# BLOG: Heidelberg Laureate Forum

Laureates of mathematics and computer science meet the next generation

This year’s HLF included a fascinating discussion between ACM AM Turing Award laureates Leslie Lamport and Whitfield Diffie, written up in full here by blog team member Andrei Mihai. As part of the conversation, when discussing how rigorous some proofs can be (and in some cases, not be), the topic of false proofs came up – with Lamport mentioning a particular example he’d seen in high school of a proof that all triangles are isosceles.

Shared a few years ago in a blog post from logician Joel David Hamkins, one such proof goes like this:

Consider an arbitrary triangle $$\triangle ABC$$. Let $$Q$$ be the intersection of the angle bisector (blue) at $$\angle A$$ and the perpendicular bisector (green) of $$BC$$ at midpoint $$P$$.

Drop perpendiculars from $$Q$$ to $$AB$$ at $$R$$ and to $$AC$$ at $$S$$. Because $$P$$ is the midpoint of $$BC$$ and $$PQ$$ is perpendicular, we deduce $$BQ\cong CQ$$ (they are congruent) by the Pythagorean theorem. Since $$AQ$$ is the angle bisector of $$\angle A$$, the triangles $$AQR$$ and $$AQS$$ are similar, and since they share a hypotenuse, they are congruent. It follows that $$AR\cong AS$$ and also $$QR\cong QS$$. Therefore the triangle $$\triangle BQR$$ is congruent to the triangle $$\triangle CQS$$ by the hypotenuse-leg congruence theorem. So $$RB \cong SC$$. And therefore,

$AB\cong AR+RB\cong AS+SC \cong AC$

and so the triangle is isosceles.

You might be wondering where the error is here – I won’t give too much away, but it hinges on the position of the point Q – and this useful Geogebra applet, created by a commenter on Joel’s blog post following the description given in the proof, might be enlightening (so don’t open it unless you want a spoiler!)

This kind of ‘proof’ is often used as an amusing diversion by mathematicians – there’s humour in the contradiction of rigorous systems of mathematical proof being misused like this. They’re a popular diversion, as evidenced by the pages and pages of examples shared whenever people ask on maths websites. It’s also amusing to consider the obviously false conclusions you can reach – for example, here’s a proof that 1=2:

Let $$a=b$$. Then

$$a^2 = ab$$ (multiplying by $$a$$ on both sides)

$$a^2 + a^2 = a^2 + ab$$ (adding $$a^2$$ to each side)

That is, $$2a^2 = a^2 + ab$$

Now subtract $$2ab$$ from both sides:

$$2a^2 – 2ab = a^2 + ab\ – 2ab$$

That is, $$2a^2 -2ab = a^2 – ab$$

Then we can take out a factor of 2 on the left:

$$2(a^2 – ab) = a^2 – ab$$

Then dividing through by a common factor of $$a^2-ab$$ on both sides, we have

$$2=1$$.

The mistake in this one comes quite near the end – hopefully you can spot it. But as well as being an amusing diversion, these proofs also provide an insight into how mistakes can happen when writing proofs. It’s often a useful exercise for students to look through these ‘joke proofs’ and try to find the error – and when they spot it, it really hammers home the importance of checking everything you’re doing follows the basic rules of mathematics, otherwise you can find yourself proving something that’s clearly false.

This one, which I saw recently on Twitter, will really annoy pure mathematicians (but engineers might not mind too much!):

$\begin{eqnarray*} x &=& \frac{(\pi+3)}{2}\\ 2x &=& \pi + 3\\ 2x(\pi-3) &=& (\pi+3)(\pi-3)\\ 2\pi x – 6x &=& \pi^2-9\\ 9-6x &=& \pi^2 -2\pi x\\ 9 – 6x +x^2 &=& \pi^2 -2\pi x + x^2\\ (3-x)^2 &=& (\pi – x)^2\\ 3-x &=& \pi – x\\ \pi &=& 3 \end{eqnarray*}$

(This one involves another of the classic easy-to-miss errors, but I won’t tell you where it is!)

Of course, the risk with sharing these might be that people are intimidated by the possibility of making a mistake in a mathematical proof. Leslie Lamport said “[When I saw the isosceles triangle proof] I realised that you could not believe a single proof that you wrote in your high school plane geometry class – none of them are rigorous”. While this is more a comment on the level of rigour employed at that level of maths education than a lack of belief in his own abilities, it’s possible that the realisation of how easy it is to make these errors might cause some mathematicians to doubt themselves.

The important thing to remember is that mistakes, while very possible and reasonably common – Lamport himself estimates that roughly a third of all published refereed mathematical papers contain an incorrect theorem – are an important part of the process of learning how to do mathematics (although I guess it would be better if they didn’t actually make it out into the literature!)

By studying these incorrect proofs, we can learn what the common pitfalls are. The Wikipedia page on mathematical fallacies lists common numerical and geometrical errors – including some evidenced in the examples given here – and I’ve run sessions with my students before which include a chance to find and correct errors in other people’s work, which have helped them develop the confidence to look for and fix errors in their own.

I’ll leave you with one final fun fallacious proof, and it’s a classic (and this time, it’s not even proving a false conclusion):

$\require{cancel} \frac{16}{64} = \frac{1\cancel{6}}{\cancel{6}4} = \frac{1}{4}$

QED.

### Posted by Katie Steckles

is a mathematician based in Manchester, who gives talks and workshops on different areas of maths. She finished her PhD in 2011, and since then has talked about maths in schools, at science festivals, on BBC radio, at music festivals, as part of theatre shows and on the internet. Katie writes blog posts and editorials for The Aperiodical, a semi-regular maths news site.

1. Proofs can also contain copy and paste errors as in the example taken from Twitter. On Twitter the fourth line reads :
2πx – 6x = π² – 9, but here the fourth line reads 2πx – 6x = π² + 9

As you can see, errors of all kinds can easily be overlooked. But people know that they are almost always wrong.

2. Katie Steckles wrote (06. Oct 2021):
> Fun with False Proofs […]

> $\begin{eqnarray*} x &=& \frac{(\pi+3)}{2}\\ 2x &=& \pi + 3\\ 2x(\pi-3) &=& (\pi+3)(\pi-3)\\ 2\pi x – 6x &=& \pi^2+9\\ \end{eqnarray*}$

… should be instead: $2\pi x – 6x &=& \pi^2 – 9$

> $\begin{eqnarray*} 9-6x &=& \pi^2 -2\pi x\\ 9 – 6x +x^2 &=& \pi^2 -2\pi x + x^2\\ (3-x)^2 &=& (\pi – x)^2\\ 3-x &=& \pi – x\\ \pi &=& 3 \end{eqnarray*}$

Here’s a variant of another false proof:

Assume that there exists a one-to-one function between the set of all natural numbers and the set of all subsets of the set of all natural numbers:
$f : \mathbb N \longleftrightarrow \text{PowerSet} \! [ \, \mathbb N \, ].$

This allows to describe the following set, whose elements (if any) are natural numbers:
$\mathcal D := \{ x \in \mathbb N : x \not\in f[ \, x \, ] \}.$

Since set $$\mathcal D$$ is necessarily one subset of the set of all natural numbers (and therefore in the range of function $$f$$), and since function $$f$$ is invertible, obtain the natural number (in domain of function $$f$$):
$d := (f^{-1})[ \, \mathcal D \, ].$

Therefore in turn
$f[ \, d \, ] \equiv \mathcal D.$

Supposing that $$d \in \mathcal D$$ is equivalent to $$d \in f[ \, d \, ]$$ and, by construction of set $$\mathcal D$$, this implies $$d \not\in \mathcal D$$.
On the other hand, supposing that $$d \not\in \mathcal D$$ is equivalent to $$d \not\in f[ \, d \, ]$$ and, by construction of set $$\mathcal D$$, this implies $$d \in \mathcal D$$.

Conclude that the initial assumption was false and that a supposed function $$f$$ cannot have existed from the outset. (q.e.d.)

• Hi Frank

Good spot – that was a non-deliberate error, unlike the deliberate error also present in that proof! 🙂 Should be fixed now.

Thanks!

3. Katie Steckles wrote (06. Oct 2021):
> Fun with False Proofs […]

> $\begin{eqnarray*} x &=& \frac{(\pi+3)}{2}\\ 2x &=& \pi + 3\\ 2x(\pi-3) &=& (\pi+3)(\pi-3)\\ 2\pi x – 6x &=& \pi^2+9\\ \end{eqnarray*}$

… should be $2\pi x – 6x = \pi^2 – 9$

> $\begin{eqnarray*} 9-6x &=& \pi^2 -2\pi x\\ 9 – 6x +x^2 &=& \pi^2 -2\pi x + x^2\\ (3-x)^2 &=& (\pi – x)^2\\ 3-x &=& \pi – x\\ \pi &=& 3 \end{eqnarray*}$

Here’s a variant of another false proof:

Assume that there exists a one-to-one function between the set of all natural numbers and the set of all subsets of the set of all natural numbers:
$f : \mathbb N \longleftrightarrow \text{PowerSet} [ \, \mathbb N \, ].$

This allows to describe the following set, whose elements (if any) are natural numbers:
$\mathcal D := \{ x \in \mathbb N : x \not\in f[ \, x \, ] \}.$

Since set $$\mathcal D$$ is necessarily one subset of the set of all natural numbers (and therefore in the range of function $$f$$), and since function $$f$$ is invertible, obtain the natural number (in domain of function $$f$$):
$d := (f^{-1})[ \, \mathcal D \, ].$

Therefore in turn
$f[ \, d \, ] \equiv \mathcal D.$

Supposing that $$d \in \mathcal D$$ is equivalent to $$d \in f[ \, d \, ]$$ and, by construction of set $$\mathcal D$$, this implies $$d \not\in \mathcal D$$.
On the other hand, supposing that $$d \not\in \mathcal D$$ is equivalent to $$d \not\in f[ \, d \, ]$$ and, by construction of set $$\mathcal D$$, this implies $$d \in \mathcal D$$.

Conclude that the initial assumption was false and that a supposed function $$f$$ cannot have existed from the outset. (q.e.d.)

4. Frank Wappler wrote (06.10.2021, 18:25 o’clock):
> […] Here’s a variant of another false proof:

> Assume that there exists a one-to-one function between the set of all natural numbers and the set of all subsets of the set of all natural numbers:
> $f : \mathbb N \longleftrightarrow \text{PowerSet} [ \, \mathbb N \, ].$

> This allows to describe the following set, whose elements (if any) are natural numbers:
> $\mathcal D_f := \{ x \in \mathbb N : x \not\in f[ \, x \, ] \}.$

Well — there’s the ((mildly) funny) mistake already:
this supposed “description of a set” does not genuinely and consistently describe a set at all.

There are however certain related genuine and consistent descriptions of sets, namely:

$\mathcal D_f^+ := \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } ((x \equiv b) \text{ and } (b_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } (x \equiv b_f) \} \, ])) \}$

or

$\mathcal D_f^- := \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } ((x \not\equiv u) \text{ and } (u_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } (x \not\equiv u_f) \} \, ])) \}.$

There the designation $$b_f$$ for the specific value $$b_f = (f^{-1})[ \, D_f^+ \, ]$$ recounts B. Russell’s reference to “the barber who’d shave all citizens who wouldn’t shave themselves, and who’d shave himself, too”;
while the designation $$u_f$$ for the specific value $$u_f = (f^{-1})[ \, D_f^- \, ]$$ correspondingly recalls “the undertaker who’d bury all citizens who wouldn’t bury themselves, but himself not”.

5. Frank Wappler wrote (06.10.2021, 18:25 o’clock):
> […] Here’s a variant of another false proof:

> Assume that there exists a one-to-one function between the set of all natural numbers and the set of all subsets of the set of all natural numbers:
> $f : \mathbb N \longleftrightarrow \text{PowerSet} [ \, \mathbb N \, ].$

> This allows to describe the following set, whose elements (if any) are natural numbers:
> $\mathcal D_f := \{ x \in \mathbb N : x \not\in f[ \, x \, ] \}.$

Well — there’s the ((mildly) funny) mistake already:
this supposed “description of a set” does not genuinely and consistently describe a set at all.

There are however certain related genuine and consistent descriptions of sets, namely:

$$\mathcal D_f^+ :=$$
$\{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } ((x \equiv b_f) \text{ and } (b_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } (x \equiv b_f) \} \, ])) \}$

or

$( \mathcal D_f^- := \) \[ \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } ((x \not\equiv u_f) \text{ and } (u_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } (x \not\equiv u_f) \} \, ])) \}.$

There the designation $$b_f$$ for the specific value $$b_f = (f^{-1})[ \, D_f^+ \, ]$$ recounts B. Russell’s reference to
“the barber who’d shave all citizens who wouldn’t shave themselves, and who’d shave himself, too”;
while the designation $$u_f$$ for the specific value $$u_f = (f^{-1})[ \, D_f^- \, ]$$ correspondingly recalls
“the undertaker who’d bury all citizens who wouldn’t bury themselves, but himself not”.

6. Frank Wappler wrote (06.10.2021, 18:25 o’clock):
> […] Here’s a variant of another false proof:

> Assume that there exists a one-to-one function between the set of all natural numbers and the set of all subsets of the set of all natural numbers:
> $f : \mathbb N \longleftrightarrow \text{PowerSet} [ \, \mathbb N \, ].$

> This allows to describe the following set, whose elements (if any) are natural numbers:
> $\mathcal D_f := \{ x \in \mathbb N : x \not\in f[ \, x \, ] \}.$

Well — there’s the ((mildly) funny) mistake already:
this supposed “description of a set” does not genuinely and consistently describe a set at all.

There are however certain related genuine and consistent descriptions of sets, namely:

$$\mathcal D_f^+ :=$$
$\{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } ((x \equiv b_f) \text{ and } (b_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ or } (x \equiv b_f) \} \, ])) \}$

or

$$\mathcal D_f^- :=$$
$\{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } ((x \not\equiv u_f) \text{ and } (u_f = (f^{-1})[ \, \{ x \in \mathbb N : (x \not\in f[ \, x \, ]) \text{ and } (x \not\equiv u_f) \} \, ])) \}.$

There the designation $$b_f$$ for the specific value $$b_f = (f^{-1})[ \, \mathcal D_f^+ \, ]$$ recounts B. Russell’s reference to
“the barber who’d shave all citizens who wouldn’t shave themselves, and who’d shave himself, too”;
while the designation $$u_f$$ for the specific value $$u_f = (f^{-1})[ \, \mathcal D_f^- \, ]$$ correspondingly recalls
“the undertaker who’d bury all citizens who wouldn’t bury themselves, but himself not”.

7. Frank Wappler wrote (08.10.2021, 14:36 o’clock):
> […] There the designation $$b_f$$ for the specific value $$b_f = (f^{-1})[ \, \mathcal D_f^+ \, ]$$ recounts B. Russell’s reference to
“the barber who’d shave all citizens who wouldn’t shave themselves, and who’d shave himself, too”;

Before this is called out as a mistake let me please quickly admit that set $$\mathcal D_f^+$$ and (consequently) the value $$(f^{-1})[ \, \mathcal D_f^+ \, ]$$ are, in some sense, not necessarily unique. Besides the (one specific) barber, we may consider by analogous construction the (likewise specific) pedicurist, manicurist, visagist, etc.

> […] while the designation $$u_f$$ for the specific value $$u_f = (f^{-1})[ \, \mathcal D_f^- \, ]$$ correspondingly recalls
“the undertaker who’d bury all citizens who wouldn’t bury themselves, but himself not”.

Similarly, set $$\mathcal D_f^-$$ and (consequently) the value $$(f^{-1})[ \, \mathcal D_f^- \, ]$$ are not necessarily unique, either. Indeed, considering (any) one specific set $$\mathcal D_f^+$$ and the specific value $$b_f = (f^{-1})[ \, \mathcal D_f^+ \, ]$$, for which by construction $$b_f \in D_f^+$$, then set $$\mathcal D_f^+$$ contains all distinct “instances” of $$u_f$$, for which by construction holds
$f[ \, u_f \, ] = D_f^- = \left( D_f^+ \setminus \{ b_f \} \right) \setminus \{ u_f \}.$