38Decimal Numbers

Solution of Problem 38.1.

1. $3$ is larger: padding, $3.000 > 2.999$. The difference is $3.000 - 2.999 = 0.001$, one thousandth.

2. Strictly between $7.4 = 7.40$ and $7.5 = 7.50$ lie

nine numbers, one per new graduation mark.

3. The same picture, ten times smaller: between $7.440$ and $7.450$ lie $7.441, 7.442, \dots, 7.449$ — nine numbers again.

4. Between $7.400$ and $7.500$, the numbers with three decimal digits are $7.401, 7.402, \dots, 7.499$: all the thousandths from $401$ to $499$, that is $99$ numbers.

5. The zoom recipe: write both numbers with the same number of decimal places, then add one more decimal place — the smaller number followed by a digit from $1$ to $9$ lands strictly between the two. Indeed $5.67 = 5.670$ and $5.68 = 5.680$, and

likewise $0.1999 = 0.19990 < 0.19995 < 0.20000 = 0.2$. Between two neighbouring graduations there is always a whole new level of nine finer graduations.

6. $3 < 3.05 < 3.1$; then $3 < 3.005 < 3.01$; then $3 < 3.0005 < 3.001$. Each candidate is beaten by one more zoom.

7. Whatever number Tom proposes — call it his candidate, strictly greater than $3$ — question 5 produces a number strictly between $3$ and the candidate. So the candidate was not the closest number to $3$: something even closer exists. Since this happens to every candidate without exception, no number can be “the number just after $3$”: it simply does not exist.

8. $2.9 < 2.95 < 3$, $2.99 < 2.995 < 3$, $2.999 < 2.9995 < 3$. And $3 - 2.999 = 0.001$. With twenty nines, the difference is a decimal point followed by nineteen zeros and a $1$ — one unit in the twentieth decimal place: tiny, but not zero. However many nines Tom writes, he never reaches $3$.

9. Whole numbers climb in steps of $1$: more than $7$ but less than $8$ would mean a whole number of units strictly between $7$ and $8$ units, and there is none. The zoom escapes this only by writing digits after the point — exactly what whole numbers do not have.

10. The smallest length rounding to $3.7$ is $3.65$ cm: its next digit is $5$, which rounds up (Definition 38.9). And $3.75$ rounds up to $3.8$ for the same reason. So “$3.7$ cm to the nearest tenth” means: at least $3.65$ cm, and strictly less than $3.75$ cm — a whole zoomed-in segment of possible true lengths hides behind one rounded value.

11. $0.0999 < 0.123 < 0.58 < 0.6$. One sentence: comparing digit by digit from the left, $0.0999$ has $0$ tenths while $0.6$ has $6$, and the comparison is settled there — the number of digits written says nothing about size (Method 38.3).

12. No: in cents, $4.99$ euros is $499$ cents and $5$ euros is $500$ cents — consecutive whole numbers, with nothing between them. Prices, having exactly two decimal places, are really whole numbers of cents in disguise: like the whole numbers of question 9, they do have next-door neighbours. The endless zoom needs the right to write ever more decimal places.

13. The candidates near $6$ are $5.82$ and $8.25$ (a number starting with $2$, or with $58$, $25$, $82$, $85$, is far from $6$). Distances: $6 - 5.82 = 0.18$ and $8.25 - 6 = 2.25$. The closest is $5.82$.

14. For instance $3.1415$, since $3.1410 < 3.1415 < 3.1420$; then $3.14159$, since $3.14150 < 3.14159 < 3.14160$. (Both are real steps in the actual hunt for $\pi = 3.14159\dots$)

15. Suppose someone names a number, as large as they like. One zoom turns every pair of neighbouring graduations between $0$ and $1$ into nine new numbers, and the zoom can be repeated forever — questions 2, 3 and 4 were only the first two levels. Repeating it enough times produces more decimal numbers between $0$ and $1$ than the named number. So no number is large enough to count them: there are infinitely many.