We want to prove that&nbsp;(Z7,⊕)&nbsp;is group. I have difficulty proving associativity axiom. The solution readsLet&nbsp;a∈Z7,b∈Z7&nbsp;and&nbsp;c∈Z7. By Theorem 3.4.10 we only need to show(a+(b+c))bmod7=((a+b)+c)bmod7.This holds since&nbsp;a+(b+c)=(a+b)+c&nbsp;for all integers a, b, and c by the associative property of the integers. Hence&nbsp;⊕&nbsp;is associative.Therorem 3.4.10. Let a and b be integers, and let m be a natural number. Then&nbsp;&nbsp;(a+b)modm=((amodm)+(bmodm))modm

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We want to prove that&amp;nbsp;(Z7,⊕)&amp;nbsp;is group. I have difficulty proving associativity axiom. The solution readsLet&amp;nbsp;a∈Z7,b∈Z7&amp;nbsp;and&amp;nbsp;c∈Z7. By Theorem 3.4.10 we only need to show(a+(b+c))bmod7=((a+b)+c)bmod7.This holds since&amp;nbsp;a+(b+c)=(a+b)+c&amp;nbsp;for all integers a, b, and c by the associative property of the integers. Hence&amp;nbsp;⊕&amp;nbsp;is associative.Therorem 3.4.10. Let a and b be integers, and let m be a natural number. Then&amp;nbsp;&amp;nbsp;(a+b)modm=((amodm)+(bmodm))modm

xmzdenisemst0 · Accepted Answer

The point of the theorem 3.4.10 is that you can take the mod additionally at any point in your calculation without changing the result. This should make it clear intuitively that the statement is true.If you want to be very formal, you can continue your calculationa⊕(b⊕c)=(a+(b+c)mod7)mod7&amp;nbsp;=(1)(amod7+((b+c)mod7)mod7)mod7&amp;nbsp;=(2)(amod7+(b+c)mod7)mod7&amp;nbsp;=(3)(a+(b+c))mod7,which is exactly the claim the solution makes (after you do the same exact calculation for&amp;nbsp;(a⊕b)⊕c). For (1) and (3), we used the theorem, for (2) notice that&amp;nbsp;mod7&amp;nbsp;is applied twice to&amp;nbsp;b+c&amp;nbsp;in the second line, which is unnecessary.

CyncgotoCancey1k6 · Accepted Answer

Step 1If what you are looking for is an intuitive approach, you can appreciate if you imagined a clock with only 7 numbers; let those numbers be 0,…,6. Adding 2 different hours h1,h2, no matter in which order, will always give you the same hour in the clock. This is what theorem 3.4.10 states.Step 2Now, from a formal definition, associativity of an operation ⋅ in a group is defined as a⋅(b⋅c)=(a⋅b)⋅c, and that is what &quot;suffices&quot; you to prove. However, if what you mean is that why does it also imply a+(b+c)mod7=a+b+cmod7 comes from extending your theorem 3.4.10 to that particular case.

We want to prove that (Z_7,\oplus) is group. I have difficulty proving associativity axiom. Th

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