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Let $$X=\mid-5,-4,-3,-2,-1,0,1,2,3,4,5\\}$$ For \(x, y \in X,\) set \(x R y\) if \(x^{2}

Short Answer

Expert verified
Relation \(R\) is a partial ordering on \(X\). The diagram shows an upward chain from the smallest \(|x|\) value to the largest.

Step by step solution

01

Understand the Definitions

To prove that \(R\) is a partial order on \(X\), we need to check if \(R\) is reflexive, antisymmetric, and transitive. Reflexivity means that \(xRx\) for all \(x\) in \(X\). Antisymmetry means that if \(xRy\) and \(yRx\), then \(x = y\). Transitivity means that if \(xRy\) and \(yRz\), then \(xRz\).
02

Check Reflexivity

For reflexivity, we need to show that \(xRx\) for all \(x\) in \(X\). By the definition of \(R\), \(xRx\) if and only if \(x = y\). Since \(x = x\) for all \(x\), \(R\) is reflexive.
03

Check Antisymmetry

For antisymmetry, assume \(xRy\) and \(yRx\). From the definitions, we have \(x^2 < y^2\) or \(x = y\). Similarly, \(y^2 < x^2\) or \(y = x\). The only way both can be true without contradiction is \(x = y\). Thus, \(R\) is antisymmetric.
04

Check Transitivity

To prove transitivity, assume \(xRy\) and \(yRz\). We must show \(xRz\). If \(xRy\), then \(x^2 < y^2\) or \(x = y\). If \(yRz\), then \(y^2 < z^2\) or \(y = z\). From \(x^2 < y^2\) and \(y^2 < z^2\), it follows that \(x^2 < z^2\). Therefore, \(xRz\). This shows \(R\) is transitive.
05

Conclude Partial Ordering

Since \(R\) is reflexive, antisymmetric, and transitive, \(R\) is indeed a partial ordering on \(X\).
06

Draw the Hasse Diagram

To draw the Hasse diagram for \(R\), note that in a partial order relation like \(R\), each element of \(X\) is related to itself and potentially to elements with larger absolute values. Start by representing each element on the vertical axis, with smaller absolute value elements lower and larger ones higher. Connect elements where \(x^2 < y^2\). Since \(x = -x\) implies \(x^2 = (-x)^2\), only consider one of each such pair per square (i.e., connect \(|x|\) values without matching negative ones).

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Key Concepts

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

Reflexivity
Reflexivity is a fundamental property of a relation where every element is related to itself. For the relation \( R \) to be reflexive on set \( X \), each element \( x \) must satisfy \( x R x \). This means that for all \( x \) in \( X \), the condition \( x = x \) must hold true.

In the exercise, since \( x R y \) if \( x^2 < y^2 \) or \( x = y \), reflexivity is satisfied because \( x = x \) is always true by definition. Therefore, every element in the set \( X \) is related to itself, ensuring reflexivity. This property is crucial in establishing that \( R \) is a partial order, as it confirms that there are no elements without self-relation.
Antisymmetry
Antisymmetry in a relation \( R \) means that if \( x R y \) and \( y R x \), it must follow that \( x = y \). This property prevents two different elements from being in a bidirectional relationship under the same criteria.

In the exercise, if \( x R y \) implies \( x^2 < y^2 \) or \( x = y \), and \( y R x \) similarly implies \( y^2 < x^2 \) or \( y = x \), the only non-contradictory scenario would be \( x = y \). Thus, it's impossible for both relations \( x^2 < y^2 \) and \( y^2 < x^2 \) to hold simultaneously unless \( x = y \), proving the antisymmetric nature of \( R \).

This is essential because it ensures that the relation does not create cycles of non-identical elements related in both directions, which would violate partial ordering.
Transitivity
Transitivity in a relation \( R \) means that if \( x R y \) and \( y R z \), then it must follow that \( x R z \). It ensures a kind of logical continuity within the set.

For the given relation \( R \), if \( x R y \) implies \( x^2 < y^2 \) or \( x = y \) and \( y R z \) implies \( y^2 < z^2 \) or \( y = z \), then combining these gives \( x^2 < z^2 \) when both initial conditions hold, leading to \( x R z \). Therefore, the chain from \( x \) to \( y \) to \( z \) logically extends across \( R \), satisfying the transitive property.

This is a key trait of partial orders, providing a consistent and coherent structure where relationships can be inferred across longer sequences. Transitivity thus helps maintain order within the set, allowing for the construction of a coherent diagram such as a Hasse diagram, which represents the partial order visually.

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Most popular questions from this chapter

Find a set \(A\) with \(n\) elements and a relation \(R\) on \(A\) such that \(R^{1}, R^{2}, \ldots, R^{t}\) are all distinct.

Which of the following relations on the set of all people are reflexive? Symmetric? Antisymmetric? Transitive? Prove your assertions. (a) \(R(x, y)\) if y makes more money than \(x\). (b) \(R(x, y)\) if \(x\) and \(y\) are about the same height. (c) \(R(x, y)\) if \(x\) and \(y\) have an ancestor in common. (d) \(R(x, y)\) if \(x\) and \(y\) are the same sex. (e) \(R(x, y)\) if \(x\) and \(y\) both collect stamps. (f) \(R(x, y)\) if \(x\) and \(y\) like some of the same music.

For the set of all people, prove that the relation "weighs no more than" is not a partial order.

(a) For \(k, n_{1}, n_{2}, m_{1}, m_{2} \in \mathbb{N},\) show that if $$ n_{1} \equiv n_{2}(\bmod k) $$ and $$ m_{1}=m_{2}(\bmod k) $$ then $$ n_{1}+m_{1} \equiv n_{2}+m_{2}(\bmod k) $$ and $$ n_{1} \cdot m_{1} \equiv n_{2} \cdot m_{2}(\bmod k) $$ (b) Part (a) says that if we take two equivalence classes \([m]\) and \([n]\), then we can unambiguously define \([m]+[n]\) and \([m] \cdot[n]\). Pick any \(m_{1} \in[m]\) and any \(n_{1} \in[n],\) and define $$ [m]+[n]=\left[m_{1}+n_{1}\right] $$ and $$ [m] \cdot[n] \equiv\left[m_{1} \cdot n_{1}\right] $$ The definition is unambiguous since it doesn't matter which \(m_{1}\) and \(n_{1}\) we pick. Find the addition and multiplication tables for the equivalence classes of \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5) .\) (Hint: For both \(\equiv(\bmod 4)\) and \(\equiv(\bmod 5),\) your answer should include $$ [0]+[0] \equiv[0],[0]+[1] \equiv[1],[0] \cdot[0] \equiv\\{0] $$ and $$ [1] \cdot[1] \equiv[1] $$ but, for \(\equiv(\bmod 4)\). $$ [2]+[2] \equiv[0] $$ whereas, that will be false for \(\equiv(\bmod 5) .\)

Let \(A=(1,2,3,4\\} .\) Find the transitive closure of the relation \(R\) defined on \(A\) as $$R=\\{(1,2),(2,1),(2,3),(3,4)\\}$$

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