## Four fun fallacies

I came across the following while browsing a collection of maths “jokes”. I’m not sure how they got into the list, but they’re an excellent way to test how well you can spot the holes in an apparently plausible mathematical argument…

Theorem 1: $n=n+1$.

$(n+1)^2 = n^2 + 2n + 1 \iff (n+1)^2 -(2n+1) = n^2.$

Subtracting $n(2n+1)$ from each side and factorising on the LHS, we have

$(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1).$

Adding $\frac{1}{4}(2n+1)^2$ to each side now yields

$(n+1)^2 - (n+1)(2n+1) + \displaystyle\frac{1}{4}(2n+1)^2 = n^2 - n(2n+1) + \displaystyle\frac{1}{4}(2n+1)^2.$

This may be written as

$\left[ (n+1) - \displaystyle\frac{1}{2}(2n+1) \right]^2 = \left[ n - \displaystyle\frac{1}{2}(2n+1) \right]^2.$

Finally, we take the square roots of both sides, obtaining

$(n+1) - \displaystyle\frac{1}{2}(2n+1) = n - \displaystyle\frac{1}{2}(2n+1),$

and cancelling the terms $\frac{1}{2}(2n+1)$ we are left with $n+1 = n$ as required. QED.

Remark. This argument can be generalised to show that $a = b$ for any numbers $a, b$.

Theorem 2: £1.00 = 1p. (A very useful theorem in modern macroeconomics!)

Proof. This is purely arithmetical. We have $\pounds 1 = 100\mathrm{p} = (10\mathrm{p})^2 = (\pounds 0.1)^2 = \pounds 0.01= 1\mathrm{p}$. QED.

Theorem 3: $\log(-1) = 0$.

Proof: By one of the standard log laws, we have

$\log\left[(-1)^2\right] = 2\log(-1).$

On the other hand, we also know that

$\log\left[(-1)^2\right] = \log(1) = 0.$

Combining these two results gives $\log\left[(-1)^2\right] = 2\log(-1) = 0$ and hence $\log(-1) = 0$ as required. QED.

Theorem 4. All positive integers are equal.

Proof. It is sufficient to show that for any two positive integers, $A$ and $B$, $A = B$. Further, it is sufficient to show that for all $N > 0$, if $A$ and $B$ (both positive integers) satisfy $\max(A, B) = N$ then $A = B$.

We proceed by induction.

(i) If $N = 1$, then $A$ and $B$, being positive integers, must each be 1. So $A = B$.

(ii) Assume that the result holds for some value $N = k$. Now take $A$ and $B$ with $\max(A, B) = k+1$. Then $\max((A-1), (B-1)) = k$, and hence $(A-1) = (B-1)$. Consequently, $A = B$.

(iii) We have shown that the result holds for $N = 1$ and that if it holds for $N = k$ then it must also hold for $N = k+1$. Hence, by induction, the theorem is true for all $n \in \mathbb{N}$. QED.