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A lot of the time, in maths, especially when I'm trying to remember a formula, I'm taught to remember it in a way that is not notationally correct, but produces the right result.

e.g. when learning the chain rule, I learnt that, if $y=f(u), u=g(x),$ then $$\frac{dy}{dx}=\require{cancel} \frac{dy}{\cancel{du}}\cdot \frac{\cancel{du}}{dx}.$$ Now, I know that this 'cancelling' of $du$s is wrong as $\frac{dy}{dx}$ is not a quotient.

Another example is the cross product of two vectors in $\mathbb{R^3}$: $$\vec u \times \vec v \equiv\begin{vmatrix} \hat \imath & \hat \jmath &\hat k \\ \vec u_1 & \vec u_2 & \vec u_3\\ \vec v_1 & \vec v_2 & \vec v_3 \\ \end{vmatrix}.$$

Finally, a trap that I certainly fell into when first learning integration by substitution:

Suppose we want to evaluate $I=\int(x+3)^{20}dx$.

Let $u=x+3\implies \overbrace{\frac{du}{dx}=1 \iff du=dx}^{\text{we're treating} \ du \ \text{and } \ dx \ \text{as fractions here, which they are} \ \bf{not}} $

As you can see, a lot of this notation is misleading, and gives the wrong idea about its topic, but it returns the correct 'result'.

I want to try and find as many examples as possible, of 'misleading' notation, so that I don't inadvertently get the wrong idea in my head about what the statement/formula actually means.

Thanks a lot!

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  • $\begingroup$ Unless you made the mistake of thinking $du=dx \iff u=x$, that doesn't look like much of a trap. That's what a lot of people think before falling into traps, though, so perhaps I'm missing something. $\endgroup$ – user2357112 Jun 6 '14 at 20:45
  • $\begingroup$ Honestly, I can't get your point. What's misleading in $\frac{du}{dx}=1\iff du=dx$ ? $\endgroup$ – Yves Daoust Jun 6 '14 at 20:52
  • $\begingroup$ You should remove the arrows inside the determinant. What you wrote is not used. $\endgroup$ – Yves Daoust Jun 6 '14 at 20:55
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    $\begingroup$ It may be a better idea to learn the ideas that make these manipulations correct. e.g. differential forms satisfy an identity $$\mathrm{d}f(x) = f'(x) dx $$ Many 'misleading' things actually don't give the wrong idea at all, just hints at deeper ideas that haven't been formally introduced. $\endgroup$ – Hurkyl Jun 6 '14 at 20:57
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    $\begingroup$ @Hurkyl I think OP's point is that you learn to manipulate $\frac{du}{dx}$ long before you actually understand that $du$ and $dx$ are mathematical concepts on their own. You just hear that sometimes, $u'$ can be written as $\frac{du}{dx}$ and then start multiplying with $dx$, although you do not know yet that you even can. $\endgroup$ – 5xum Jun 7 '14 at 6:47
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Another prominent example is the way we solve ODEs with separable variables:

$$y'=f(x)g(y)\\\frac{dy}{dx} = f(x)g(y)\\ dy=f(x)g(y)dx\\\frac{dy}{g(y)} = f(x)dx\\ \int\frac{dy}{g(y)} = \int f(x) dx.$$

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Blatantly replacing a variable with its limit and "knowing what the (wrong) expression means" comes to my mind, like in

$$\lim_{n\to0} \cfrac{1}{n} = \cfrac{1}{0} = \infty$$

or

$$\lim_{n\to\infty} \cfrac{1}{n} = \cfrac{1}{\infty} = 0$$

Not sure if this counts (it fails pretty often, obviously)

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  • $\begingroup$ $1/x$ is an everywhere continuous function of the projective reals, so plugging in the values is a correct way to compute the limit in such a context. If you're using the extended reals instead (which is more typical in calculus), $1/x$ is still a continuous function at $\pm \infty$, although it is discontinuous (and undefined) at $0$. $\endgroup$ – Hurkyl Jun 6 '14 at 20:54

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