# User Contributed Dictionary

### Verb

figurate- Form of Second-person plural present tense, figurare
- Form of Second-person plural imperative, figurare#Italian|figurare

# Extensive Definition

A figurate number is a number that can be
represented as a regular and discrete geometric pattern (e.g. dots).
If the pattern is polytopic, the figurate is
labeled a polytopic number, and may be a polygonal
number or a polyhedral number.

The first few triangular
numbers can be built from rows of 1, 2, 3, 4, 5, and 6 items:
The n-th regular r-topic number is given by the formula:

P_r(n) = = \quad \mbox \ n \ge 1

Polytopic numbers for r = 2, 3, and 4 are:

- P2(n) = 1/2 n(n + 1) (triangular numbers)
- P3(n) = 1/6 n(n + 1)(n + 2) (tetrahedral numbers)
- P4(n) = 1/24 n(n + 1)(n + 2)(n + 3) (pentatopic numbers)

Our present terms square
number and cubic number
derive from their geometric representation as a square
or cube.

## Gnomon

Figurate numbers were a concern of Pythagorean geometry, since Pythagoras is credited with initiating them, and the notion that these numbers are generated from a gnomon or basic unit. The gnomon is the piece added to a figurate number to transform it to the next bigger one.For example, the gnomon of the square number is
the odd
number, of the general form 2n + 1, n = 0, 1, 2, 3, ... . The
square of size 8 composed of gnomons looks like this:

8 8 8 8
8 8 8 8 8 7
7 7 7 7 7
7 8 7 6 6 6
6 6 6 8 7 6
5 5 5 5 5 8
7 6 5 4 4
4 4 8 7 6 5
4 3 3 3 8 7
6 5 4 3 2
2 8 7 6 5 4
3 2 1

To transform from the n-square (the square of
size n) to the (n + 1)-square, one adjoins 2n + 1 elements: one to
the end of each row (n elements), one to the end of each column (n
elements), and a single one to the corner. For example, when
transforming the 7-square to the 8-square, we add 15 elements;
these adjunctions are the 8s in the above figure.

Note that this gnomonic technique also provides a
proof
that the sum of the first n odd numbers is n2; the figure
illustrates 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 82.

## Square roots

Conversely, one can calculate the square root of any number by subtracting odd numbers. Thus, 64 − 1 = 63; 63 − 3 = 60; 60 − 5 = 55; 55 − 7 = 48; 48 − 9 = 39; 39 − 11 = 28; 28 − 13 = 15; 15 − 15 = 0. The subtraction of the first 8 odd numbers from 64 yields 0; hence, the square-root of 64 is 8.The tedium of increasing number of subtractions
as the number grows is bypassed by a method similar to the standard
way of square-rooting taught in school. For example: 1225 = 35 ×
35, Note the sum of the digits of this square root: 3 + 5 = 8. This
square-root shortcut reduces 35 subtractions to only 8
subtractions. The shortcut involves two "tricks": a markoff trick,
and resumptive trick.

The markoff trick is already known from the
familiar square root algorithm. One marks off the target number in
pairs of digits, from the right, as in marking 1225 as 12′25; then,
calculation begins with the first digit-pair to the left. The
reason is that squaring a one-digit number results in a 1- or
2-digit square. Thus, 1, 2, 3 have, respectively, the 1-digit
squares of 1, 4, 9. But 4 has the 2-digit square of 16; and numbers
5, 6, 7, 8, 9 have 2-digit squares. To allow for this, one begins
with two digits to provide one digit at each process-stage.

The resumptive trick (unique to this present
algorithm) shifts from one pair of target number digits to its next
(rightward) two digits, explained in calculating the square root of
1225.

- Mark off 1225 as 12′25; begin calculation with left pair of digits, namely, 12.
- Begin subtracting odd numbers: 12 − 1 = 11; 11 − 3 = 8; 8 − 5 = 3; but the next odd number, 7, cannot be subtracted from difference 3, so the resumptive trick is needed.
- The left-most digit of the square root, 3, actually representing 30, because the second digit from the right in decimal numeration is the "tens digit".
- To difference 3 (= 8 − 5), adjoin next two marked off digits (25), obtaining 325, and resume odd number subtraction.
- The last "successful" subtrahend was 5; but the next odd number, 7, cannot be subtracted, so interpolate between 5 and 7 for number 6. (This is "first part" of the resumptive trick.)
- Since (noted above) the successful 3 subtractions actually represent the 2-digit 30, treat the interpolated 6 as 60; resume odd number subtraction with the first odd number in the sixties, namely, 61. (This is "second and final part" of the resumptive trick or subalgorithm.)
- Result: 325 − 61 = 264; 264 − 63 = 201; 201 − 65 = 136; 136 − 67 = 69; 69 − 69 = 0.
- Having passed from 325 to 0 by five subtractions, the second digit is 5: and 30 + 5 = 35, that is, the square root of 1225 is 35, obtained in exactly 3 + 5 = 8 subtractions by applying the markoff and resumptive tricks or subalgorithms.

To see again, consider 144 = 122. The square-root
is easily calculated by twelve subtractions: 144 − (1 + 3 + 5 + 7 +
9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) = 144 − 144 = 0. However, the
mark-off and resumptive tricks reduce this to 1 + 2 = 3 odd number
subtractions.

- Mark off 144 as 1′44.
- Starting with left-most pair, start subtraction 1 − 1 = 0; so leftmost digit of square-root is 1, representing 10.
- Bring down second pair of digits: 0 + 44 = 44 and begin subtracting odd numbers.
- Interpolate between "successful" 1, "failed" 3, namely, 2, to represent the twenties, whose first odd number is 21, resuming the subtraction with 21.
- 44 − 21 = 23; 23 − 23 = 0, resulting in two subtractions, so second digit is 2: 10 + 2 = 12 as square-root of 144, obtained by 1 + 2 = 3 subtracting of odd numbers.

A special case involves a "zero difference",
illustrated in 102 = 100. Marking off 100 as 1′00, then 1 − 1 = 0
(difference), which combines with the second pair of digits as 000,
But this in decimal notation is simply 0, resulting in 10 as square
root, achieved in 1 + 0 = 1 subtraction. Another example is 202 =
400. Marking off 400 as 4′00. then 4 − (1 + 3) = 0 (difference),
which combines with second pair of digits as 0000, representing 0,
yielding root 20, achieved in 2 + 0 = 2 subtractions, i.e., 4 − (1
+ 3).

## Cubes and cube roots

Cubes of natural numbers or positive integers can be generated from S = 1, 3, 5, 7, 9,..., 2n − 1, ...; n = 1, 2, 3, ..., by "moving sums", similar to the "moving averages" of statistics:- First member of S: 1 = 13.
- next two members of S: 3 + 5 = 8 = 23.
- Next three members of S: 7 + 9 + 11 = 27 = 33.
- Next four members of S: 13 + 15 + 17 + 19 = 64 = 43.
- Next five of S: 21 + 23 + 25 + 27 + 29 = 125 = 53.
- Next six of S: 31 + 33 + 35 + 37 + 39 + 41 = 216 = 63.
- Next seven of S: 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 = 73.

Thus, "moving differences" of S yield
cube-roots.

This procedure (taking many words to explain, but
quickly executed) is not restricted to calculating square roots of
natural numbers or positive integers. It can even be applied toward
calculating the irrational square root of 2, to any number of
decimal places.

## Demonstration of mathematical properties

School children construct figurate numbers from pebbles, bottle caps, etc. As a bonus, children can use figurate numbers to discover the commutative law and associative law for addition and multiplication — laws usually dictated to them — by building rows and tables of dots.For example, the additive commutativity of 2 + 3
= 3 + 2 = 5 becomes:

And the multiplicative commutativity of 2 × 3 = 3
× 2 = 6 becomes:

Besides the subtractive method, the additive
method can also approximate square roots of positive integers and
solve quadratic equations.

The concepts of figurate numbers and gnomon
implicitly anticipate the modern concept of recursion.

## See also

## References

- Gnomon: from pharaohs to fractals

## External links

figurate in German: Figurierte Zahl

figurate in Esperanto: Figuriga nombro

figurate in Finnish: Kuvioluku

figurate in French: Nombre figuré

figurate in Italian: Numero figurato

figurate in Russian: Фигурные числа

figurate in Vietnamese: Số hình học

figurate in Chinese: 有形數