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Chapter 12: Problem 1

Show that \(t^{n}-1\) is divisible by \(t-1\).

Short Answer

Expert verified
Using mathematical induction, we have shown that the expression \(t^n - 1\) is divisible by \(t - 1\) for all positive integer values of n. We proved the base case when n = 1, assumed the statement to be true for n = k, and then showed it holds true for n = k + 1. Therefore, the divisibility of \(t^n - 1\) by \(t - 1\) is established for all positive integer values of n.

Step by step solution

01

Prove the Base Case (n = 1)

When n = 1, the given expression \(t^n - 1\) becomes \(t^1 - 1\), which is simply \(t - 1\). Since \(t - 1\) is clearly divisible by \(t - 1\), our base case holds.
02

Assume the Statement is True for n = k

Assume that the statement is true for n = k, which means that the expression \(t^k - 1\) is divisible by \(t - 1\). Mathematically, this can be represented as: \(t^k - 1 = (t-1)P\), where P is an integer.
03

Prove the Statement for n = k + 1

We now need to prove that the expression \(t^{k+1} - 1\) is also divisible by \(t - 1\). Let's first manipulate the expression: \(t^{k+1} - 1 = t^{k+1} - t^k + t^k - 1\) Now, we add and subtract \(t^k\) to the expression. Next, we factor the expression to make it easier to work with: \(t^{k+1} - 1 = t^k(t - 1) + (t^k - 1)\) We know from our assumption in Step 2 that \(t^k - 1 = (t-1)P\), so we can substitute this into the expression: \(t^{k+1} - 1 = t^k(t - 1) + (t-1)P\) Now, we can factor out a \(t - 1\) from the expression: \(t^{k+1} - 1 = (t-1)(t^k + P)\) Since \((t^k + P)\) must be an integer, we have shown that \(t^{k + 1} - 1\) is also divisible by \(t - 1\).
04

Conclusion

Through mathematical induction, we have shown that the expression \(t^n - 1\) is divisible by \(t - 1\) for all positive integer values of n.

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Key Concepts

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

Polynomial Division
When we talk about polynomial division, we are dealing with the process of dividing one polynomial by another. In this particular exercise, we want to show that \(t^n - 1\) is divisible by \(t - 1\). This means that when dividing \(t^n - 1\) by \(t - 1\), the remainder should be zero.
  • The divisor here is \(t - 1\).
  • The dividend is \(t^n - 1\).
  • If the division has no remainder, \(t - 1\) is a factor of \(t^n - 1\).
Polynomial division works similarly to numerical division. But instead of numbers, we divide terms and powers of the variable \(t\). When the remainder is zero, it confirms that the division is exact, confirming the relationship between the dividend and the divisor.
Factorization
Factorization in this context refers to the breakdown of the polynomial \(t^n - 1\). The idea is to express it in a way that makes division clearer. Factorization turns \(t^n - 1\) into a product of simpler expressions: we showed that \(t^{k+1} - 1 = (t-1)(t^k + P)\).
  • Breaking down the expression reveals hidden factors like \(t-1\).
  • The process simplifies the arithmetic operations.
  • This gives us a clearer path to prove divisibility.
By factorizing \(t^{k+1} - 1\), we can more easily manipulate the terms and ultimately highlight how \(t-1\) acts as a factor throughout the process.
Base Case
The base case is an essential first step in mathematical induction. Here, it's where we start proving that our statement is true. For this problem, the base case occurs when \(n = 1\). At this stage, we confirmed that \(t - 1\) divides itself neatly, which seems quite obvious.
  • A base case typically involves substituting the smallest number in the range.
  • In our scenario, \(n = 1\) transformed \(t^1 - 1\) into \(t - 1\).
  • Observing this case verifies the viability of the statement.
Establishing the base case is crucial as it sets the stage for proving the inductive step, laying a foundation for subsequent claims.
Inductive Step
The inductive step is where mathematical induction takes form. After proving the base case, we assume the statement is true for \(n = k\), called the "inductive hypothesis." This means \(t^k - 1\) is divisible by \(t - 1\).
  • The inductive step involves two smaller tasks:
  • Substitute \(n = k + 1\) into the polynomial to form \(t^{k+1} - 1\).
  • Manipulate the expression to show it remains divisible by \(t - 1\).
With our expression \(t^{k+1} - 1 = (t-1)(t^k + P)\), we've successfully shown maintaining this pattern confirms the truth for all \(n\). Therefore, the entire polynomial is confirmed to be divisible for all integer \(n\). This reinforces the power and beauty of the inductive step in proving general statements.

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Most popular questions from this chapter

Show that the polynomial \(t^{n}-1\) has no multiple roots in C. Can you determine all the roots and give its factorization into factors of degree 1 ?

Let \(f(t)\) be a polynomial with real coefficients. Let \(\alpha\) be a root of \(f\), which is complex but not real. Show that \(\bar{\alpha}\) is also a root of \(f\).

Let \(f(t)\) be an irredueible polynomial with leading coefficient 1 over the real numbers. Assume \(\operatorname{deg} f=2\). Show that \(f(t)\) can be written in the form $$ f(t)=(t-a)^{2}+b^{2} $$ with some \(a, b \in \mathbf{R}\) and \(b \neq 0 .\) Conversely, prove that any such polynomial is irreducible over \(\mathbf{R}\).

Let \(V\) be a finite dimensional vector space over the field \(K\), and let \(S\) be the set of all linear maps of \(V\) into itself. Show that \(V\) is a simple \(S\) -space.

Let \(V=\mathbf{R}^{2}\), let \(S\) consist of the matrix \(\left(\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right)\) viewed as linear map of \(V\) into itself. Here, \(a\) is a fixed non-zero real number. Determine all \(S\) -invariant subspaces of \(V\).
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