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Assume \( {P Q}\) equals \( {R S} .\) Does it follow that \( {P R}\) is equal to \( {Q S}\) ? Prove your conclusion.

Short Answer

Expert verified
Answer: No, \(PQ = RS\) does not imply that \(PR = QS\). We have found a counterexample where \(PQ = RS\) but \(PR \neq QS\), proving the statement false.

Step by step solution

01

(Step 1: Consider Counterexamples)

To prove the statement false, we can find a specific case where \(PQ = RS\) but \(PR \neq QS\). Let's experiment with numerical values: - Let \(P = 1\) - Let \(Q = 5\) - Let \(R = 3\) - Let \(S = 7\) Now, we calculate the given conditions and see if they hold: - \(PQ = 1 \cdot 5 = 5\) - \(RS = 3 \cdot 7 = 3 \cdot 7 = 21\) And then calculate the required conditions: - \(PR = 1 \cdot 3 = 3\) - \(QS = 5 \cdot 7 = 35\) In this case, \(PQ = RS = 5 \neq 21\), but \(PR = 3 \neq 35 = QS\). Therefore, the given statement is false, and we have found a counterexample to show that \(PQ = RS\) does not imply \(PR = QS\).
02

(Final Conclusion)

Since we have found a counterexample where \(PQ = RS\) but \(PR \neq QS\), it proves that \(PQ = RS\) does not imply \(PR = QS\).

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Key Concepts

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

Counterexample
A counterexample is a specific instance or case that shows a statement to be false. In mathematical proofs, they are an effective method for disproving universal statements, which claim that something is true for all cases. By finding just one instance where the statement does not hold, the entire statement is proven incorrect.

When faced with the claim that if lines segments PQ and RS are equal then PR must equal QS, the best approach to disprove it may be to find a counterexample. As seen in the original exercise, assigning numerical values to the points and testing the given conditions delivers a clear result. Despite PQ and RS being equal in length (which they are not in the provided example), the lengths of PR and QS are different. This single instance successfully refutes the general statement and demonstrates the power of counterexamples in logical deduction.
Geometric Properties
The realm of geometry is rich with various properties that define the relationships between different shapes, sizes, angles, and distances of points, lines, and figures. It is crucial to understand these properties when solving geometric problems or proving certain geometric claims. For instance, the congruence and equality of line segments, which are at the heart of the textbook problem, are foundational concepts in geometry.

However, it is also necessary to recognize that equal lengths between some line segments do not automatically determine the lengths of others. It relies on the arrangement and properties of the specific geometric figures involved. The misuse of geometric properties can lead to incorrect conclusions, as shown in the misunderstanding that equal segments (PQ and RS) would imply other segments (PR and QS) are also equal.
Logical Reasoning
Logical reasoning is the process of using a structured, methodical approach to deduce conclusions from premises or facts. It involves the principles of logic and critical thinking to solve problems or prove statements. Within mathematics, especially in proofs and problem-solving, logical reasoning is applied to reach a conclusion whether a statement is true or false.

In the context of the original exercise, logical reasoning helps discern that congruence in one pair of line segments does not infer congruence in another pair without further geometric relationships or properties. The error in the initial claim arises from an assumption not supported by the strict logic of geometric principles. It is vital for students to practice logical reasoning in order to not only follow through the steps of a proof but also to understand the implications of assertions within a mathematical framework.

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Most popular questions from this chapter

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In contrast to the proof in Exercise \(83,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane, and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\Delta P Q R\) a. Let \(M_{1}\) be the midpoint of the side \(P Q .\) Find the coordinates of \(M_{1}\) and the components of the vector \(R M_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\hat{R} \vec{M}_{1}\) c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(P M_{2}\) to obtain the vector \(\overline{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(Q M_{3}\) to obtain the vector \(\overline{O Z}_{3}\) e. Conclude that the medians of \(\Delta P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\Delta P Q R\) intersect.

Intersecting planes Find an equation of the line of intersection of the planes \(Q\) and \(R\) $$Q: x+2 y-z=1 ; R: x+y+z=1$$

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