Higher only · recurring decimals

Recurring decimal
to a fraction.
Subtract it away.

x, 10x, subtract

the trick behind the method —

Line up the tails

multiply so the repeating parts match

then subtract and they disappear

step 1 of 4

One repeating digit

1. One digit repeats, so multiply by 10.

2. 10x − x: the decimals cancel exactly.

✓ 0.4 = 49

x = 0.4444…

10x = 4.4444… × 10

9x = 4 subtract

x = 49

step 2 of 4

Two repeating digits

1. Two digits repeat, so multiply by 100.

2. 100x − x = 99x.

✓ Simplify at the end

x = 0.181818…

100x = 18.181818… × 100

99x = 18

x = 1899 = 211

step 3 of 4

A non-repeating digit

1. Get two lines with the same tail.

2. 10x and 100x both end .666…

✓ Subtract those two instead

x = 0.1666…

10x = 1.666… tails now match

100x = 16.666…

90x = 15 100x − 10x

x = 1590 = 16

step 4 of 4

Which multiplier?

1. Count the digits in the repeating block.

2. That many zeros in the multiplier.

✓ The tails must line up exactly

1 repeating digit: × 10

2 repeating digits: × 100

3 repeating digits: × 1000

recurring to fraction —

Let x = …

write out the decimal

Multiply

so the tails match

Subtract

solve, then simplify

Asked to “prove” it?
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