Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find all \(a\) in \(\mathbb{Z}_{10}\) such that: a. \(a^{2}=a\) b. \(a\) has an inverse (and find the inverse). c. \(a^{k}=0\) for some \(k \geq 1\). d. \(a=2^{k}\) for some \(k \geq 1\). e. \(a=b^{2}\) for some \(b\) in \(\mathbb{Z}_{10}\).

Short Answer

Expert verified
a = 0, 1 for (a); a = 1, 3, 7, 9 for (b); a = 0 for (c); a = 2, 4, 6, 8 for (d); a = 0, 1, 4, 5, 6, 9 for (e).

Step by step solution

01

Find solutions for a² = a

In modulo 10, solve the equation \(a^2 = a\). This simplifies to \(a(a - 1) = 0\). Since we are working modulo 10, \(a\) must be such that \(a(a-1)\equiv0 \pmod{10}\). Hence, the possible solutions are \(a = 0\) or \(a = 1\).
02

Find elements with inverses

In order for an integer \(a\) in \(\mathbb{Z}_{10}\) to have a multiplicative inverse, \(a\) must be coprime to 10. Check each element 0 through 9 to determine which are coprime: 1, 3, 7, and 9. Their respective inverses can be found using the extended Euclidean algorithm: \(1^{-1} = 1\), \(3^{-1} = 7\), \(7^{-1} = 3\), \(9^{-1} = 9\).
03

Find solutions for a^k = 0

Find all \(a\) such that \(a^k \equiv 0 \pmod{10}\) for some \(k \geq 1\). Only \(a = 0\) satisfies \(a^k \equiv 0\) as \(0^k = 0\) for any \(k \geq 1\).
04

Find solutions for a = 2^k

For \(a = 2^k\) in \(\mathbb{Z}_{10}\), compute powers of 2: \(2^1 \equiv 2\), \(2^2 \equiv 4\), \(2^3 \equiv 8\), \(2^4 \equiv 6\), \(2^5 \equiv 2\) (and it cycles through now). Thus, possible values of \(a\) are 2, 4, 6, and 8.
05

Find solutions for a = b^2

Find \(b\) such that \(b^2 \equiv a \pmod{10}\). The squares are: \(0^2 = 0\), \(1^2 = 1\), \(2^2 = 4\), \(3^2 = 9\), \(4^2 = 6\), \(5^2 = 5\), \(6^2 = 6\), \(7^2 = 9\), \(8^2 = 4\), \(9^2 = 1\). The possible values for \(a\) are 0, 1, 4, 5, 6, and 9.

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 Vaia!

Key Concepts

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

Inverse Elements in Modular Arithmetic
Understanding inverse elements in modular arithmetic can be quite intriguing. An inverse element for a number \( a \) in modular arithmetic exists if there is another number \( b \) such that their product is congruent to 1 modulo \( n \). Essentially, \( a \cdot b \equiv 1 \pmod{n} \).

In
  • this article, we consider the modular arithmetic over \( \mathbb{Z}_{10} \), which means we are working with numbers 0 through 9.
  • The key requirement for an inverse is that the number must be coprime to the modulus (in this case, 10).
  • This means the number shares no common factors with 10 other than 1.
Taking the elements of \( \mathbb{Z}_{10} \), it's found that:
  • \(1\), \(3\), \(7\), and \(9\) are coprime with \(10\).
  • Their respective inverses, calculated using techniques like the extended Euclidean algorithm, are \(1\), \(7\), \(3\), and \(9\) respectively.
This insight reveals a fundamental aspect of modular arithmetic and how it can be applied to find inverses within different modulus systems.
Congruences
The concept of congruences is central in modular arithmetic. A congruence relation means that two numbers leave the same remainder when divided by a particular modulus. We express this by saying \( a \equiv b \pmod{n} \).

Consider the scenario
  • where you need to solve quadratic congruences such as \( a^2 \equiv a \pmod{10} \).
  • Simplifying, we have \( a(a - 1) \equiv 0 \pmod{10} \).
  • This equation is satisfied if \( a(a-1) \) is divisible by 10, leading us to the solutions \( a = 0 \) or \( a = 1 \).
The beauty of congruences is that they allow us to work with divisions and multiplications without having to solve lengthy computations directly involving large numbers or complex calculations. They help in understanding and solving modular equations efficiently.
Powers in Modular Arithmetic
Exploring powers in modular arithmetic demonstrates how numbers behave under repeated multiplication. When raising a number to higher powers, the result is reduced modulo \( n \). This process can reveal interesting patterns and cycles.

Using the powers of
  • 2 as an example within \( \mathbb{Z}_{10} \):
  • \(2^1 = 2\), \(2^2 = 4\), \(2^3 = 8\), \(2^4 = 6\), and\( 2^5 \equiv 2 \pmod{10}\).
  • Notice the cycle repeating when the fifth power shows the same remainder as the first power, starting the cycle anew.
This cyclical nature is crucial, as it helps in simplifying power calculations and understanding modular patterns. The analysis of powers in modular arithmetic goes beyond simple calculations; it helps in predicting outcomes, finding patterns and cycles which are essential in cryptography, computer algorithms, and advanced mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(P\) be an orthogonal matrix. a. Show that \(\operatorname{det} P=1\) or \(\operatorname{det} P=-1\). b. Give \(2 \times 2\) examples of \(P\) such that \(\operatorname{det} P=1\) and \(\operatorname{det} P=-1\) c. If \(\operatorname{det} P=-1\), show that \(I+P\) has no inverse. \(\left[\right.\) Hint \(\left.: P^{T}(I+P)=(I+P)^{T} .\right]\) d. If \(P\) is \(n \times n\) and det \(P \neq(-1)^{n}\), show that \(I-P\) has no inverse. \(\left[\right.\) Hint: \(\left.P^{T}(I-P)=-(I-P)^{T} .\right]\)

a. If \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\), show that \(U^{-1} A U\) is not diagonal for any invertible complex matrix \(U\). b. If \(A=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\), show that \(U^{-1} A U\) is not upper triangular for any real invertible matrix \(U\).

Show that the following are equivalent for an \(n \times n\) matrix \(P\). a. \(P\) is orthogonal. b. \(\|P \mathbf{x}\|=\|\mathbf{x}\|\) for all columns \(\mathbf{x}\) in \(\mathbb{R}^{n}\). c. \(\|P \mathbf{x}-P \mathbf{y}\|=\|\mathbf{x}-\mathbf{y}\|\) for all columns \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n}\) d. \((P \mathbf{x}) \cdot(P \mathbf{y})=\mathbf{x} \cdot \mathbf{y}\) for all columns \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n}\). [Hints: For \((\mathrm{c}) \Rightarrow(\mathrm{d}),\) see Exercise \(5.3 .14(\mathrm{a}) .\) For (d) \(\Rightarrow\) (a), show that column \(i\) of \(P\) equals \(P \mathbf{e}_{i}\), where \(\mathbf{e}_{i}\) is column \(i\) of the identity matrix. \(]\)

Normalize the rows to make each of the following matrices orthogonal. a. \(A=\left[\begin{array}{rl}1 & 1 \\ -1 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{rr}3 & -4 \\ 4 & 3\end{array}\right]\) c. \(A=\left[\begin{array}{rr}1 & 2 \\ -4 & 2\end{array}\right]\) d. \(A=\left[\begin{array}{rr}a & b \\ -b & a\end{array}\right],(a, b) \neq(0,0)\) e. \(A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 2\end{array}\right]\) f. \(A=\left[\begin{array}{rrr}2 & 1 & -1 \\ 1 & -1 & 1 \\ 0 & 1 & 1\end{array}\right]\) g. \(A=\left[\begin{array}{rrr}-1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1\end{array}\right]\) h. \(A=\left[\begin{array}{rrr}2 & 6 & -3 \\ 3 & 2 & 6 \\ -6 & 3 & 2\end{array}\right]\)

Show that \(\langle A \mathbf{x}, \mathbf{y}\rangle=\left\langle\mathbf{x}, A^{H} \mathbf{y}\right\rangle\) holds for all \(n \times n\) matrices \(A\) and for all \(n\) -tuples \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{C}^{n}\).

See all solutions

Recommended explanations on Economics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free