In a 2015 Rolling Stone interview, actor Terrence Howard introduced what he calls "Terryology", built around the claim that \(1\times1=2\). His reasoning, in his own words: "One times one equals two because the square root of four is two, so what's the square root of two? Should be one, but we're told it's two, and that cannot be." The claim resurfaced in a 2024 appearance on The Joe Rogan Experience, where he also attempted to debunk the Pythagorean theorem, drawing a public rebuttal from Neil deGrasse Tyson.
Buried in the internet folklore that grew up around these claims is a small, genuinely true observation about \(\sqrt2\) that gets dressed up as something mysterious. Stated plainly, the iteration
\[ x_0=\sqrt2, \qquad x_{n+1}=\frac{x_n^3}{2} \]
never leaves \(\sqrt2\): every \(x_n\) equals \(\sqrt2\) again. That is really all there is to it, no hidden number theory, no contradiction with anything, just one specific number that happens to reproduce itself under one specific cubic map. It is a fun fact, not a discovery.
Proof by Exponent Rules
Write \(\sqrt2=2^{1/2}\) and use ordinary exponent laws. If \(x_n=2^{1/2}\), then
\[ x_{n+1} = \frac{x_n^3}{2} = \frac{\left(2^{1/2}\right)^3}{2} = \frac{2^{3/2}}{2^1} = 2^{3/2-1} = 2^{1/2} = x_n. \]
Feeding \(x_n=\sqrt2\) into the map returns exactly \(x_n\) again, and the same computation applies once more to that output, so every subsequent term stays put: \(x_0=x_1=x_2=\dots=\sqrt2\).
Proof by Direct Computation
The exponent-law version hides how elementary this really is. Cubing \(\sqrt2\) just means multiplying it by itself three times,
\[ (\sqrt2)^3 = \sqrt2\cdot\sqrt2\cdot\sqrt2 = 2\cdot\sqrt2, \]
using \(\sqrt2\cdot\sqrt2=2\) for the first two factors. Dividing by \(2\) then cancels exactly the factor that cubing introduced,
\[ \frac{(\sqrt2)^3}{2} = \frac{2\sqrt2}{2} = \sqrt2. \]
Cubing \(\sqrt2\) multiplies it by an extra factor of \(2\) (since \((\sqrt2)^2=2\)), and dividing by \(2\) immediately removes that same factor again; the two steps are built to cancel for this one particular starting value.
Why Only \(\sqrt2\)?
The interesting question the folklore version skips entirely: for which \(x\) does cubing and halving return the same number? Setting up the fixed-point equation and solving it directly answers this:
\[ \frac{x^3}{2}=x \quad\Longleftrightarrow\quad x^3-2x=0 \quad\Longleftrightarrow\quad x(x^2-2)=0. \]
A product is zero exactly when one of its factors is, giving three solutions,
\[ x=0 \qquad\text{or}\qquad x=\pm\sqrt2. \]
So the loop is not unique to \(\sqrt2\): \(0\) trivially reproduces itself (\(0^3/2=0\)), and \(-\sqrt2\) works for the same reason \(\sqrt2\) does, cubing a negative number keeps it negative, \((-\sqrt2)^3=-2\sqrt2\), and halving gives back \(-\sqrt2\). These three numbers are the only fixed points of the map \(f(x)=x^3/2\); every other starting value moves under iteration instead of looping.
In fact, \(\sqrt2\) and \(-\sqrt2\) are repelling fixed points rather than attracting ones: since \(f'(x)=\tfrac32x^2\), the derivative at \(x=\pm\sqrt2\) is \(f'(\pm\sqrt2)=3\), whose absolute value exceeds \(1\). Starting even slightly away from \(\sqrt2\) and iterating \(x\mapsto x^3/2\) drives the sequence away from \(\sqrt2\) rather than toward it; the fixed point only "holds" for the exact value, and ordinary rounding error is enough to break the loop in practice.