Linear conqruence and relations problem

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The relation R defined on the set Z, where aRb means a = ±b, is confirmed as an equivalence relation because it satisfies reflexivity, symmetry, and transitivity. The solutions to the linear congruences are derived as follows: for a) x ≡ 3 (mod 5), the complete solution set is x = 3 + 5k, where k ∈ Z. For b) 2x ≡ 5 (mod 9), the solution involves finding the multiplicative inverse of 2 modulo 9, leading to the solution x ≡ 5y (mod 9) after determining y.

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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
 

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