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Chapter 4: 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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Most popular questions from this chapter

L.ct \(F\) be a field. Show that the ideal \(\left(t_{1}, t_{2}\right)\) is not principal in the ring \(F\left[t_{1}, t_{2}\right]\). Similarly, show that \(\left(t_{1}, \ldots, t_{2}\right)\) is not principal in the ring \(F\left[t_{1+\cdots}, t_{n}\right]\)

Let \(f(X)=X^{3}+a_{1} X^{2}+a_{2} X+a_{3} .\) Show that the discriminant of \(f\) is $$ a_{1}^{2} a_{2}^{2}-4 a_{2}^{3}-4 a_{1}^{3} a_{3}-27 a_{3}^{2}+18 a_{1} a_{2} a_{3} . $$ [Reduce the question to the case of a polynomial \(Y^{3}+a Y+b\), and use the formula for this special case.]

Let \(R\) be a rational function over the field \(K\), and express \(R\) as a quotient of polynomials, \(R=g / f\). Define the derivative $$ R^{\prime}=\frac{f g^{\prime}-a f^{\prime}}{f^{2}} $$ where the prime means the formal derivative of polynomials as in the text. (a) Show that this derivative is independent of the expression of \(R\) as a quotient of polynomials, i.e. if \(R=g_{1} / f_{1}\) then $$ \frac{f g^{\prime}-a f^{\prime}}{f^{2}}=\frac{f_{1} g_{1}^{\prime}-\theta_{1} f_{1}}{f_{1}^{2}} $$ (b) Show that the derivative of rational functions satisfies the same rules as before, namely for rational functions \(R_{1}\) and \(R_{2}\) we have $$ \left(R_{1}+R_{2}\right)^{\prime}=R_{1}^{\prime}+R_{2}^{\prime} \quad \text { and } \quad\left(R_{1} R_{2}\right)^{\prime}=R_{1} R_{2}^{\prime}+R_{1}^{\prime} R_{2} $$ (c) Let \(x_{1} \ldots \ldots \alpha_{n}\) and \(a_{1} \ldots, a_{n}\) be elements of \(K\) such that $$ \frac{1}{\left(t-x_{1}\right) \cdots\left(t-x_{n}\right)}=\frac{a_{1}}{t-x_{1}}+\cdots+\frac{a_{n}}{t-x_{n}} $$ Let \(f(t)=\left(t-\alpha_{1}\right) \cdots\left(t-x_{n}\right)\) and assume that \(\alpha_{1}, \ldots, \alpha_{n}\) are distinct. Show that $$ a_{1}=\frac{1}{\left(x_{1}-\alpha_{2}\right) \cdots\left(\alpha_{1}-\alpha_{n}\right)}=\frac{1}{f^{\prime}\left(\alpha_{1}\right)} . $$

Let \(R=Z[t]\). Let \(p\) be a prime number. Show that \(t-p\) is a prime element in \(R\). Is \(t^{2}-p\) a prime element in \(R ?\) What about \(t^{3}-p ?\) What about \(t^{n}-p\) where \(n\) is a positive integer?

Let \(F\) be a field and let \(a: F[t] \rightarrow F[t]\) be an automorphism of the polynomial ring such that \(\sigma\) restricts to the identity on \(F\). Show that there exists elements \(a \in F, a \neq 0\), and \(b \in F\) such that \(\sigma t=a t+b\).
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