Discrete Math: Proving p(x)|(p1(x)-p2(x)) is Equiv. Rel.

In summary, we need to show that p1(x)≈p2(x) if and only if p(x)|(p1(x)-p2(x)) holds true, which means that p1(x) is equivalent to p2(x) if they share a common divisor p(x). Additionally, for p1(x) to be an equivalence relation, it must be symmetric, transitive, and reflexive, which means that p1(x) must be equivalent to itself, the sum of p1(x) and p2(x) must also be equivalent to p1(x), and if p1(x) is equivalent to p2(x) and p2(x) is equivalent to p3(x), then p1(x) must
  • #1
hammonjj
33
0

Homework Statement


Let p(x) be a polynomial in F[x].

Show that p1(x)≈p2(x) if and only if p(x)|(p1(x)-p2(x)) is an equivalence relation

The Attempt at a Solution


To be completely honest, I have no idea where to begin. This class has been a nightmare and this has been, by far, the worst professor I have ever had. No one in the class has any idea what is going on. I don't even really understand and to show the the second part of this is an equivalence relation.

Thanks in advance for the help. I won't be offended if you speak to me like I'm a small child as I am so lost in this class.
 
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  • #2
Well what are the three properties of an equivalence relation?
 
  • #3
tt2348 said:
Well what are the three properties of an equivalence relation?

In order for an equivalence relation to exist it must be symmetric, transitive and reflexive, but I don't know how to apply those.
 
  • #4
Start out with showing p1~p1... That is, p|(p1-p1)... p1~p2 => p|(p1-p2) => p|-(p2-p1) (assuming p=/=-1) => p|(p2-p1)
Also p1~p2 and p2~p3.. What would p1-p2+(p2-p3) look like?
 

Related to Discrete Math: Proving p(x)|(p1(x)-p2(x)) is Equiv. Rel.

1. What is discrete math?

Discrete math is a branch of mathematics that deals with discrete, separated, or distinct quantities. It involves the study of mathematical structures and objects that are countable and can be enumerated, such as graphs, integers, and sets.

2. What does it mean to prove p(x)|(p1(x)-p2(x)) is equiv. rel.?

Proving p(x)|(p1(x)-p2(x)) is equiv. rel. means to show that the expression p(x) divides the difference between p1(x) and p2(x) without leaving any remainder. In other words, p1(x) and p2(x) are equivalent modulo p(x).

3. Why is it important to prove that p(x)|(p1(x)-p2(x)) is equiv. rel.?

Proving that p(x)|(p1(x)-p2(x)) is equiv. rel. is important because it allows us to understand the relationship between different polynomials and how they can be divided without leaving any remainder. This is a fundamental concept in algebra and is used in various fields of mathematics and computer science.

4. What are some common techniques used to prove that p(x)|(p1(x)-p2(x)) is equiv. rel.?

Some common techniques used to prove p(x)|(p1(x)-p2(x)) is equiv. rel. include using the division algorithm, the Euclidean algorithm, and the Bezout's identity. These techniques involve breaking down the polynomials into smaller parts and showing that the remainder is equal to 0.

5. Can you provide an example of a proof for p(x)|(p1(x)-p2(x)) is equiv. rel.?

One example of a proof for p(x)|(p1(x)-p2(x)) is equiv. rel. is as follows: Let p(x) = x+2, p1(x) = x^2+3x+2, and p2(x) = x^2+2. We can rewrite p1(x)-p2(x) as (x^2+3x+2)-(x^2+2) = 3x. Since p(x) = x+2, we can rewrite 3x as 3(x+2). Therefore, p(x) divides p1(x)-p2(x) without leaving any remainder, and we can conclude that p(x)|(p1(x)-p2(x)) is equiv. rel.

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