# Why does 0.999… = 1? (Part 2)

In this series, I discuss some ways of convincing students that $0.999\dots = 1$ and that, more generally, a real number may have more than one decimal representation even though a decimal representation corresponds to only one real number. This can be a major conceptual barrier for even bright students to overcome. I have met a few math majors within a semester of graduating — that is, they weren’t dummies — who could recite all of these ways and were perhaps logically convinced but remained psychologically unconvinced.

Methods #2 and #3 are indirect methods. We start with a decimal representation that we know and end with $0.999\dots$.

Method #2. This technique should be accessible to any student who can do long division. With long division, we know full well that

$\displaystyle \frac{1}{3} = 0.333\dots$

Multiply both sides by $3$:

$\displaystyle 3 \times \frac{1}{3} = 3 \times 0.333\dots$

$\displaystyle 1 = 0.999\dots$

Though not logically necessary, this method could be reinforced for students by also considering

$\displaystyle 1 = 9 \times \frac{1}{9} = 9 \times 0.111\dots = 0.999\dots$

Method #3. With long division, we know full well that

$\displaystyle \frac{1}{3} = 0.333\dots \quad$ and $~ \quad \displaystyle \frac{2}{3} = 0.666\dots$

Add them together:

$\displaystyle \frac{1}{3} + \frac{2}{3} = 0.333\dots + 0.666\dots$

$\displaystyle 1 = 0.999\dots$

Though not logically necessary, this method could be reinforced for students by also considering any (or all) of the following:

$1 = \displaystyle \frac{1}{9} + \frac{8}{9} = 0.111\dots + 0.888\dots = 0.999\dots$

$1 = \displaystyle \frac{2}{9} + \frac{7}{9} = 0.222\dots + 0.777\dots = 0.999\dots$

$1 = \displaystyle \frac{4}{9} + \frac{5}{9} = 0.444\dots + 0.555\dots = 0.999\dots$

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