Chapter 15: Problem 7
Prove by induction that if \(r \geq 1\) and each \(t_{i},\) is a transposition, then \(\left(t_{1} \circ t_{2} \circ \cdots \circ t_{r}\right)^{-1}=t_{r} \circ \cdots \circ t_{2} \circ t_{1}\)
Short Answer
Expert verified
By induction, \((t_1 \circ t_2 \circ \cdots \circ t_r)^{-1} = t_r \circ \cdots \circ t_2 \circ t_1\) for \(r \geq 1\).
Step by step solution
01
Base Case
First, consider the case where \( r = 1 \). This means our sequence contains just one transposition, \( t_1 \). By definition of inverse, the inverse of a transposition is itself. Thus, \( (t_1)^{-1} = t_1 \), which means \( t_1^{-1} = t_1 \). This matches the expression \( t_1 \), so the statement holds for \( r = 1 \).
02
Inductive Hypothesis
Assume that for some integer \( k \geq 1 \), the statement holds true. That is, assume \( (t_1 \circ t_2 \circ \cdots \circ t_k)^{-1} = t_k \circ \cdots \circ t_2 \circ t_1 \). This is known as the inductive hypothesis.
03
Inductive Step
We want to prove the statement for \( r = k + 1 \). Consider the sequence with \( k+1 \) transpositions: \( t_1 \circ t_2 \circ \cdots \circ t_k \circ t_{k+1} \). The inverse is \[(t_1 \circ t_2 \circ \cdots \circ t_k \circ t_{k+1})^{-1} = t_{k+1}^{-1} \circ (t_1 \circ t_2 \circ \cdots \circ t_k)^{-1}\]Using the fact that the inverse of a transposition is itself, and by the inductive hypothesis, the expression becomes:\[t_{k+1} \circ (t_k \circ \cdots \circ t_2 \circ t_1)\]This expression evaluates to \( t_{k+1} \circ t_k \circ \cdots \circ t_2 \circ t_1 \), which shows that the statement holds for \( r = k + 1 \).
04
Conclusion of Induction
Since the base case holds and the truth of the statement for \( r = k \) implies the truth for \( r = k + 1 \), by the principle of mathematical induction, the statement is true for all \( r \geq 1 \). Therefore, \( (t_1 \circ t_2 \circ \cdots \circ t_r)^{-1} = t_r \circ \cdots \circ t_2 \circ t_1 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transpositions in Permutations
Transpositions are a fundamental concept in the study of permutations. A transposition is a permutation that swaps exactly two elements while leaving the rest unchanged. For example, the transposition \((1\ 3)\) swaps the positions of the numbers 1 and 3. In this case, 1 goes to the position of 3, and 3 goes to the position that 1 occupied, while all other positions remain fixed. Transpositions are particularly important because any permutation can be expressed as a combination of transpositions.
- The inverse of a transposition is the transposition itself. This is because swapping the same two numbers back reverses the swap.
- Mathematically, if \(t\) is a transposition, then \(t^{-1} = t\).
Permutations
In mathematics, a permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Permutations are used to describe the reordering of elements in a particular set. For example, if you have a set with three elements \( \{a, b, c\} \), some permutations of this set would be \((a, c, b)\) and \((b, a, c)\). Each arrangement is distinct from the others in terms of the sequence of items.
- A single permutation might consist of several transpositions, which makes understanding transpositions essential to understanding permutations.
- Permutations can be represented in cycle notation, which effectively expresses how elements in the positions are permuted or cycled.
Inverse Functions within Permutations
Inverse functions play a pivotal role in the realm of permutations, allowing us to 'reverse' a permutation to restore the original order of elements. The concept of inverses is straightforward: if \(f\) is a function that maps set \(A\) to set \(B\), then the inverse function \(f^{-1}\) maps \(B\) back to \(A\). For permutations, this concept translates to rearranging elements back to their initial positions.
- In permutations involving transpositions, since each transposition is its own inverse as discussed earlier, the inverse function is applied by reversing the sequence of transpositions.
- If you have a composite permutation consisting of multiple transpositions, the inverse permutation is created by applying the inverse of each transposition in the reverse order.