Self-Similar Expressions: Practice

Let $x = 0.\overline{9}$. Then $10x = 9.\overline{9} = 9 + x$.

$9x = 9 \quad\Longrightarrow\quad x = 1$

The repeating decimal $0.\overline{9}$ is exactly $1$, not a hair less.

The algebra is correct, so the flaw is assuming a value exists. With $r = 3 \ge 1$ the terms grow.

$1, \ 4, \ 13, \ 40, \ \ldots \ \text{grow without bound}$

There is no number $S$, so $-\tfrac{1}{2}$ is meaningless.

Move the constant and linear terms to the other side.

$\varphi^2 = \varphi + 1$

So squaring the golden ratio is the same as adding $1$ to it.

Each new value is the square root of $2$ plus the previous value.

$\text{value} < 2 \ \Longrightarrow \ \sqrt{2 + \text{value}} < \sqrt{4} = 2$

Starting below $2$, the tower stays below $2$, so it rises toward a ceiling and closes in on it.

Self-similarity gives $x = 3 + \tfrac{1}{x}$, so $x^2 - 3x - 1 = 0$. The quadratic formula gives

$x = \frac{3 \pm \sqrt{13}}{2}.$

The fraction is positive, so $x = \tfrac{3 + \sqrt{13}}{2} \approx 3.30$.

The first three all settle on numbers ($2$, $\tfrac{3}{2}$, and $\tfrac{1}{3}$). The last has ratio $2$.

$1 + 2 + 4 + \cdots \ \Longrightarrow \ S = 1 + 2S \ \Longrightarrow \ S = -1$

A sum of positive numbers cannot be $-1$, so the move lied: the series has no value.

The innermost visible factor already gives $\sqrt{2} \approx 1.41$, and the tower only grows from there.

$x \ \ge \ \sqrt{2} \ > \ 1$

Since the value clearly exceeds $1$, the candidate $x = 0$ cannot be it, so $x = 2$.

The formula $\tfrac{1}{1-r}$ only applies when $|r| < 1$. Here $r = \tfrac{3}{2} > 1$, so the terms grow.

$1, \ \tfrac{5}{2}, \ \tfrac{19}{4}, \ \ldots \ \text{grow without bound}$

The series has no finite sum, so the $-2$ is a false answer from the same self-similarity trap.

Let $x = 0.\overline{27}$. The block is two digits, so $100x = 27 + x$, giving $99x = 27$.

$x = \frac{27}{99} = \frac{3}{11}$

Reducing $\tfrac{27}{99}$ by $9$ gives $\tfrac{3}{11}$.

Set $x = \sqrt{110 + x}$, then square and factor.

$x^2 - x - 110 = 0 \quad\Longrightarrow\quad (x - 11)(x + 10) = 0$

The value is the nonnegative candidate, $x = 11$.

Square both sides and rearrange into a quadratic.

$x^2 = 2 - x \quad\Longrightarrow\quad x^2 + x - 2 = 0 \quad\Longrightarrow\quad (x + 2)(x - 1) = 0$

The candidates are $-2$ and $1$; the value is nonnegative, so $x = 1$.

Every algebraic step is valid; only the premise fails.

$\text{write } S = \ \Longrightarrow \ \text{assumes a value exists}$

The series has none, so the flaw is committed before any algebra, at the moment of naming $S$.