Recurring Decimals at GCSE - A Flawed Proof

When questions on recurring decimals arise, they commonly test a student's ability to convert them to fractions, potentially using algebraic proof.

This often looks like the following.

Let \(x=0.8\dot{2}\) then \begin{align} x &= 0.8222\dots \\ 10x &= 8.2222\dots \\ 100x &= 82.2222\dots \end{align} Using the difference of the last two equations \begin{align} 90x &= 74 \\ x &= \frac{74}{90} \\ &= \frac{37}{45} \end{align}

While acceptable for GCSE standard, it's technically incorrect!

To highlight the flaw, using the same structure I can "prove" another interesting equivalence.

Let 5 be a recurring digit to the left of the decimal point (that is, 5 ones, 5 tens, 5 hundreds and so on) \(x=\dots5555.\) then \begin{align} x &= \dots5555 \\ 10x &= \dots5550 \end{align} Using the difference of these two equations \begin{align} -9x &= 5 \\ x &= -\frac{5}{9} \end{align} So \(\dots5555 = -0.555\dots\)

That's clearly false but both proofs followed the same logical steps, so there's something else causing a problem.

It's the original claim that \(x=\dots5555.\), which assumes that the number actually exists and can be written in decimal form. That caused the problem.

Similarly, the first proof starts with \(x=0.8\dot{2}\) which assumes we can write the number in that form. As it happens, we can, but a proof shouldn't hinge on an assumption.

What's the point in bringing this up? For GCSE, the proof is fine, but it's a nice discussion point for those doing A-Level Maths or Further Maths.

Mathematics is the underlying language supporting other subjects and it needs to be used with precision and careful thought, otherwise we can "prove" wildly inaccurate claims, or worse, incorrect yet believable claims. So how should we prove \(0.8\dot{2}=\frac{37}{45}\)?

Let's start by saying \(x\) is a series, with a common ratio \(\frac{1}{10}\) which proves it's convergent as \(\mid\frac{1}{10}\mid < 1\). \begin{align} x &= 0.6 + \sum_{i=1}^{\infty}{\frac{2}{10^i}} \\ &= 0.6 + 0.2 + 0.02 + 0.002 + \dots \\ &= 0.6 + 0.\dot{2} \\ &= 0.8\dot{2} \end{align}

We've shown that the series is equivalent to \(0.8\dot{2}\) but we can also evaluate the series with \(a = \frac{2}{10}\) and \(r=\frac{1}{10}\). \begin{align} x &= 0.6 + \frac{a}{1-r} \\ &= 0.6 + \frac{\frac{2}{10}}{1-\frac{1}{10}} \\ &= 0.6 + \frac{(\frac{2}{10})}{(\frac{9}{10})} \\ &= 0.6 + \frac{2}{9} \\ &= \frac{37}{45} \end{align}

And that's our proof! It's a subtle change, but we started from an expression (with no assumption about the number existing) and showed it can be written in two different forms. Those forms must be equivalent so \(0.8\dot{2}=\frac{37}{45}\).

Attempting this proof for \(x=\dots5555.\) results in a non-convergent series, meaning we never reach \(-\frac{5}{9}\).

I've found this to be a nice investigation problem for a year 13 Maths class to explore. The mathematical content isn't too complicated and it really shows the importance of precision in mathematical proofs.