I came across many people who believe the below false implication. I don't know why people believe it true in high school and middle school and also students in university level. Really I would like to know why they believe: $\displaystyle c>0\implies \frac{1}{c}<0$ is true.

Note: $C$ is a real number

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    $\begingroup$ I haven't seen this particular error so much, but I would guess it is due to approaching inequalities entirely procedurally (used in contradistinction with conceptually) and students are thinking that $\frac{1}{c}$ involves a sort of inversion (i.e., finding the multiplicative inverse of $c$) so there must be some sort of inversion of the inequality sign (something like: flip the number, flip the sign). I suspect there are plenty who believe squaring makes things bigger and would assert $c > 0 \implies c^2 > c$, too. $\endgroup$ Dec 17, 2016 at 22:25

1 Answer 1


I don't think anyone believes this in the way you have stated. Perhaps you should be more concrete in how the student is actually being presenting the misconception. Perhaps they think $\frac{1}{2} < 0$ or something similar?

I find students will be unsure rather than actively wrong about the relationship between $\frac{1}{2}$ and $0$. But it's very easy to correct. Would you rather have half ($\frac{1}{2}$) a pizza or no ($0$) pizza at all?

This will likely not be the end of the development in the student's understanding of the number line, but it's a strong starting point that you can call upon later.

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    $\begingroup$ RE: "the way you have stated" ... But the OP has stated matters without concrete numbers. Rather, the statement above is for a generic positive real, and I wouldn't be surprised if that (the use of a variable, quantifying over all positive reals, etc) is how the misconception is arising. (As you suggest, I doubt students are under the impression $1/2$ is negative.) $\endgroup$ Dec 18, 2016 at 0:11
  • $\begingroup$ Right. What I really meant was that OP should be explicit about how the actual misconception is grounded. $\endgroup$
    – Tac-Tics
    Dec 18, 2016 at 0:20

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