Does this ever end?

Try 10 ÷ 3 by long division. Regroup 1 ten as 10 ones: 10 ones ÷ 3 gives 3 ones with 1 one left. Regroup that as 10 tenths: 3 tenths, 1 tenth left. Regroup as 10 hundredths: 3 hundredths, 1 hundredth left.

Something is happening. Each time we divide, the remainder is 1 again, so the next step is identical to the last. The process will NEVER end: 10 ÷ 3 = 3.333…

So 10 ÷ 3 cannot be expressed using a finite number of digits in decimal form. Some divisions simply do not terminate, and you will study such numbers more closely in a later class. Try 10 ÷ 9 and 100 ÷ 11 and watch the same thing happen.

Now divide 1 by 7, and instead of watching the quotient, watch the REMAINDERS. They run 1, then 3, then 2, then 6, then 4, then 5 — and then 1 again, and the whole chain repeats.

Because the remainders repeat in a cycle, the digits of the quotient must repeat in the same cycle: 1 ÷ 7 = 0.142857 142857 14… The block 142857 goes round for ever. And now you can see WHY a division must either end or repeat: there are only finitely many possible remainders, so one must eventually come back.

That block 142857 has a startling property. Multiply it by 1, 2, 3, 4, 5 and 6 and look at the answers: you get the same digits back, cycled around. (Then multiply it by 7 and see what you get — the answer is worth the surprise.)

Are there other such numbers? Yes — work out 1 ÷ 17 as a decimal and use its repeating block. Are there infinitely many of these 'cyclic' numbers? In 1927 the Austrian mathematician Emil Artin CONJECTURED that there must be infinitely many. Nearly a century later, despite a great deal of research, that conjecture remains unsolved.