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Areas and Ratios

Do you have enough information to work out the area of the shaded quadrilateral?

Shape and Territory

If for any triangle ABC tan(A - B) + tan(B - C) + tan(C - A) = 0 what can you say about the triangle?

Napoleon's Hat

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

Dodgy Proofs

Age 16 to 18
Challenge Level



Thuong from International School Manila in the Philippenes and Soham from Lampton Academy explained what's wrong with proof S1: A pound equals a penny. Click to see the original proof again. Soham's explanation is below the proof.

£$1 = 100 p = (10p)^2 = ($ £$ 0.1)^2 = $ £$0.01 = 1p$

Soham wrote:
$100p$ isn't equal to $10p^2$
$(10p)^2 = 100p^2$

Yazan from Dubai International Academy in the UAE, Thuong and Aryansh explained what's wrong with proof S2: ${\mathbf{2 = 3}}$. Click to see the proof again.

We have $2\times 0 =0$ and also $3\times 0 = 0$. 
Therefore we have $2 \times 0 = 3 \times 0$. 
Dividing both sides by $0$ gives $2=3$.

Yazan wrote:
Multiply any number by zero and you get 0, but it doesn’t mean they’re all equal. If you multiply them by 1 or higher, all the products will be different.

Aryansh wrote:
To divide by 0 you will need a multiplicative inverse to get back the answer by multiplying, but as zero has multiplicative inverse because multiplying by 0 will give you 0, it is impossible to determine the value of anything that is to be divided by 0.

Soham and Yazan found the problem in proof S3: The perimeter of a square is four times its area. Click to see the proof again:

Choose units so that the side of the square is length 1.
Then the perimeter equals 4 units and the area equals $1\times 1=1$ unit.
Thus, the perimeter of the square is four times its area.

Soham wrote:
Area and perimeter have different units. Perimeter is a linear measurement. Area is squared measurement. 

Paolo from Twynham in England and Thuong thought about proof S4: ${\mathbf{0 = 1}}$. Click to see the proof again.

$0=0+0+0+\cdots$. But $0=1-1$, so $$ 0=(1-1)+(1-1)+(1-1)+\cdots $$ So, by rearranging the brackets, we have $$ 0=1+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+0+\cdots = 1 $$

Paolo wrote:
The problem is that using the distributive law over an infinite sum doesn't work.

However, note that $$\tfrac12+\tfrac14+\tfrac18+\tfrac1{16}+\tfrac1{32}+\tfrac1{64}+... = 1\\
\begin{align}\tfrac12 + \left(\tfrac14+\tfrac18 \right) + \left(\tfrac1{16}+\tfrac1{32}\right) + \left(\tfrac1{64}+\tfrac1{128}\right) + ... &= \tfrac12 + \tfrac38 + \tfrac3{32} +\tfrac3{128} + ... \\&= \tfrac12 + \tfrac38\times\frac1{1-\tfrac14}\\&=\tfrac12 + \tfrac38\times\frac43 \\&= \tfrac12+\tfrac12=1\end{align}$$ so over some infinite sums, this does work.

Thuong explained what the problem is:
The series {1, -1, 1, -1, 1, -1} does not converge on a single value.


Thuong also found the issues in proofs M1 and M2. Click to see proof M1: ${\mathbf{\infty = -1}}$.

Let $$x=1+2+4+8+\dots $$ Thus, $$1+2x = 1+2(1+2+4+\cdots) = 1+2+4+8+\cdots = x$$ Thus, $1+2x=x$. Rearranging this gives $x=-1$. However, $x$ is also obviously infinite. Thus, $\infty = -1$.

Thuong wrote:
If $x$ is infinite, then $2x$ is also infinite. If $2x$ is infinite, then $1+2x$ is infinite. Both sides are equal.

Click to see proof M2: Any two real numbers are the same.

Pick any three real numbers $a$, $b$ and $c$.
If $a^b = a^c$, then $b = c$.
Therefore, since $1^x = 1^y$, we may deduce $x = y$ for any two real numbers $x$ and $y$..

Thuong wrote: 
$a^b=a^c$ implies $b=c$ for all values of $a$ except $0$ and $1$.

Nishad from England and Thuong found the problem with proof M3. All numbers are equal. Click to see the proof.

Suppose that all numbers were not the same. Choose two numbers $a$ and $b$ which are not the same. Therefore one is bigger; we can suppose that $a> b$. Therefore, there is a positive number $c$ such that $a=b+c$. Therefore, multiplying sides by $(a-b)$ gives $$a(a-b) = (b+c)(a-b)$$ Expanding gives $$a^2-ab = ab-b^2+ac -bc $$ Rearranging gives $$a^2-ab-ac= ab-b^2-bc$$ Taking out a common factor gives $$ a(a-b-c) = b(a-b-c) $$ Dividing throughout by $(a-b-c)$ gives $a=b$, therefore $a$ and $b$ could not have been different after all, hence all numbers are equal.

Nishad wrote:
On the final step we have $a(a-b-c)=b(a-b-c)$ but recall that $a=b+c \Rightarrow a-b-c=0$ so in fact both the LHS (left hand side) and RHS are 0, giving us a division by 0 error.

Ruoyan corrected proof M4. All numbers are equal - version 2. Clich to see the proof.

Choose any two numbers $a$ and $b$, where $a \neq b$, and let $a+b=s$.

Thus, $(a+b)(a-b) = s(a-b)$

Thus, $a^2-b^2 = sa - sb$

Thus, $a^2 -sa = b^2-sb$

Thus, $a^2-sa+s^2/4 = b^2-sb+s^2/4$

Thus, $(a-s/2)^2 = (b-s/2)^2$

Thus, $a-s/2 = b-s/2$

Thus, $a=b$, and so all numbers are equal

Ruoyan wrote:
The problem of this proof appears here:
Thus, $(a−s/2)^2=(b−s/2)^2\hspace{10mm} (*)$
Thus, $a−s/2=b−s/2 \hspace{19mm} (**)$

In fact, we can get two possible answers from $(*)$
One is $(**)$, and the other one is $a−\dfrac s2=\dfrac s2-b\text{   }(***).$ We need to consider both cases. 
As we assumed at the beginning of the proof, $a≠b.$ Therefore $(**)$ shoud be ignored and we can find $(***)$ leads to $a+b=s$, which is right. So we should accept this one. But as we can see, this proof cannot tell us anything new. What we got is just the assumption we made (that $a+b=s$).

Thuong rewrote $a-s/2$ and $b-s/2$ in a way that makes Ruoyan's explanation even clearer:
$a-\dfrac s2 = \dfrac a2-\dfrac b2$ and $b-\dfrac s2=\dfrac b2-\dfrac a2.$ Because $a$ is not equal to $b$, either $\dfrac{a-b}2$ or $\dfrac{b-a}2$ is negative.


Syed from Wilson's School in the UK and Hans explained what's wrong with proof P1: $\mathbf{3 = 0}$. Click to see the proof again.

Consider the quadratic equation $ x^2+x+1=0 $. Then, we can see that $ x^2=-x-1 $. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give $$ x= -1-\frac{1}{x} $$ Substitute this back into the $x$ term in the middle of the original equation, so $$ x^2 +\left(-1-\frac{1}{x}\right)+1=0 $$ This reduces to $$x^2=\frac{1}{x}$$ So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives $$ 1^2+1+1=0 $$ Therefore, $3=0$. 

Click here to see Syed's explanation.

Clare found the trick in proof P2. The smallest positive number is 1. Click to see the proof.

Suppose that $x$ is the smallest positive number. Clearly $x\le 1$ and also $x^2\gt 0$. Since $x$ is the smallest positive number, $x^2$ can't be smaller then $x$, so we must have $x^2\geq x$. We can divide both sides of this by the positive number $x$ to get $x \geq 1$. Since $x$ is both less than or equal to $1$ and greater than or equal to $1$, $x$ must equal $1$. Thus the smallest positive number is $1$. 

Clare wrote:
There is no smallest positive number. If $0\lt x\lt1,$ then we know $0\lt x^2 \lt x\times1 = x.$ Therefore for any positive number $x$ below $1$ that you might think is the smallest positive number, there is always a smaller positive number $x^2.$ $\frac x2$ will also be a positve number less than $x.$ So we can never find the smallest positive number - it doesn't exist. 

This means that proof P2 begins with an invalid assumption, so the proof is invalid. In fact, you could consider P2 to be a proof by contradiction that there is no smallest positive number. It begins by assuming there is a smallest positive number, $x,$ and leads to nonsense.

Hans found the problem with P3: ${\mathbf{1=-1}}$. Click to see the proof.

Clearly, $-1=-1$ and $\frac{1}{1} = \frac{-1}{-1}$
Therefore, $-1\times \frac{1}{1}=-1\times \frac{-1}{-1}$
Therefore, $\frac{-1\times 1 }{1}=\frac{-1\times -1}{-1}$
Therefore, $\frac{-1}{1}=\frac{1}{-1}$
Therefore, $\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$
Therefore, $\frac{\sqrt{-1}}{1}=\frac{1}{\sqrt{-1}}$
Multipliying both sides by $\sqrt{-1}\times 1$ gives $\sqrt{-1}\times \sqrt{-1} = 1\times 1$
Therefore, $-1=1$

Hans wrote:
The error is in the step where $\sqrt{\frac1{-1}}$ is substituted by $\frac1{\sqrt{-1}}.$
Reason:  The equation $\sqrt{\frac xy} = \frac{\sqrt x}{\sqrt y}$ is not valid for $y$ smaller than $0$ and at the sme time $x$ larger than $0.$ 
In general for imaginary numbers the multiplication and division rules are no longer valid: $\sqrt x \times \sqrt y$ is not equal to $\sqrt{xy}$ if $x$ and $y$ are negative.