MHB Linear conqruence and relations problem

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The relation R defined on the set Z, where aRb means a = ±b, is established as an equivalence relation because it is reflexive, symmetric, and transitive. For the linear congruence x ≡ 3 (mod 5), the solutions are expressed as 3 + 5k, where k is an integer. For the congruence 2x ≡ 5 (mod 9), the solution involves finding the multiplicative inverse of 2 modulo 9, leading to the expression for x in terms of this inverse. The discussion emphasizes the importance of understanding equivalence relations and the methods for solving linear congruences. Overall, the thread provides insights into both theoretical and practical aspects of these mathematical concepts.
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Suppose that the relation R is defined on the set Z where aRb means a = ±b. Establish whether R is an equivalence relation giving your justifications.

Find the set of solutions of each of the linear congruence:
a) x ≡ 3 (mod 5).
b) 2x ≡ 5 (mod 9).(please write the full solutions thanks)
 
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mehdi98 said:
Suppose that the relation R is defined on the set Z where aRb means a = ±b. Establish whether R is an equivalence relation giving your justifications.
R is an equivalence relation if R is reflexive, symmetric and transitive. The fact that R is symmetric means $\forall x,y\in\mathbb{Z}\,(x,y)\in R\implies (y,x)\in R$. For this definition of $R$ this means that whenever $x=\pm y$, we also have $y=\pm x$. Do you think this is true?

mehdi98 said:
a) x ≡ 3 (mod 5).
It's easy to see that 3, 8, 13, 18, ... give 3 as a remainder when divided by 5. Therefore, solutions are $3+5k$, $k\in\mathbb{Z}$.

mehdi98 said:
b) 2x ≡ 5 (mod 9).
We need to divide both sides by 2. Note that 2 has an inverse modulo 9, i.e., there exists a number $y$ such that $2y$ gives the remainder 1 when divided by 9. Then $2xy\equiv x(2y)\equiv x\equiv 5y\pmod{9}$.

For the future, please read the http://mathhelpboards.com/rules/, especially rules 8 and 11.
 
thank you:D and sorry
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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