Mistake Museum

Diagnosis When $a=2$, both elements $a+2$ and $a^2$ equal $4$, and the set reads $\{4,4\}$. The elements of a set must be distinct, so this is not a two-element set at all.

“When a set contains a parameter, always substitute back and check that the elements are distinct”

Diagnosis $\varnothing$ has no elements at all; $\{\varnothing \}$ has one element, which happens to be the empty set. Think of an empty box versus a box containing one empty box.

“Count the elements first, then ask whether the sets are equal”

Diagnosis Take $A=B=C=\{1\}$. On the left, $B\setminus C=\varnothing$, so the left side is $\{1\}$; on the right, $A\setminus B=\varnothing$, and removing more still leaves $\varnothing$. The two sides disagree on the region $A\cap B\cap C$.

“Only union and intersection let you move brackets freely; difference does not”

Diagnosis An element outside $A\cap B$ is only missing from at least one of them: it may be in $A$ but not in $B$. For example, with $U=\{1,2\}$, $A=\{1\}$, $B=\{2\}$, the left side is $U$ and the right side is $\varnothing$.

“When the complement goes inside the brackets, intersection and union swap”

Diagnosis Out of the $2^n$ subsets, two have to go: the empty set (not non-empty) and the whole set itself (not proper).

“Exclude as many as you name, and name every one you exclude”

Diagnosis Different subsets can have the same sum: $\{1,4\}$ and $\{2,3\}$ both sum to $5$. This correspondence is “many-to-one”, so the two sides do not have the same size (there are actually $16$ subsets).

“Only a one-to-one correspondence lets you count the other side instead”

Diagnosis Two of the $16$ colorings use a single color for everything, and then the other class is the empty set. Every piece of a partition must be non-empty, so these two do not count.

“No piece of a partition may be empty”

Diagnosis Some students may be in both clubs. “Math club” and “physics club” are not a partition: the two pieces intersect, and anyone in both is counted twice.

“Check that the pieces are pairwise disjoint before you use the addition principle”

Diagnosis Counterexample: $\{1,2\},\{1,3\},\{2,3\}$ together with all $4$ three-element subsets and $\{1,2,3,4\}$. That is $8$ members, every two of which intersect (a three-element subset misses only one element, so it meets every two-element subset), yet no element is shared by all of them.

“A construction that reaches the bound is often not unique”

Diagnosis Subsets in one layer really do not contain one another, but the layers have different sizes: there are only $5$ one-element subsets, but ${\mathrm{C}}_5^2=10$ two-element subsets.

“Take the widest layer”

Diagnosis What follows is $5-3=2$, $3-2=1$, and from there every integer is in $S$, that is, $S\supseteq \mathbf{Z}$. But the problem never said $S$ contains only integers: $S=\mathbf{R}$ is also closed under subtraction and also contains $3$ and $5$.

“What you derive is a lower bound, not the whole set”

Diagnosis $a=\frac{1}{1-a}$ becomes $a^2-a+1=0$, whose discriminant is $1-4<0$: no real solution. So in a set $S$ of real numbers, $a$ and $\frac{1}{1-a}$ are always two different elements.

“Before you make the rule “stand still”, solve the equation”

Diagnosis $x=0$ and $x=2$ belong to no case: something is left out. And this equation holds at every point of $0\le x\le 2$, including both endpoints, so leaving them out gives a wrong answer.

“Nothing left out (the union is everything); nothing counted twice (the classes are pairwise disjoint)”

Diagnosis $6,12,18$ are both even and multiples of $3$, so they are counted twice; and $1,5,7,\dots$ are in neither class. This is not a partition.

“One classification, one criterion”

Diagnosis $S$ has no smallest element: for any $m>0$, $\frac{m}{2}$ is smaller. The well-ordering principle only promises a smallest element for a finite non-empty set of reals, or a non-empty set of natural numbers.

“Make sure the extreme object exists before you pick it”

Diagnosis $B$ beating $A$ only means $B$ won one game that $A$ did not; $A$ may also have won many games that $B$ did not. And several players may be tied for the most wins.

“Use “nobody is larger”; do not swap it for “it is larger than everyone else””

Diagnosis Look at the middle region $A\cap B\cap C$: the first three terms count it $3$ times, the last three subtract it $3$ times, leaving $0$ times. It has been left out.

“If you subtract too much, add it back, until every region is counted exactly once”

Diagnosis The $5$ numbers $6,12,\dots ,30$ are multiples of both $2$ and $3$, so they were subtracted twice. In fact there are $10$: $1,5,7,11,13,17,19,23,25,29$.

“To count a complement, first get the union right”

Diagnosis The elements of $A$ are the ones making choices: each picks an image from the $3$ elements of $B$, and $2$ choices are made. That is $3\times 3$, not $2\times 2\times 2$.

“Whoever makes the choices is the exponent”

Diagnosis A mapping needs every element of $A$ to have a unique image. Here $1$ goes to both $3$ and $4$, so its image is not unique.

“On the A side, none missing and none doubled; on the B side, anything goes”

Diagnosis The point of the new graph at $x$ has the same height as the point of the old graph at $x+1$. So what the old graph looked like at $x+1$ now appears at $x$: everything has moved $1$ to the left.

“Write it as f(x − h), then read off h”

Diagnosis That is $y=|f(x)|$: the absolute value sits on the function value, so it acts up and down. In $f(|x|)$ the absolute value sits on the variable: $x$ and $-x$ give the same value, so it is an even function. Keep the part to the right of the $y$-axis and replace the left part by the mirror image of the right.

“The absolute value acts on whatever it wraps”

Diagnosis $f(x)=\frac{1}{x}$ is odd, but $0$ is not in its domain, so the graph does not pass through the origin at all. “$f(0)=0$” assumes that $0$ is in the domain.

“Before you use f(0), ask whether 0 is in the domain”

Diagnosis Simplifying threw away a restriction on the domain: the original function needs $x\ne 1$. So $x=-1$ is in the domain but $1$ is not. The domain is not symmetric about the origin, so the function can be neither odd nor even.

“No symmetric domain, no parity”

Diagnosis Take $x_1=-1<x_2=1$: $f(-1)=-1<f(1)=1$, so the right-hand point is higher. Monotonicity must hold for any two points of the interval, and a pair straddling $0$ breaks it.

“Do not glue monotone intervals together carelessly”

Diagnosis Three points say nothing about what happens in between: $f(x)=(x-1.2{)}^2$ has $f(1)=0.04<f(2)=0.64<f(3)=3.24$, yet it is decreasing on $[1,1.2]$.

“Examples can only disprove; a proof needs “any””

Diagnosis After a shift by $2$ the graph does not land on itself: it lands upside down. A period needs $f(x+T)=f(x)$, and here there is an extra minus sign.

“Flip twice to get back to where you started”

Diagnosis The constant function $f(x)=c$ has every positive number as a period, so there is no smallest one. The Dirichlet function ($1$ when $x$ is rational, $0$ when $x$ is irrational) has every positive rational as a period, and also has no least positive period.

“Having a period does not mean having a least positive period”

Diagnosis $f(1)>0$ only says that $1$ does not lie between the roots: both roots could be to the right of $1$, or both to the left. Here the roots are $-1$ and $-3$, both on the left.

“The sign at an end point only tells “same side” or “opposite sides”; which side comes from the axis”

Diagnosis $\Delta =9-12<0$: there are no real roots at all. The whole parabola floats above the $x$-axis; the end-point and axis conditions hold, but there are no roots to place.

“First ask whether there are roots, then ask where they are”

Diagnosis The argument of a logarithm must be positive: $x^2-2x-3>0$, that is, $x<-1$ or $x>3$. Points of $[-1,1)$ are not in the domain at all; at $x=0$, for instance, the argument is $-3$.

“With a logarithm, settle the argument first, then talk about monotonicity”

Diagnosis Equality in $x+\frac{4}{x}\ge 4$ holds at $x=2$, and $2\notin [3,6]$. On this interval the value is always greater than $4$, so $4$ is not the minimum.

“When you use AM–GM to find an extreme value, first ask whether equality can actually be reached”

Diagnosis Equality needs $\sqrt{x^2+5}=1$, which is impossible. Let $t=\sqrt{x^2+5}$; then $t\ge \sqrt{5}>1$, which is on the increasing part of $t+\frac{1}{t}$.

“After a substitution, write down the range of the new variable first”

Diagnosis $x+\frac{9}{x}=6$ only when $x=3$, and $3$ is not in $[1,2]$. So $6$ is only a lower bound. It is never reached, so it is not the minimum.

“Before you write “minimum”, find the $x$ where equality holds”

Diagnosis When $y=2$ the equation becomes $0\cdot x^2+0\cdot x+3=0$, which has no solution at all. $\Delta$ only means something for a quadratic equation; here $\Delta =0$ tells you nothing.

“Last step of the discriminant method: test separately the value that makes the leading coefficient 0”

Diagnosis “The two ends are enough” is true only for a linear function, whose segment lies between its two ends. $h$ is a parabola opening upward, and it dips in the middle: $h\left(\frac{1}{2}\right)=-\frac{1}{2}<0$.

“The endpoint method needs “linear” or “monotonic””

Diagnosis $f(t)=t^3-t$ is not monotonic on $\mathbb{R}$: $f(0)=f(1)=f(-1)=0$, so different inputs can give the same value. For example $x=0$ gives $f(1)+f(0)=0$, so it is a solution too.

“Oddness makes the two sides match; monotonicity makes the inputs equal”

Diagnosis $f^{(3)}(x)$ means composing three times: $f(f(f(x)))=((x^2{)}^2{)}^2=x^8$. $[f(x){]}^3$ is a power of the value. They are completely different operations.

“Iterating means “do it again”, not “multiply again””

Diagnosis The first three terms fitting a formula does not mean every term fits it. “Spotting the pattern” is only a guess; the proof is still missing.

“A guessed formula needs a proof by induction, or a method that derives it directly”

Diagnosis $a_n$ is $1,3,7,15,\dots$, which grows without bound and has no limit at all. Solving $L=f(L)$ only says: if there is a limit, it must be a fixed point.

“A fixed point is only a candidate; whether the orbit goes there needs its own proof”

Diagnosis $x$ can be positive or negative, but $t=x^2$ can only be non-negative, and both $x=\pm \sqrt{t}$ must satisfy the equation. $x=2$ gives $f(4)=3$, while $x=-2$ gives $f(4)=-1$: a contradiction.

“The substitution must be reversible, and the range must be stated”

Diagnosis Elimination only proves: “if $f$ satisfies the original equation, then $f$ can only be this formula.” That is a necessary condition. Sometimes the formula is undefined at some points, or does not satisfy the original equation at all.

“Elimination gives a candidate; substituting back makes it an answer”

Diagnosis $f(0)=0$ and “even” are only necessary conditions. Plenty of functions satisfy them (for example $f(x)=x^4$), and $x^4$ does not satisfy the original equation; on the other hand $f(x)=3x^2$ does satisfy it too. Plugging in values by itself cannot give “$f(x)=x^2$”.

“Plugging in gives clues, not the answer”

Diagnosis The equation only “strings together” the rational points. An irrational number like $\sqrt{2}$ can never be reached by finitely many additions, subtractions and integer multiples. Without an extra condition, there really are solutions that are not $cx$ (it is just impossible to write one down).

“The rung from ℚ to ℝ needs an extra condition”

Diagnosis $c_1\cdot 2^n+c_2\cdot 2^n=(c_1+c_2)2^n$ really has only one unknown constant, which cannot match two initial values $f(1)$ and $f(2)$. For example, with $f(1)=2$ and $f(2)=12$, $2c=2$ contradicts $4c=12$.

“Two initial values need two independent constants”

Diagnosis ${\sin }^2\alpha +{\cos }^2\alpha =1$ only tells you that ${\cos }^2\alpha =\frac{144}{169}$, and that has two square roots, one positive and one negative. An angle with $\sin \alpha >0$ can be in quadrant I or in quadrant II: the line $y=\frac{5}{13}$ meets the unit circle at two points.

“The Pythagorean identity gives the size; ask the unit circle for the sign”

Diagnosis At $90^{\circ }$ the terminal side $OP$ points straight up and is parallel to the tangent $x=1$, so it never meets it: $\tan 90^{\circ }$ is undefined. It is not “the number infinity”. And if you approach from the $91^{\circ }$ side, the tangent line points down and $\tan$ heads off in the negative direction.

“No intersection, no function value”

Diagnosis A shift adds to or subtracts from $x$ itself. Shifting $y=\sin 2x$ left by $a$ gives $y=\sin 2(x+a)=\sin (2x+2a)$. For $2a=\frac{\pi }{3}$ we need $a=\frac{\pi }{6}$.

“The shift is what gets added to x, so factor out ω first”

Diagnosis $y=\sin 3x$ already reaches its maximum at $x=\frac{\pi }{6}$, while $y=\sin x$ only gets there at $\frac{\pi }{2}$: the same height comes earlier, so the graph has been squeezed. The period $\frac{2\pi }{3}$ is shorter than $2\pi$.

“The larger ω, the denser the waves”

Diagnosis For $x⩾0$ it is just $\sin x$, but for $x<0$ the graph is the mirror image of the right half. Take $x=-\frac{\pi }{2}$: $f\left(-\frac{\pi }{2}\right)=1$, while $f\left(-\frac{\pi }{2}+2\pi \right)=\sin \frac{3\pi }{2}=-1$. One counterexample is enough.

“A period must work for every x; one x is enough to break it”

Diagnosis The inner function $u=\frac{\pi }{4}-2x$ is decreasing in $x$. While $\sin u$ increases, $y$ actually decreases as $x$ grows, so this gives exactly the decreasing intervals.

“Make ω positive first, then substitute the whole block”

Diagnosis The right-hand side is about $1.207$, which is bigger than $1$: it cannot be a sine. The picture shows it clearly: the two pieces that make up the height of $P$ are not $\sin \alpha$ and $\sin \beta$; each one is multiplied by a cosine, giving $\cos \beta \sin \alpha$ and $\sin \beta \cos \alpha$.

“sin is not multiplication, so you cannot just open the brackets”

Diagnosis If $\alpha$ is in $(270^{\circ },360^{\circ })$, then $\frac{\alpha }{2}$ is in $(135^{\circ },180^{\circ })$, an angle in quadrant II, whose cosine is negative. The formula only gives the size; the sign depends on where $\frac{\alpha }{2}$ lands.

“Before taking the root in a half-angle formula, work out the range of the half angle”

Diagnosis Cosine is decreasing on $(0^{\circ },90^{\circ })$, so $\cos 40^{\circ }>\cos 80^{\circ }$ and the difference must be positive. The formula is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$, with a minus sign in front. In the picture: $u-v$ is perpendicular to the angle bisector, and turning the bisector direction by $90^{\circ }$ turns its $\cos$ into $-\sin$.

“Of the four formulas, only “cos minus cos” carries a minus sign”

Diagnosis Sum-to-product needs the same function in both terms: both vectors must be projected onto the same axis. $\sin \alpha$ is a projection onto the $y$-direction and $\cos \beta$ onto the $x$-direction, so they do not combine into one diagonal.

“Same function first, then the product”

Diagnosis Expand: $2\sin \left(x-\frac{\pi }{3}\right)=\sin x-\sqrt{3}\cos x$, off by exactly a minus sign. $\tan \phi$ only fixes the direction of a line, so $\phi$ could still be off by $\pi$. The auxiliary angle must satisfy two conditions, $\cos \phi =\frac{a}{R}$ and $\sin \phi =\frac{b}{R}$: the point $(a,b)=(-1,\sqrt{3})$ is in quadrant II.

“φ is the direction of the point (a, b), not just a slope”

Diagnosis $-2$ would need $x+\frac{\pi }{3}=\frac{3\pi }{2}$, that is $x=\frac{7\pi }{6}$, which is not in the interval. The L may only turn through $90^{\circ }$, so the hypotenuse never gets to point straight down.

“After combining, find the range of x + φ first”

Diagnosis In $(0^{\circ },180^{\circ })$ there are two angles whose sine is $\frac{\sqrt{3}}{2}$: $60^{\circ }$ and $120^{\circ }$. The law of sines only tells you $\sin B$, not whether $B$ is acute or obtuse. Here $b>a$, so $B>A$, and neither angle clashes with $A=30^{\circ }$.

“When you get an angle from its sine, ask: does the obtuse one work too?”

Diagnosis Two angles in $(0,2\pi )$ with the same sine are either equal or supplementary. Both $2A$ and $2B$ lie in $(0,2\pi )$, and $2A+2B=\pi$ is perfectly possible, for example $A=30^{\circ }$, $B=60^{\circ }$.

“Equal sines? Think “equal or supplementary” first”

Diagnosis Angles with the same tangent differ by an integer multiple of $\pi$, so the value of the tangent alone does not pin down the angle. Both $\arctan 2$ and $\arctan 3$ lie in $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$, so their sum lies in $\left(\frac{\pi }{2},\pi \right)$: it cannot possibly be negative.

“Estimate the range of the angle first, then use the trig value to pin it down”

Diagnosis Dividing by $\sin x$ silently assumes $\sin x\ne 0$. But when $\sin x=0$ both sides are $0$, so $x=k\pi$ are solutions too, and they have been quietly thrown away.

“A factor containing the unknown may be factored out, never cancelled”

Diagnosis Sine is not monotonic. $x=\pi$ satisfies $x>\frac{\pi }{6}$, yet $\sin \pi =0$; $x=-\frac{11\pi }{6}$ does not satisfy $x>\frac{\pi }{6}$, yet $\sin x=\frac{1}{2}$… Nothing matches. The solution set comes in separate pieces, and it repeats every $2\pi$.

“Draw one period first, then add the period”

Diagnosis $\sqrt{1-x^2}$ is non-negative, but $\sin t$ can be negative. Only with $t\in [0,\pi ]$ is $\sqrt{1-x^2}=\sin t$ true; without that restriction the equation is wrong whenever $t$ is on the lower half of the circle, and the minimum $-\sqrt{2}$ is never actually reached.

“A substitution must match both ways: x covers its whole domain, and t ranges over exactly enough”

Diagnosis For $m=1$ we get $z=0+0i=0$, which is a real number, not a purely imaginary one. A purely imaginary number needs real part $0$ and imaginary part not $0$; the second condition was dropped.

“Purely imaginary = real part zero + imaginary part nonzero: you need both conditions”

Diagnosis $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for $a,b\ge 0$. Write the square root of a negative number as a multiple of $i$ first: $\sqrt{-4}=2i$, $\sqrt{-9}=3i$.

“When you see the square root of a negative number, bring out the i before you calculate”

Diagnosis Trigonometric form requires $r>0$. The modulus is a length and can never be negative. The minus sign in front amounts to a further turn of $180^{\circ }$.

“Trigonometric form has three requirements: r positive, cos first, a plus sign in the middle”

Diagnosis There is a minus sign in the middle, so this is not trigonometric form. The point $(\cos 40^{\circ },-\sin 40^{\circ })$ is in the fourth quadrant; its argument is $-40^{\circ }$ plus a full turn.

“Rewrite in standard trigonometric form first, then read off the argument”

Diagnosis The distance between the two fixed points is $2$, and the sum of the distances is also $2$. Two sides of a triangle add up to more than the third side, with equality only when $P$ lies on the segment, so the locus is just a segment.

“Compare the constant with the distance between the two fixed points before you conclude anything”

Diagnosis “If $\Delta \ge 0$ there are real roots” was derived for equations with real coefficients. When the coefficients contain $i$, the discriminant cannot even be “greater than zero”. In fact, substitute $x=2$: $4-4-2i+2i=0$, so there is a real root.

“The discriminant is only for real coefficients; with complex coefficients, turn the complex into the real first”

Diagnosis For two vectors with the same starting point, the difference points from the head of the vector being subtracted to the head of the vector it is subtracted from: $\overrightarrow{AB}-\overrightarrow{AC}$ goes from $C$ to $B$. Check: $\overrightarrow{AC}+\overrightarrow{CB}=\overrightarrow{AB}$.

“After a subtraction, add back to check it”

Diagnosis Adding vectors combines directions. You get $7$ only when $\vec{a},\vec{b}$ point the same way; $1$ when they point opposite ways, and $5$ when they are perpendicular. In general there is only the triangle inequality: $||\vec{a}|-|\vec{b}||\le |\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|$.

“When vectors add, their lengths usually do not”

Diagnosis $\vec{a}\cdot \vec{b}$ only sees the shadow of $\vec{b}$ in the direction of $\vec{a}$, and infinitely many vectors cast the same shadow: with $\vec{a}=(1,0)$, both $\vec{b}=(2,1)$ and $\vec{c}=(2,-5)$ have dot product $2$.

“A dot product is a number: the direction information is lost, so you cannot cancel”

Diagnosis $\vec{a}\cdot \vec{b}<0$ only says $\cos \theta <0$, that is, $90^{\circ }<\theta \le 180^{\circ }$. When $\theta =180^{\circ }$ (the vectors point in opposite directions) the dot product is negative too, but that is not an obtuse angle.