State and prove chinese remainder theorem

I'm currently going through Harvard's Abstract Algebra lectures. I was doing one of the homework's and wanted to make sure that my thinking was correct. The problem states Using the Lemma that $m\mathbb$. This completes the proof. So, I feel like the general outline is there, but I don't know if every step can be properly justified. Any and all help would be greatly appreciated. Just follow the enouncement of the theorem. Given $n,m,a,b\in\mathbbk_in_i$$ and the term of right satisfices what we want. Then, taking $x=\sum a_ix_i$ we've prove the theorem. HINT : There is an easy way to prove the CRT by considering the map $\Bbb Z \rightarrow \Bbb Z_m \oplus \Bbb Z_n: x\mapsto (x \mod(m),x \mod(n)) $, and proving it is a surjection. Make use of the facts that this map is a ring homomorphism and that $\Bbb Z_ \cong \Bbb Z_m \oplus \Bbb Z_n$ (under the assumption that $(m,n) = 1$). We can make a proof of this theorem by "Analysis-Synthesis" : Analysis : We want to build $x \pmod$ otherwise we will obtain the wrong congruences. We have proof the existence of $x$. In general with this method in most case there is automatically uniqueness. Howewer we can check that if there exist two different solutions $x$ and $y$ which satisfy all the properties then they are the same.

\( \newcommand \ \ j\neq k,\] thus we see that \[x\equiv b_kN_ky_k(mod \ n_k).\] Also notice that \(N_ky_k\equiv 1(mod \ n_k)\). Hence \(x\) is a solution to the system of t congruences. We have to show now that any two solutions are congruent modulo \(N\). Suppose now that you have two solutions \(x_0,x_1\) to the system of congruences. Then \[x_0\equiv x_1(mod \ n_k)\] for all \(1\leq k\leq t\). Thus by Theorem 23, we see that \[x_0\equiv x_1(mod \ N).\] Thus the solution of the system is unique modulo \(N\). We now present an example that will show how the Chinese remainder theorem is used to determine the solution of a given system of congruences. Example \(\PageIndex \ \ y_3\equiv 1(mod \ 5).\] As a result, we get \[x\equiv 1.15.1+2.10.1+3.6.1\equiv 53\equiv 23 (mod \ 30).\] Exercises • Find an integer that leaves a remainder of 2 when divided by either 3 or 5, but that is divisible by 4. • Find all integers that leave a remainder of 4 when divided by 11 and leaves a remainder of 3 when divided by 17. • Find all integers that leave a remainder of 1 when divided by 2, a remainder of 2 when divided by 3 and a remainder of 3 when divided by 5.

My note states that we can prove Chinese remainder theorem as the way shown: Let $m$ and $n$ be coprime natural numbers. Then $C_$ if we know $(a\pmod 7,a\pmod 5)$ for positive integer $a$? Edit: I know the chinese remainder theorem as If one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1) Let $m$ and $n$ be coprime natural numbers. Then $C_$. Now using the CRT (as stated by me above) we get one element in $C_Q$ by the isomorphism, where $Q=q_1\cdot \dots\cdot q_k$. This element is precisely the remainder ( $n$ mod $Q$) $\begingroup$ Yeah I see it in another form in wikipedia and it states that “if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1)” . And I don’t understand how the isomorphism shows this statement. $\endgroup$ The thing to note is not just that there is a group isomorphism $f : C_$ to compute $a \mod nm$. $\begingroup$ To my last statement: Starting with an arbitrary element $km+r \in C_$. $\endgroup$

\( \newcommand\) • A key step in many proofs consists of showing that two possibly different values are in fact the same. The Pigeonhole principle can sometimes help with this. Theorem Suppose that \(n+1\) (or more) objects are put into \(n\) boxes. Then some box contains at least two objects. Proof Suppose each box contains at most one object. Then the total number of objects is at most \(1+1+\cdots+1=n\), a contradiction. This seemingly simple fact can be used in surprising ways. The key typically is to put objects into boxes according to some rule, so that when two objects end up in the same box it is because they have some desired relationship. Example \(\PageIndex\): Suppose 5 pairs of socks are in a drawer. Picking 6 socks guarantees that at least one pair is chosen. Label the boxes by "the pairs'' (e.g., the red pair, the blue pair, the argyle pair,…). Put the 6 socks into the boxes according to description. Some uses of the principle are not nearly so straightforward. Example \(\PageIndex+\cdots+a_j\), as desired. A similar argument provides a proof of the Chinese Remainder Theorem. Theorem If \(m\) and \(n\) are relatively prime, and \(0\le ai\), so \(r_i=r_j=r\). This means that \[a+im=q_1n+r\quad\hbox(j-i)\). This contradiction finishes the proof. More general versions of the Pigeonhole Principle can be proved by essentially the same method. A natural generalization would be something like this: If \(X\) objects are put into \(n\) boxes, some box contains at lea...