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For the set of all people, prove that the relation "weighs less than" is not a linear order.

Short Answer

Expert verified
"Weighs less than" is not a linear order due to failing antisymmetry and being non-total.

Step by step solution

01

Understand the Properties of a Linear Order

A relation qualifies as a linear order if it is transitive, antisymmetric, and total. That means for any elements \(a\), \(b\), and \(c\):- **Transitive**: If \(aRb\) and \(bRc\), then \(aRc\).- **Antisymmetric**: If \(aRb\) and \(bRa\), then \(a = b\).- **Total**: For any \(a\) and \(b\), either \(aRb\) or \(bRa\) (or both if \(a = b\)).
02

Evaluate Transitivity and Antisymmetry

The relation "weighs less than" is transitive. For example, if person \(A\) weighs less than person \(B\), and person \(B\) weighs less than person \(C\), then person \(A\) weighs less than person \(C\). However, it is not antisymmetric because if person \(A\) weighs less than person \(B\), it cannot be true that \(B\) weighs less than \(A\) for distinct \(A\) and \(B\). Hence, antisymmetry is invalid here.
03

Assess the Total Property

A total relation must allow for any pair (\(a\), \(b\)) such that either \(a\) weighs less than \(b\), \(b\) weighs less than \(a\), or both \(a = b\). However, in the real world, two distinct people may weigh the same. Thus, "weighs less than" is not a total relation, as it cannot compare people of equal weight.
04

Conclusion

Since the relation "weighs less than" is not antisymmetric and not total, it cannot be a linear order. Even though it satisfies the transitive property, failing any of the required criteria disallows it from being a linear order.

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Key Concepts

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

Linear Order
In the world of discrete mathematics, a linear order is a type of binary relation that is crucially defined by several specific properties: transitivity, antisymmetry, and totality. Think of a linear order as a well-organized line-up or sequence, where each element is positioned according to defined rules. This concept helps in arranging or categorizing objects systematically.
  • To qualify as a linear order, a relation must meet all three properties: transitive, antisymmetric, and total.
  • If any one of these properties is not met, the relation cannot be considered a linear order.
Linear orders are fundamental in many areas of computer science and mathematics, such as sorting algorithms, where items need to be compared and arranged uniquely.
Transitive Property
The transitive property is among the core logical rules, indicating a kind of connection among elements. In mathematics, for a relation to be transitive, if an element \(a\) is related to \(b\) and \(b\) is related to \(c\), then \(a\) must be related to \(c\) as well.
  • Imagine a scenario involving weights: If person A weighs less than person B, and person B weighs less than person C, then it logically follows that person A weighs less than person C. This exemplifies the transitive property of the relation "weighs less than."
  • Transitivity is a critical feature in sorting algorithms where elements need to be arranged in a particular order based on relationships.
Without transitivity, the logical flow across relationships would be disrupted, making it difficult to form a coherent sequence or order.
Antisymmetric Property
Antisymmetry is an essential property that helps in distinguishing different elements within the context of a relation. A relation is antisymmetric when, for any two distinct elements \(a\) and \(b\), if \(a\) is related to \(b\) and \(b\) is related to \(a\), then \(a\) and \(b\) must be the same element.
  • Consider the relation "weighs less than" among people. If person A weighs less than person B, it's impossible that B weighs less than A unless A and B are actually the same person, which is why "weighs less than" doesn't satisfy antisymmetry.
  • This property distinguishes elements clearly, which is vital for correctly ordering or identifying items uniquely.
Antisymmetry establishes a clear hierarchy or sequence among elements, without ambiguity, preventing cyclical or contradictory comparisons.
Total Property
The total property ensures that every pair of elements in a set is comparable under a relation. For a relation to be total, given any two elements \(a\) and \(b\), you should be able to say (without exception) that either \(a\) is related to \(b\), \(b\) is related to \(a\), or both.
  • In the relation "weighs less than," consider two distinct individuals M and N. The total property fails here because M and N could weigh the same, making it impossible to establish either "M weighs less than N" or "N weighs less than M."
  • In contexts requiring decisions or rankings, a total property ensures a robust framework where comparisons between every possible pair of elements are definite.
The lack of the total property leads to incompleteness, as not all elements can be directly compared, which is crucial for creating fully ordered structures.

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Most popular questions from this chapter

Challenge: Find a partial ordering with exactly one minimal element but where that element is not a minimum element.

(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 \(\equiv(\bmod 5) .\)

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.

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}

Which of the following relations on the set of all people are reflexive? Symmetric? Antisymmetric? Transitive? Explain why your assertions are true. (a) \(R(x, y)\) if \(x\) and \(y\) either both like German food or both dislike German food. (b) \(R(x, y)\) if (i) \(x\) and \(y\) either both like Italian food or both dislike it, or (ii) \(x\) and \(y\) either both like Chinese food or both dislike it. (c) \(R(x, y)\) if \(y\) is at least two feet taller than \(x\).

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