11.2.3. Uniqueness Proof Part for the Chinese Remainder Theorem

Now, coming back to the uniqueness proof part for the Chinese remainder theorem in the last lecture, we showed that there exists at least one solution in the range 0 to M – 1. How do I prove that there is no other solution possible satisfying the same system of linear congruences and which is also in the range 0 to M - 1, we have to refute the existence of a second solution.

In this part of the proof, the lecturer emphasizes that we've already established the existence of at least one solution within the range from 0 to M – 1. The next step is to demonstrate that there cannot be another distinct solution in this range. If there were another solution, it would contradict the earlier finding. Essentially, the proof will eliminate the possibility of a second solution existing by refuting its existence.

Imagine trying to find your assigned seat in a classroom where no two students can sit in the same place. You have one spot, and if someone else claims to have a different seat, it creates confusion. The goal is to prove no other person could 'claim' that second seat by showing there’s only one valid position for each student.

So, again we will take the help of a helping lemma and this helping lemma will be useful later as well. So, what this helping lemma says is the following: imagine you are given n modulus which are pairwise relatively prime that means, you take any pair of modulus m and they are co-prime to each other. And suppose you know that you are given 2 numbers a and b which are congruent with respect to all the n individual modulus...

The helping lemma states that if we have multiple moduli that are pairwise co-prime, and two numbers a and b are congruent to each other under all these moduli, then a and b must also be congruent with respect to the product of these moduli, denoted as M. This lemma serves as a foundational principle essential for ensuring unique solutions in modular arithmetic.

Consider that each modulus represents a different time zone, and a and b are two appointments in those time zones. If both appointments occur at the same local time across every time zone, they effectively reflect the same rendezvous time, even when seen from the perspective of the larger global time.

Now, let us consider an arbitrary prime factor of the bigger modulo M and suppose it is appearing with power e in the prime power factorization that means so, M = 2e1, 3e2 and so on and some pe so on. So, I am taking an arbitrary prime which is appearing with some power in the prime power factorization of M...

Here, the proof examines how if a prime factor appears in the prime factorization of M, it must also appear in the prime factorization of the differences a-b. The proof follows through logical deductions using earlier established principles like Euclid’s lemma, asserting that if such a factor divides M, it must divide at least one of the smaller moduli. This connection ultimately confirms that the difference a - b must be divisible by M.

This can be likened to ensuring that each house (representing prime factors) on a block (representing M) has direct access to the main road. If all homes are linked well enough to the main street, it becomes evident that each house must share some common service line with the route connected to M. Just as vehicles (the factors) must also align with the road (M) to traverse to other areas, thus ensuring they connect back through one main route.

So, we will prove it as follows. On contrary imagine you have 2 solutions, 2 solutions in the range 0 to M - 1 satisfying this system of linear congruences and those solutions be let x and y. That means x satisfies this system of n linear congruences that when x is congruent to a modulo m...

In this step, we assume for contradiction that there are two different solutions, x and y. By showing that both these solutions lead to the same results across all linear congruences, we then derive that x must be equal to y. This contradiction indicates that our original assumptions (that there were two distinct solutions) must be false, leading us to conclude that there is a unique solution.

Picture a race where two runners mistakenly claim to finish at different times, yet their times recorded at each checkpoint (moduli) are identical. When reviewed, you realize both must have crossed the finish line at the same moment, proving only one valid finishing time exists, hence reaffirming that they cannot be two separate winners.

And since both x and y were in the range 0 to M - 1, that means they were strictly less than M, and both of them are congruent, then that is possible only when x = y that shows that there exists a unique solution modulo M satisfying your system of linear congruences.

In concluding the uniqueness proof, it is established that since both solutions x and y are confined within the bounds of 0 to M - 1 and are congruent, they must necessarily be equal. This final step reinforces that the earlier proofs and assumptions hold true, solidifying the assertion of a singular unique solution under the system of linear congruences.

Think of tracking attendance in a classroom—if two students claim they attended the same class and are accounted as present at every roll call, if they remain in the same class and timeframe, they are, without a doubt, the same individual rendering only one valid attendee for each session.

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