# Math Help - Equivalence classes

1. ## Equivalence classes

Given the following words {rot, tot, root, toot, roto, toto, too, to, otto}

Let f be the function that maps a word to its bag of letters. For the kernel relation of f, describe the equivalence classes.

I understand the first part of the question but not the second.

Next question

Let R be a relation on a set S such that R is symmetric and transitive and for each x c- S there is an element y c- S ( those are supposed to be the E rounded) such that x R y. Prove that R is an equivalence relation in other words prove that it is reflexive.

2. Originally Posted by cj's mom
Given the following words {rot, tot, root, toot, roto, toto, too, to, otto}

Let f be the function that maps a word to its bag of letters. For the kernel relation of f, describe the equivalence classes.

I understand the first part of the question but not the second.
I have seen a similar problem (on equivalence classes) posted by you, before. Unfortunately I fail to understand the language. Can you tell me the definition of "the kernel relation of a mapping"??

Let R be a relation on a set S such that R is symmetric and transitive and for each x c- S there is an element y c- S ( those are supposed to be the E rounded) such that x R y. Prove that R is an equivalence relation in other words prove that it is reflexive.
Pick any element $x \in S$, then by hypothesis $\exists y \in S$ such that $x \text{ R } y$. But since the relation is symmetric we get $y \text{ R } x$.

So $x \text{ R } y$ and $y \text{ R } x$.... what do you think transitivity will do?

3. Hello, cj's mom!

Let $R$ be a relation on a set $S$ such that $R$ is symmetric and transitive
and for each $x \in S$, there is an element $y \in S$ such that $x\:R\:y.$

Prove that $R$ is an equivalence relation.
Iin other words, prove that $R$ is reflexive.

We are told that: "every $x$ has a $y$": . $x \:R\:y$

We are told that $R$ is symmetric: .If $x\:R\:y$, then $y\:R\:x$

We are told that $R$ is transitive: .If $(x\:R\:y) \wedge (y\:R\:z)$, then $x\:R\:z.$

Hence, we have: . $x\:R\:y$ and $y \:R\:x$

By the transitive property: . $(x\:R\:y) \wedge (y\:R\:x) \;\Rightarrow\;x\:R\:x$

Therefore, $R$ is reflexive.

4. Originally Posted by cj's mom
Given the following words {rot, tot, root, toot, roto, toto, too, to, otto}

Let f be the function that maps a word to its bag of letters. For the kernel relation of f, describe the equivalence classes.
Do you mean that two words are equivalent if they are mapped to the same thing? That's not the way I would understand the word "kernel".
You also say "bag" rather than "set". That's more a computer term than math term- it's like a set but multiples of the same thing are allowed. In that case, "toot", "toto", and "otto" are mapped to the same thing and "roto" and "root" are mapped to the same thing but none of the others are mapped to the same thing. If that is what you mean then the equivalence classes are {rot}, {tot}, {root, roto}, {toot, toto, otto},{too}, {to}.

I understand the first part of the question but not the second.

Next question

Let R be a relation on a set S such that R is symmetric and transitive and for each x c- S there is an element y c- S ( those are supposed to be the E rounded) such that x R y. Prove that R is an equivalence relation in other words prove that it is reflexive.