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 $1 = 2$.

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