Question

If \(\mathrm{aN}=\\{\mathrm{ax} / \mathrm{x} \in \mathrm{N}\\}\) and \(\mathrm{bN} \cap \mathrm{cN}=\mathrm{d} \mathrm{N}\) Where \(\mathrm{b}, \mathrm{c} \in \mathrm{N}\) are relatively prime then (a) \(\mathrm{d}=\mathrm{bc}\) (b) \(c=b d\) (c) \(b=c d\) (d) \(a=b d\)

Short answer

In conclusion, based on the given conditions and our analysis, the correct relationship among a, b, c, and d is (a) d = bc.

Step-by-step solution

Find the intersection of bN and cN.

Since b and c are relatively prime, their lowest common multiple (LCM) is equal to their product bc. So, the first common multiple of b and c is bc. Any multiple n of bc will be a common multiple of both b and c, and we have the intersection of bN and cN which is the elements in dN. Therefore, we can write: Intersection of bN and cN (which is dN) = {bcm | m ∈ N}. Here, d = bc.

Compare the results with the options given

Now that we have determined that d = bc, we can compare it to the options given in the exercise. (a) d = bc: This matches our result and should be the correct option. (b) c = bd: If we compare it to our result d = bc, this is false. (c) b = cd: Similarly, this is also false, according to our result d = bc. (d) a = bd: This does not match our result and is false.

Make a conclusion.

Based on our analysis, we found that d = bc, which matches option (a). Therefore, the answer is (a) d = bc.

Key concepts

Intersection of Sets

When learning about sets and their properties in mathematics, one of the core concepts is the 'Intersection of Sets'. Simply put, the intersection of two sets is a new set containing all the elements that both original sets have in common. Think of it as the overlapping region when you draw two sets as circles in a Venn diagram.

For instance, if one set represents all students who play basketball and another set represents all students in a chess club, the intersection of these sets would consist of students who both play basketball and are in the chess club. Mathematically, if you have two sets, say Set A and Set B, their intersection is denoted by Set A ∩ Set B.

In the textbook exercise mentioned above, the intersection of two sets, bN and cN, where N denotes the set of natural numbers, is crucial to understanding the solution. Since b and c are relatively prime numbers, which we'll discuss more in a bit, any number that is a multiple of both b and c must be a multiple of their product, bc. This leads to the conclusion that the intersection of bN and cN is dN, where d is the product of b and c.

Relatively Prime Numbers

Relatively prime numbers, also known as coprime or mutually prime numbers, are a pair of numbers that have no other common divisors other than 1. In other words, their greatest common divisor (GCD) is 1. An easy way to remember this is if you can't evenly divide both numbers by the same number (other than 1), they're relatively prime.

For example, consider the numbers 8 and 15. They do not share any divisors other than 1, making them relatively prime to each other. However, 8 and 16 would not be considered relatively prime since they are both divisible by 8.

In the context of the problem from the textbook, knowing that b and c are relatively prime is a key part of the puzzle. It informs us that the least common multiple of b and c is simply their product. This property drives the entire solution process and ensures that when looking for the intersection of the sets bN and cN, we are essentially looking for multiples of their product, which is critical for finding the correct value of d.

Lowest Common Multiple (LCM)

Diving into another fundamental concept, the 'Lowest Common Multiple (LCM)' is the smallest number that is a multiple of two or more numbers. To find the LCM of two numbers, you're looking for the smallest number that both original numbers will divide into without leaving a remainder.

Let's consider an example. The multiples of 4 are 4, 8, 12, 16, and so on, while the multiples of 6 are 6, 12, 18, 24, etc. The LCM of 4 and 6 would be 12 since it's the smallest number that appears in both lists of multiples.

In the given exercise, the LCM of b and c becomes simple to compute once we know that b and c are relatively prime: it's just their product, bc. This is because there are no smaller common multiples due to the lack of any other common divisors between relatively prime numbers. Hence, the LCM concept ties in directly with the solution to the exercise, reinforcing that the intersection of the two sets, bN and cN, is represented by the set dN, where d is indeed defined to be the LCM of b and c, or simply bc.