Prove that if \(s \in[m]_{n}\) and \(t \in[k]_{n}\) then \(s \times t \in[m\) \(\times k] n\).

Short Answer

Expert verified
The product of two elements from equivalence classes modulo-n will also be an element of an equivalence class modulo-n. Specifically, if \(s \in[m]_{n}\) and \(t \in[k]_{n}\), then their product \(s \times t\) will belong to the equivalence class modulo-n represented by the product of the representatives of the original two classes, \(m \times k\).

Step by step solution

01

Introduction of arbitrary elements

Let's introduce two arbitrary elements from the equivalence classes \(s\) and \(t\). These can be represented as \(s = m + nx\) and \(t = k + ny\) where \(x\) and \(y\) are integers.
02

Multiply the elements

Now we multiply these elements together. \(s \times t = (m + nx) \times (k + ny) = mk + n(mx + ky + nxy)\). We can see the product can be written in the form \(a + nb\) where \(a = mk\) and \(b = mx + ky + nxy\), which are both integers since \(m\), \(k\), \(x\), and \(y\) are integers and \(n\) is a natural number.
03

Conclude the proof

Since the product \(s \times t = mk + n(mx + ky + nxy)\) is of the form \(a + nb\) where \(a\) and \(b\) are integers, this implies that \(s \times t\) is an element of the equivalence class modulo n, specifically the \(mk modulo-n\) class. Thus, it is proven that if \(s \in[m]_{n}\) and \(t \in[k]_{n}\) then \(s \times t \in[m \times k]_{n}\).

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