11.2.2. Euclid’s Lemma

Now we prove another property which we often call as the Euclid’s Lemma, which is also very useful while proving the uniqueness part of the CRT theorem. So, the Euclid’s Lemma, is as follows, it says that if p is a prime number and if it is given that p divides the product of n numbers. Then definitely, it has to be the case that p divides at least one of those n numbers, it cannot be the case that p is a prime number and p does not divide a_1, a_2, ..., a_n but still it automatically divides the product of those n numbers.

Euclid’s Lemma states a fundamental property concerning prime numbers and their divisibility. If you have a prime number p and a set of n integers such that p divides the product of these integers, then at least one of those integers must be divisible by p. This lemma is crucial for understanding properties in number theory, especially when working with primes.

Imagine you have a box of apples that can be divided evenly among several friends. If you know that the total number of apples (the product) can be evenly divided by the number of friends (the prime number), it implies that at least one friend must receive an apple. In this analogy, the apples represent the integers, and the friends represent the prime number.

And there are again multiple ways to prove this. I will prove it using induction because it is convenient to prove it using induction, since it is a universally quantified statement for all. So, the base case will be for n = 1 and which is trivial, because if it is given to you that p divides a_1 that means p divides a_1.

The proof of Euclid's Lemma can be elegantly shown using mathematical induction. The base case is simple; if there is only one number (n=1), and p divides that number, then it follows directly. Now, assuming the hypothesis is true for k integers—that if p divides their product, at least one is divisible by p—we can show it's true for k + 1 integers by considering the GCD (greatest common divisor). If the GCD of p and the product of the first k integers is 1, it implies p must divide the last integer.

Think of a chain of friends sharing candies. If you know that each friend has at least one candy that can be divided evenly among them, you can prove that if one more friend joins the group, at least one friend must also have an extra candy by following the current distribution logic.

Now let us do the inductive step and take a new number which is the product of k + 1 numbers and imagine you have a prime number p which divides the product of those k + 1 number. My goal is to show that there is at least 1 number out of this k + 1 number which is completely divisible by p.

The inductive step involves showing that if p divides the product of k + 1 numbers, then it also divides at least one of those numbers. This involves examining the GCD of p with the product of the previous k numbers. If the GCD is 1, we can use the lemma established earlier to conclude that p must divide the last number in the product.

Consider a situation where a group of towns is sharing a number of logs, and each town has an equal chance of getting logs based on their population levels. If it turns out the total logs can be evenly divided, the inductive argument helps us ascertain that at least one town must have received logs. By testing each town’s GCD with the total logs, we can determine which town benefited.

And I have shown the existence of one such number this is when the GCD(p, A) = 1. Now, consider the case when the GCD(p, A) = p and then in that case, if the GCD(p, A) = p then p of course divides A.

Ultimately, you can conclude that whether the GCD is 1 or p itself, there is always some number in the set that must be divisible by p. This is crucial for establishing the foundations of number theory and is often used in further mathematical proofs and problems.

Returning to our candy example, if we know some friends share candies in such a way that maintains a ratio based on their personal contributions, we can always determine who received at least one candy. Even if the distribution changes (the GCD being p instead of 1), it remains clear that someone received an even distribution derived directly from the total pool.

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