- Ways to decompose natural numbers
- Decomposition as a product of prime factors
- Decomposition as a sum of powers of 2
- Additive decomposition
- Exercises and solutions
- Decomposition into a product of prime numbers
- Decomposition
- Additive decomposition
- References
The decomposition of natural numbers can be given in different ways: as a product of prime factors, as the sum of powers of two and additive decomposition. They will be explained in detail below.
A useful property of powers of two is that they can convert a number from the decimal system to a number from the binary system. For example, 7 (number in the decimal system) is equivalent to the number 111, since 7 = (2 ^ 2) + (2 ^ 1) + (2 ^ 0).
Natural numbers are used to count
The natural numbers are the numbers with which objects can be counted and enumerated. In most cases, natural numbers are considered to start from 1. These numbers are taught in school and are useful in almost all activities of daily life.
Ways to decompose natural numbers
As mentioned before, here are three different ways to decompose natural numbers.
Decomposition as a product of prime factors
Every natural number can be expressed as a product of prime numbers. If the number is already prime, its decomposition is itself multiplied by one.
If not, it is divided by the smallest prime number by which it is divisible (it can be one or several times), until obtaining a prime number.
For example:
5 = 5 * 1.
15 = 3 * 5.
28 = 2 * 2 * 7.
624 = 2 * 312 = 2 * 2 * 156 = 2 * 2 * 2 * 78 = 2 * 2 * 2 * 2 * 39 = 2 * 2 * 2 * 2 * 3 * 13.
175 = 5 * 35 = 5 * 5 * 7.
Decomposition as a sum of powers of 2
Another interesting property is that any natural number can be expressed as a sum of powers of 2. For example:
1 = 2 ^ 0.
2 = 2 ^ 1.
3 = 2 ^ 1 + 2 ^ 0.
4 = 2 ^ 2.
5 = 2 ^ 2 + 2 ^ 0.
6 = 2 ^ 2 + 2 ^ 1.
7 = 2 ^ 2 + 2 ^ 1 + 2 ^ 0.
8 = 2 ^ 3.
15 = 2 ^ 3 + 2 ^ 2 + 2 ^ 1 + 2 ^ 0.
Additive decomposition
Another way to decompose natural numbers is by considering their decimal numbering system and the place value of each digit.
This is obtained by considering the figures from right to left and starting with unit, ten, hundred, unit thousand, ten thousand, hundred thousand, unit million, etc. This unit is multiplied by the corresponding numbering system.
For example:
239 = 2 * 100 + 3 * 10 + 9 * 1 = 200 + 30 + 9.
4893 = 4 * 1000 + 8 * 100 + 9 * 10 + 3 * 1.
Exercises and solutions
Consider the number 865236. Find its decomposition into a product of prime numbers, sum of powers of 2, and its additive decomposition.
Decomposition into a product of prime numbers
-As 865236 is even, you can be sure that the smallest prime that it is divisible by is 2.
-Dividing by 2 you get: 865236 = 2 * 432618. Again you get an even number.
-It continues dividing until an odd number is obtained. Then: 865236 = 2 * 432618 = 2 * 2 * 216309.
-The last number is odd, but it is divisible by 3 since the sum of its digits is.
-So, 865236 = 2 * 432618 = 2 * 2 * 216309 = 2 * 2 * 3 * 72103. The number 72103 is a prime.
-Therefore the desired decomposition is the last one.
Decomposition
-The highest power of 2 that is closest to 865236 is sought.
-This is 2 ^ 19 = 524288. Now repeat the same for the difference 865236 - 524288 = 340948.
-The closest power in this case is 2 ^ 18 = 262144. Now we continue with 340948-262144 = 78804.
-In this case the closest power is 2 ^ 16 = 65536. Continue 78804 - 65536 = 13268 and we get that the closest power is 2 ^ 13 = 8192.
-Now with 13268 - 8192 = 5076 and you get 2 ^ 12 = 4096.
-Then with 5076 - 4096 = 980 and we have 2 ^ 9 = 512. We continue with 980 - 512 = 468, and the closest power is 2 ^ 8 = 256.
-Now comes 468 - 256 = 212 with 2 ^ 7 = 128.
-Then 212 - 128 = 84 with 2 ^ 6 = 64.
-Now 84 - 64 = 20 with 2 ^ 4 = 16.
-And finally 20 - 16 = 4 with 2 ^ 2 = 4.
Finally you have to:
865 236 = 2 ^ 19 + 2 ^ 18 + 2 ^ 16 + 2 ^ 13 + 2 ^ 12 + 2 ^ 9 + 2 ^ 8 + 2 ^ 7 + 2 ^ 6 + 2 ^ 4 + 2 ^ 2.
Additive decomposition
Identifying the units, we have that the unit corresponds to the number 6, the ten to 3, the hundred to 2, the unit from one thousand to 5, the ten from one thousand to 6 and the hundred from one thousand to 8.
Then, 865236 = 8 * 100,000 + 6 * 10,000 + 5 * 1,000 + 2 * 100 + 3 * 10 + 6
= 800,000 + 60,000 + 5,000 + 200 + 30 + 6.
References
- Barker, L. (2011). Leveled Texts for Mathematics: Number and Operations. Teacher Created Materials.
- Burton, M., French, C., & Jones, T. (2011). We Use Numbers. Benchmark Education Company.
- Doudna, K. (2010). No One Slumbers When We Use Numbers! ABDO Publishing Company.
- Fernández, JM (1996). Chemical Bond Approach project. Reverte.
- Hernández, J. d. (sf). Math notebook. Threshold.
- Lahora, MC (1992). Mathematical activities with children from 0 to 6 years old. Narcea Editions.
- Marín, E. (1991). Spanish grammar. Editorial Progreso.
- Tocci, RJ, & Widmer, NS (2003). Digital systems: principles and applications. Pearson Education.