On 10/25/09, ...but here's a question wrote:
> OK, so I define a rational number as a number that can be represented
> as a ratio of two integers.
>
> I also tell my kids that irrational numbers (as decimals) go on
> forever without repeating.
>
> That means any ratio of two integers, converted to decimal, must
> either terminate or repeat.
>
> How do you prove that there is no ratio of two numbers that goes on
> forever without repeating?

A fraction is a division problem. When you do a long division problem, in
each cycle there is either a remainder or not. If there is no remainder
then the decimal terminates.

But lets say there is never no remainder, i.e., the decimal never terminates.

If the denominator is n then there are at most n-1 possible remainders.
For example, if the denominator is 7 then the only possible remainders are
1, 2, 3, 4, 5, or 6. Once a remainder repeats a previous remainder the
decimal will repeat from that point forward.

So at the latest, when you get to the 7th cycle of the long division, you
will have to have a repeat of the remainder. And the decimal repeats.

This also proves a corollary, that the length of the repeating portion
cannot exceed the denominator minus 1.

Try it with 1/7. In the first cycle the division result is 1 and the
remainder is 3. In the next cycle the division result is 4 and the
remainder is 2. In the third cycle the division result is 2 and the
remainder is 6. In the fourth cycle the division result is 8 and the
remainder is 4. In the fifth cycle the division result is 5 and the
remainder is 5, In the sixth cycle the division result is 7 and the
remainder is 1. And that essentially repeats the one that you were
originally dividing into. So the pattern of remainders will repeat,
3,2,6,4,5,1 and as they do the division results: 1, 4, 2, 8, 5, 7 will
repeat as well.

I think this will be accessible to your brighter students.

This can actually also be used as on introduction to the pigeon-hole principle.

Share This Post | Report This Post

Next Post >>

Posts on this thread, including this one