Problem 41 - Discrete Mathematics with Applications

Question

Let \(a, b, c,\) and \(n\) be any positive integers and \(p\) be any prime. Prove each. \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\}=\operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\}\)

Step-by-Step Solution

Answer

Since the min function is associative and commutative, we can see that the LHS and RHS are equal: \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\} = p_1^{\min\\{m_1, n_1, r_1\\}} p_2^{\min\\{m_2, n_2, r_2\\}} \cdots p_k^{\min\\{m_k, n_k, r_k\\}} = \operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\}\) Thus, we have proven the given equation: \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\} = \operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\}\)

Step 1: Understanding the gcd

The greatest common divisor (gcd) is the largest positive integer that divides each of the numbers in the given set without a remainder. It is denoted by \(\operatorname{gcd}\\{a, b\\}\) for integers a and b.

Step 2: Breaking the gcd relations in prime factors

To break the given relations into prime factors, we will use the given information. We have three integers a, b, and c, and the prime p. Let each integer have their prime factor representation: \(a = p_1^{m_1} p_2^{m_2} \cdots p_k^{m_k}\) \(b = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}\) \(c = p_1^{r_1} p_2^{r_2} \cdots p_k^{r_k}\) where \(p_i\) represents the prime factors, and \(m_i, n_i, r_i\) are their respective exponents.

Step 3: Expressing gcd relations using prime factorizations

Using the prime factorizations, we can express the gcd of any two numbers as: \(\operatorname{gcd}\\{a, b\\} = p_1^{\min\\{m_1,n_1\\}} p_2^{\min\\{m_2,n_2\\}} \cdots p_k^{\min\\{m_k,n_k\\}}\) Similarly, for other combinations: \(\operatorname{gcd}\\{b, c\\} = p_1^{\min\\{n_1,r_1\\}} p_2^{\min\\{n_2,r_2\\}} \cdots p_k^{\min\\{n_k,r_k\\}}\) \(\operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\} = p_1^{\min\\{m_1, (\min\\{n_1,r_1\\})\\}} p_2^{\min\\{m_2, (\min\\{n_2,r_2\\})\\}} \cdots p_k^{\min\\{m_k, (\min\\{n_k,r_k\\})\\}}\)

Step 4: Finding LHS of the equation

Now, let's find the LHS: \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\} = p_1^{\min\\{(\min\\{m_1,n_1\\}), r_1\\}} p_2^{\min\\{(\min\\{m_2,n_2\\}), r_2\\}} \cdots p_k^{\min\\{(\min\\{m_k,n_k\\}), r_k\\}}\)

Step 5: Observing the min function

It can be observed that the min function in both LHS and RHS is associative and commutative, i.e., \(\min\\{a, \min\\{b, c\\}\\} = \min\\{ \min\\{a, b\\}, c\\}\) for any positive integers a, b and c.

Step 6: Equating LHS and RHS using properties of min function

Since the min function is associative and commutative, we can see that the LHS and RHS are equal: \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\} = p_1^{\min\\{m_1, n_1, r_1\\}} p_2^{\min\\{m_2, n_2, r_2\\}} \cdots p_k^{\min\\{m_k, n_k, r_k\\}} = \operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\}\) Thus, we have proven the given equation: \(\operatorname{gcd}\\{\operatorname{gcd}\\{a, b\\}, c\\} = \operatorname{gcd}\\{a, \operatorname{gcd}\\{b, c\\}\\}\)

Key Concepts

Prime Factorization

Prime factorization is a process in which a number is broken down into a product of prime numbers, where each prime number is called a factor. For instance, if we take the number 18, its prime factorization would be \(2 \times 3 \times 3\), or \(2 \cdot 3^2\). When we compare the prime factorizations of two or more numbers, we can determine the greatest common divisor (gcd) by taking the lowest power of common primes. This method simplifies finding the gcd and is fundamental to many areas of mathematics, including the problem solving approach shown in our given exercise.

To better grasp the concept, let's consider two numbers, like 28 and 35. Their prime factorizations are \(28 = 2^2 \cdot 7^1\) and \(35 = 7^1 \cdot 5^1\). The gcd is found by taking the lowest exponent of common primes, which here is just \(7^1\), making the gcd 7. Understanding how to break down numbers into their prime factors is crucial for recognizing patterns in numbers and solving complex mathematical problems.

Associative Property

The associative property is a principle that applies to many operations in mathematics, such as addition and multiplication. It states that the grouping of the numbers does not affect the result of the operation. In other words, when adding or multiplying several numbers, it doesn't matter how we pair them up. The sum or product will be the same.

For example, in addition, \((2 + 3) + 4 = 2 + (3 + 4)\), illustrating that the sum remains 9 regardless of how the numbers are grouped. Similarly, in multiplication, \((2 \times 3) \times 4 = 2 \times (3 \times 4)\), and the product is 24 in both cases. This property is instrumental in simplifying expressions and is a core principle when working with the gcd. When working with gcd, understanding the associative property allows us to rearrange and group the numbers in a way that makes computation more straightforward, just as we see in our exercise where the grouping of factors can be changed without affecting the final gcd.

Commutative Property

The commutative property, another fundamental mathematical principle, refers to the ability to change the order of the numbers involved in an operation without changing the result. This property applies to addition and multiplication but not to subtraction or division. It's easy to remember as the 'order doesn't matter' rule.

For instance, with addition, \(5 + 8\) yields the same result as \(8 + 5\), both equalling 13. For multiplication, \(7 \times 4\) is the same as \(4 \times 7\), with both expressions equaling 28. In the context of gcd, the commutative property means that \(\text{gcd}(a, b)\) is the same as \(\text{gcd}(b, a)\). This property can simplify the task of finding the greatest common divisor by allowing us to rearrange terms as needed for ease of calculation, which is particularly beneficial when working with several numbers, as demonstrated in the step-by-step solution to our exercise.