$.

If we choose any $x$ in $(-\infty,0)$, it is true that $x$ is in $(-\infty,1/4)$. The latter condition does not contradict the first condition. Does this help?

 

[RChristenk]

If we choose any $x$ in $(-\infty,0)$, it is true that $x$ is in $(\infty,1/4)$. The latter condition does not contradict the first condition. Does this help?

Yes I understand better now. Thank you.

 

[docnet]

Yes I understand better now. Thank you.

The second interval should be $(-\infty,1/4)$, not $(\infty,1/4)$! Sorry about the typo, I'm very tired and it somehow got past me.

 

[FactChecker]

I'm getting confused from assuming $x<0$, because doing the calculation results in $x<\dfrac{1}{4}$, which by itself means $x$ is positive in $[0,\dfrac{1}{4})$. But I think I'm starting to understand I must keep in mind the starting assumption $x$ is negative, so $x<0$ is a fixed condition, despite the calculation ending up showing $x<\dfrac{1}{4}$.

That's good. The equation tells you that AT LEAST $x\lt \dfrac{1}{4}$. But that was with the additional assumption that $x \lt 0$. So both must be true. So $x \lt 0$ for this case.

 

[haruspex]

Another thing to be careful about with inequalities is reversibility of the inference.
Showing that $1/x<4$ and $x<0$ leads to $x<1/4$ does not of itself prove that $x<0$ and $x<1/4$ implies $1/x<4$.