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Chapter 9: Problem 11

Show that the relation \(R\) consisting of all pairs \((x, y)\) such that \(x\) and \(y\) are bit strings of length three or more that agree in their first three bits is an equivalence relation on the set of all bit strings of length three or more.

Short Answer

Expert verified
The relation is an equivalence relation because it is reflexive, symmetric, and transitive.

Step by step solution

01

Identify Reflexivity

To show that a relation is reflexive, we need to confirm that every element is related to itself. In this case, any bit string of length three or more will agree with itself in its first three bits.
02

Identify Symmetry

To prove symmetry, if \(x R y\), then \(y R x\) must also hold. If two bit strings x and y agree in their first three bits, then y and x also agree in their first three bits.
03

Identify Transitivity

To establish transitivity, we need to show that if \(x R y\) and \(y R z\), then \(x R z\) must hold. If x and y agree in their first three bits, and y and z also agree in their first three bits, then x and z have the same first three bits as well.
04

Conclusion

Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation on the set of all bit strings of length three or more.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reflexivity
Reflexivity in a relation means that every element is related to itself.
To understand why the relation consisting of pairs \(x, y\) where \(x\) and \(y\) are bit strings of length three or more is reflexive, let's break it down.
Take any bit string of length three or more. For reflexivity, we need to confirm that each bit string is related to itself.
Since a bit string will always agree with itself on its own bits, it will definitely agree with itself on its first three bits.
So yes, every bit string of length three or more is related to itself in this relation.
This shows that the relation is reflexive.
Reflexivity is usually straightforward to establish. If you ever get confused, just remember that you're checking if each element maps back to itself!
Symmetry
Symmetry in a relation requires that if one element is related to another, then the second is related back to the first.
For the relation \(R\) consisting of pairs \(x, y\) where \(x\) and \(y\) agree in their first three bits, we need to see if \(x R y\) implies \(y R x\).
If \(x R y\), it means the first three bits of \(x\) and \(y\) are the same.
By this logic, the first three bits of \(y\) and \(x\) must also be the same since bit comparison works both ways.
Therefore, if two bit strings agree in their first three bits, it holds true in both directions.
So, \(x R y\) implies \(y R x\), confirming the relation is symmetric.
Symmetry forms the mutual respect in mathematics: if I relate to you, you surely relate back to me.
Transitivity
Transitivity in a relation means that if an element \(x\) is related to an element \(y\), and \(y\) is related to \(z\), then \(x\) must also relate to \(z\).
For the relation where pairs \(x, y\) have matching first three bits, let鈥檚 see if \(x R y\) and \(y R z\) implies \(x R z\).
If \(x R y\), \(x\) and \(y\) agree on their first three bits.
If additionally \(y R z\), \(y\) and \(z\) also agree on their first three bits.
Thus, the first three bits of \(x\), \(y\), and \(z\) are all the same, ensuring \(x R z\).
This proves the relation is transitive.
Transitivity is like a logical chain in a relationship: if A connects to B and B to C, A connects to C without a doubt!

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Most popular questions from this chapter

Exercises \(34-38\) deal with these relations on the set of real numbers: \(\begin{aligned} R_{1}=&\left\\{(a, b) \in \mathbf{R}^{2} | a>b\right\\}, \text { the greater than relation, } \\ R_{2}=&\left\\{(a, b) \in \mathbf{R}^{2} | a \geq b\right\\}, \text { the greater than or equal to relation, } \end{aligned}\) \(\begin{aligned} R_{3}=\left\\{(a, b) \in \mathbf{R}^{2} | a < b\right\\}, \text { the less than relation, } \\ R_{4}= \left\\{(a, b) \in \mathbf{R}^{2} | a \leq b\right\\}, \text { the less than or equal to relation, } \end{aligned}\) \(R_{5}=\left\\{(a, b) \in \mathbf{R}^{2} | a=b\right\\},\) the equal to relation, \(R_{6}=\left\\{(a, b) \in \mathbf{R}^{2} | a \neq b\right\\},\) the unequal to relation. Find $$ \begin{array}{lll}{\text { a) } R_{2} \cup R_{4}} & {\text { b) } R_{3} \cup R_{6}} \\ {\text { c) } R_{3} \cap R_{6}} & {\text { d) } R_{4} \cap R_{6}} \\\ {\text { e) } R_{3}-R_{6}} & {\text { f) } R_{6}-R_{3}} \\ {\text { g) } R_{2} \oplus R_{6}} & {\text { h) } R_{3} \oplus R_{5}}\end{array} $$

Show that the relation \(R\) on the set of all bit strings such that \(s R t\) if and only if \(s\) and \(t\) contain the same number of 1 \(\mathrm{s}\) is an equivalence relation.

Let \(R_{1}=\\{(1,2),(2,3),(3,4)\\}\) and \(R_{2}=\\{(1,1),(1,2)\) \((2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4) \\}\) be relations from \(\\{1,2,3\\}\) to \(\\{1,2,3,4\\} .\) Find $$ \begin{array}{ll}{\text { a) } R_{1} \cup R_{2}} & {\text { b) } R_{1} \cap R_{2}} \\ {\text { c) } R_{1}-R_{2}} & {\text { d) } R_{2}-R_{1}}\end{array} $$

Suppose that the function \(f\) from \(A\) to \(B\) is a one-to-one correspondence. Let \(R\) be the relation that equals the graph of \(f .\) That is, \(R=\\{(a, f(a)) | a \in A\\} .\) What is the inverse relation \(R^{-1} ?\)

Show that the relation \(R=\emptyset\) on the empty set \(S=\emptyset\) is reflexive, symmetric, and transitive.
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