This afternoon at Mathemati-Con, James Tanton proved to us 12 different ways that 1 = 2. Here are just a few of his arguments:

1. Since $i^2 = -1$,

\[ 1 = {i^2 + 3 \over 2} = {(\sqrt{-1})(\sqrt{-1}) + 3 \over 2} = {\sqrt{(-1)(-1)} + 3 \over 2} = 2 \]

2. Take a unit square. The diagonal has length $\sqrt{2}$ by the Pythagorean theorem. But we can also approximate the diagonal using a staircase. The lengths of the horizontal segments of the staircase must add to 1, as must the lengths of the vertical segments. So the staircase, no matter how closely it approximates the diagonal, always has length 2. As we make the approximation better, this means that $\sqrt{2} = 2$ and therefore $2 = 4$ implies $1 = 2$.

3. Write $ 0 = 0 + 0 + 0 + \dots$. Since $0 = 1 – 1$, we have

\[ 0 = (1-1) + (1-1) + (1-1) + \dots \]

But if we rearrange the parentheses, we get

\[ 0 = 1 + (-1+1) + (-1+1) + (-1 +1) + \dots = 1 + 0 + 0+ \dots = 1. \]

Thus $0 = 1$; adding 1 to both sides gives $1 = 2$.

See if you can spot the mistakes in each of the arguments. Tanton ended with the following irrefutable “proof by shopping”: At a store holding a 2-for-1 sale, he asked the sales associate how much it would cost to buy 1 item. She responded that it was the regular price–that is, the same price as buying 2 items. Hence it must follow that $1 = 2$.

I’m really glad that I decided to take notes of the Tanton presentation because I would have forgotten most of them by the time I got home. Tanton is just as exciting and dynamic live as he is in his videos, albeit he starts talking too fast when he gets going. I’m also really glad that he agreed to take a photo with us after his presentation.