While listening to recordings of Calculus $I$ lectures, I noticed that some students get confused between showing that "some object $x$ is a solution", and showing that "every (possible) solution must be of the form of $x$",

e.g. the difference between proving "$x=6$ is a solution" and "if there is a solution, then it must be $6$".

The problem seems to be that most equations seen in school are usually analyzed using reversible logical manipulations.

I would like to have some nicely motivating and simple examples for why the students should be careful to check that all their manipulations/logical steps are bi-directional.

In particular, good examples for equations would be nice. For inequalities, that's easy:

$x>0 \Rightarrow x^2>0$ is not reversible in $\mathbb{R}$.

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    $\begingroup$ Well, $x=1\Rightarrow x^2=1$ is not reversible in $\mathbb R$ either. You can apply any function that is either not injective or extends the natural domain after the computation to both sides to get a whole bunch of such examples. I'm not sure what you are looking for beyond that though. Can you clarify a bit? $\endgroup$
    – fedja
    Oct 28, 2022 at 23:40
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    $\begingroup$ How about going for broke: "If $x=1,$ then $0\cdot x = 0$" (i.e. $1$ is a solution to $0 \cdot x = 0)$ and "If $0 \cdot x = 0,$ then $x = 1$" (oops!) $\endgroup$ Oct 29, 2022 at 19:32
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    $\begingroup$ I mean trying to solve $\sqrt{1+2x}=\sqrt{2+3x}$ by squaring. The operation itself is OK, because we are working on the non-negative reals now, but the result $x=-1$ is outside the original domain. $\endgroup$
    – fedja
    Oct 29, 2022 at 20:29
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    $\begingroup$ @Vandermonde: I don't see how a proper understanding (or a lack thereof) of the direction of logical implications would be related to emphasizing inequalities or equalities. Most students will have the same (or even more) problems as described in the question when they manipulatie inequalities rather than equalities. $\endgroup$ Oct 30, 2022 at 11:29
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    $\begingroup$ @Vandermonde: Apart from this, claiming that "fashionable" political notions would influence the emphasis of certain concepts in math education just because of terminological or etymological similarity, seems to be pretty far-fetched. $\endgroup$ Oct 30, 2022 at 11:31

1 Answer 1


I think calling some operations bi-directional is not the clearest way to put it. What I think you mean is that while for all functions $f$ and $x,y$ in the domain of $f,$ $$x=y \implies f(x)=f(y),$$ but $$f(x)=f(y) \;\not\!\!\!\implies x=y.$$ I think a better name would be invertible operations. If your students have been introduced to the concept of function inverses (I would expect so in a calculus class), then operation inverses should be an easy extension.

The inverse of adding one is subtracting one. The inverse of multiplying by three is dividing by three. However, not all operations have inverses--at least, not always.

Operations with restricted domains eliminate solutions

$$x^2 = x$$ $$\frac{x^{2}}{x} = \frac{x}{x}$$ $$x = 1$$

Most students will be able to look at the original equation and see that $x=0$ is a solution as well. Ask them what happened to zero in the worked out solution. One way to explain this is that since division by zero is not defined, dividing by $x$ eliminates zero as a possible value for $x$. Multiplying by zero does not have an inverse operation because dividing by zero is undefined.

Operations that are not one-to-one fabricate solutions

$$x = 3$$ $$x^2 = 9$$ $$x^2 - 9 = 0$$ $$(x - 3)(x + 3) = 0$$ $$x = 3 \quad \textrm{or} \quad x = -3$$ Where did the extra solution come from? You could say that since $x^2 = (-x)^2,$ squaring both sides of an equation mixes up answers with different signs. Squaring has no inverse for negative inputs.

Operations that aren't quite inverses

$$x^2 = 9$$ $$\sqrt{x^2} = \sqrt{9}$$ $$x = 3$$ Where did the solution $x = -3$ go? The square root operation only gives positive results, so a square root is not an inverse operation of squaring when the input is negative. This is related to the previous example. A non one-to-one function cannot have an inverse for all inputs. One can use $\sqrt{x^2} = |x|$ to get all of the correct answers.

Other thoughts

I wonder if the above is related to the trouble I and others had with the proving trigonometric identities back in high school. When problems were of the form: $$\textrm{Prove that: } \textrm{A mess of trig functions} = \textrm{Another mess of trig functions}$$ a common strategy was to manipulate the equation until it resulted in something like $1 = 1$ and declare that, "Since the original equation is equivalent to $1=1,$ it must be true." This is obviously fallacious since one could prove that $\sin\theta=\cos\theta$ just by multiplying both sides of the equation by zero to get the obviously true $0=0.$ However, if all of the operations were invertible, then we would have a valid proof just be reversing the order of the lines: start with $1=1$ and end up with the identity.


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