Five maths mistakes that look right

Some mistakes are easy to spot — a dropped sign, a number copied wrong — and others are not spotted at all, because the step that produces them looks like a real rule. They are not careless slips: they are shortcuts that resemble a legitimate property too closely, pass a quick visual check, and calmly hand you a wrong number.

The five below cut across years and chapters: I see them from the first two years of high school all the way to the maturità (Italy’s school-leaving exam), in algebra as well as in trigonometry. For each one: the numerical counterexample that takes it apart, and the correct rule written out in positive form.

1. The square root of a square does not give back the number you started with

From $x^2=9$ it does not follow that $x=3$: it follows that $x=3$ or $x=-3$, because both of them, squared, give 9. The reason is that $\sqrt{x^2}=|x|$, not $x$: the square root of a square is always an absolute value, never negative.

The same slip, in a sneakier form, shows up in $x^2=5x$: you do not cancel the $x$ on both sides. Doing so means dividing by $x$ without knowing whether it can be zero — and $x=0$ is precisely one of the solutions. Moving everything to one side and factorising:

Rule: the equation $x^2=k$ has two opposite solutions if $k>0$, only $x=0$ if $k=0$, and no real solution if $k<0$. And when an unknown is a common factor on both sides, you factorise it — you never cancel it.

2. Functions do not distribute over a sum

Square roots, logarithms, sines and reciprocals do not behave like multiplication: none of them distributes over a sum. With $a=9$ and $b=16$:

Five against seven: the equality $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$ is false in general. The same happens with $\ln (1+1)=\ln 2\approx 0.693$ against $\ln 1+\ln 1=0$, with $\sin (30°+60°)=1$ against $\sin 30°+\sin 60°\approx 1.366$, and with $\frac{1}{2+3}=0.2$ against $\frac{1}{2}+\frac{1}{3}\approx 0.833$.

Rule: what is true concerns the product, not the sum — $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (with $a,b\ge 0$) and $\ln (ab)=\ln a+\ln b$ (with $a,b>0$). The only apparent exception is the sine, which really does have a formula for the sum of two angles, but it is an identity of its own with two terms — $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ — not a distribution.

3. You cancel factors, not terms

$\frac{x+3}{x}$ is not 3: the $x$ in the numerator is added to 3, not multiplied by it, so it is not a factor in common with the $x$ in the denominator. With $x=2$ you see it straight away: $\frac{2+3}{2}=2.5$, not 3.

$\frac{x^2+3x}{x}$ is different, because here the numerator factorises as $x(x+3)$: the $x$ is a common factor, and cancelling it gives $x+3$ — but only for $x\ne 0$, because you are dividing by that letter.

Rule: you cancel a factor common to the whole numerator and the whole denominator, never a term that appears inside a sum. And every time you cancel a letter, the condition that the letter is not zero still holds afterwards.

4. You don’t flip an inequality sign at random

Multiplying or dividing by a negative number flips the inequality; by a positive number it keeps it — so far it is mechanical. The problem is multiplying by something whose sign you don’t know, like $x$ itself. From $\frac{1}{x}>1$ it does not follow that $x<1$: with $x=-1$, which satisfies $x<1$, you get $\frac{1}{-1}=-1$, which is not greater than 1. The forbidden step has introduced a solution that does not exist.

The correct way is to move everything to one side and study the sign:

Numerator and denominator have the same sign (both positive) only for $0<x<1$.

Rule: if the factor you would like to multiply by contains $x$, don’t multiply — move everything to one side and study the sign of the result.

5. Write the domain conditions first, not last

In an equation with logarithms or radicals, combining terms can introduce solutions that the original equation could not even be written for. Take $\log (x-1)+\log (x-3)=\log 3$: combining the logarithms leads to $x(x-4)=0$, that is $x=0$ or $x=4$. But the conditions — $x-1>0$ and $x-3>0$, that is $x>3$ — must be written before combining, not deduced at the end by checking whether the result “looks odd”: $x=0$ does not satisfy them and is discarded from the start, leaving $x=4$ as the only solution.

Rule: every argument of a logarithm must be positive, every radicand of an even-index root non-negative — conditions written at the start of the exercise, before any step that hides them.

The common thread

None of these five is a wrong calculation: they are all steps that seem legitimate because they resemble a real operation. The defence is always the same — when a step looks too convenient, stop and check it with a concrete number before trusting the algebra. It is the same reflex you need when simplifying radicals or choosing the right method for a linear system: the shortcut that looks obvious is almost always the one to check twice. A quick way to find out which of the five affect you is to sit a mixed second-year test against the clock and reread your rough work with fresh eyes.

If you’d like to go over these steps on your own exercises, with maths tutoring, write to me: we start exactly from where you get stuck.