MHB Linear conqruence and relations problem

  • Thread starter Thread starter mehdi98
  • Start date Start date
  • Tags Tags
    Linear Relations
Click For Summary
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.
mehdi98
Messages
2
Reaction score
0
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)
 
Physics news on Phys.org
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
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

Similar threads

Replies
8
Views
3K
  • · Replies 28 ·
Replies
28
Views
6K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K