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## The Mundane Becomes Repeatingly Unmundane

Sometimes interesting mathematical problems can be found in the mundane. For example, the 1987 edition of McGraw-Hill Mathematics has this "Think & Try" problem set...

Evaluate each expression. Express the answer as a repeating decimal:
1. 0.099999... - 0.09090909... =
2. 0.099999... + 0.09090909... =
3. 0.999999... x 0.9 =
4. 0.999999... ÷ 0.9 =

Some concerns or observations. First, when working the problems, do not resort to converting to common fraction format before doing the computations. Second, remember that 0.99999... = 1, so that in #3 above, the product should end up being 0.9, though you may get a repeating decimal as an answer.

But now, push a little further. How would respond to these three revised problems:

1. 0.999999... + 0.999999... =
2. 0.999999... x 0.999999... =
3. 0.999999... ÷ 0.999999... =
That is, even knowing the answers are 2, 1, and 1 respectively, can you figure out how to perform the well-known algorithms to get those answers?

And, once you figure that out, try these to test yourself further:

1. 0.999999... + 0.888888... =
2. 0.999999... x 0.888888... =
3. 0.999999... ÷ 0.888888... =
Or even these:
1. 0.909009000900009000009... + 0.9008000090000008000000009... =
2. 0.909009000900009000009... - 0.9008000090000008000000009... =
3. 0.909009000900009000009... x 0.9008000090000008000000009... =
4. 0.909009000900009000009... ÷ 0.9008000090000008000000009... =
You can now devise your own extensions to explore....

Hint: Set up the numbers to perform the algorithm the way you normally do it, then jump in and figure out what to do...looking for patterns.

Solution Commentary: The first three sets of problems can be checked by converting the decimals (repeating and nonrepeating) to their equivalent fractions and then perform the given operation.

However, the situation is different for the last set of four problems, as they involve arithmetic with numbers that are irrational...but they do exhibit patterns. For example, the following set may be a good starting set (to build a bit of confidence):

1. 0.909009000900009000009... + 0.08008000900008000008... =
2. 0.909009000900009000009... - 0.08008000900008000008... =
3. 0.909009000900009000009... x 0.08008000900008000008... =
4. 0.909009000900009000009... ÷ 0.08008000900008000008... =