# Solving Quadratic Equations Pure Imaginary Numbers

For y = x 2 , as you move one unit right or left, the curve moves one unit up. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm to calculate "small" solutions of relative Thue equations (where "small" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadratic equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw solver unblocked. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure imaginary numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions taken from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = √a. A number of the form bi, where 𝑏≠0, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots. 2 Problem 101E. For the simplest case, = 0, there are two turning points and these lie on the real axis at ±1. See full list on intmath. use the discriminant to determine the number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math·e·mat·ics (măth′ə-măt′ĭks) n. An obvious choice for x(0) is a turning point. For a method of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (p. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete detail. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers in the standard form bi: 1 5 , 11, , 7, 18. 5 + 4i A) real B) real, complex C) imaginary D) imaginary, complex Ans: D Section: 2.  Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. So tricky, in fact, that it’s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. • Perform operations with pure imaginary numbers • Perform operations with complex numbers • Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of x when you set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers • pure imaginary number: square root of a negative number • complex numbers • i2 = -1 i99 = 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part + bi Imaginary part Sec. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic. If you move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving standard quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product and sum of roots. Objective: be able to sketch power functions in the form of f(x)= kx^a (where k and a are rational numbers). complex numbers are required to be covered. Use the relation i 2 = −1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a ≠ 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers were developed as a result of the need to solve some types of quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative numbers, finding squares, cubes and their roots; the rule of False Position (see History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential equations worked out step by step. Imaginary Part. " Although there are two possible square roots of any number, the square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it is impossible to solve one equation for two unknowns. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with Modeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we ﬁnd x = 2± p (−2)2 −(4)(1)(10) 2. Videos are created by fellow teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Imaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard form. Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of symmetry will intersect a parabola in one point called the _____. Division of a complex number by a complex number; Division of a complex number by a complex number (example) Argand diagrams; Modulus and argument; Equating real and imaginary parts to solve equations; Square roots of a complex number; Solving quadratic equations with complex roots; Solving cubic equations; Solving quartic equations; Reflection. When I became a student at the. Complex numbers are built on the idea that we can define the number i (called "the imaginary unit") to be the principal square root of -1, or a solution to the equation x²=-1. 3 x 2 = 100 - x 2 Solution: Step 1. verb) The study of the measurement, properties, and relationships of quantities and sets, using. The Unit Imaginary Number, i, has an interesting property. Now you will solve quadratic equations with imaginary solutions. Real numbers. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. Consider the pure quadratic equation: x 2 = a , where a – a known value. The imaginary number i=sqrt(-1), i. What was most perplexing was that in using these subtle and imaginary numbers it was possible to solve cubic equations. A complex number is a number of the form where. Its use was prompted by the need to deal with algebraic expressions such as $$x^2+1$$ that have no root in the real numbers. Complex numbers; Non-real roots of quadratic equations. SOLVING QUADRATIC EQUATIONS. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Textbook solution for Precalculus: Mathematics for Calculus (Standalone… 7th Edition James Stewart Chapter 1. Simplifying Roots Of Negative Numbers Khan Academy. The special case corresponding to two squares is often denoted simply (e. Obviously when you get one root of a cubic equation, you can get the other two by dividing the original cubic equation by minus the first root and then use the quadratic formula in order to obtain the other two roots. 4 (1992): 824-842. complex number system The complex number system is made up of both the real numbers and the imaginary numbers. Comparing real and imaginary parts. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If b 0, then the complex number is called an imaginary number (Figure 2. It is well known that is perpendicular to iff is a pure imaginary number. An imaginary number is an even root of a negative number. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we ﬁnd x = 2± p (−2)2 −(4)(1)(10) 2. But suppose some wiseguy puts in a teensy, tiny minus sign: Uh oh. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. 156 complex number, p. Beware that in some cases the. Be able to find complex roots for quadratic equations. All non-imaginary numbers are real. SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE. Plug values into the quadratic formula. x2 + 9 = 0 b. Both hyperbolas are of relatively simple form. There's also a bunch of ways to solve these equations! Watch this tutorial and get introduced to quadratic equations!. Real numbers. I make note of which method needs the most reinforcement (likely completing the square) to that I can provide more practice when we get to imaginary numbers, later in the unit. The value of the discriminant of a quadratic equation can be used to describe the number of real and complex solutions. Core Vocabulary quadratic equation in one variable, p. This script is nothing extraordinary I just put it up so someone trying to do something similar with imaginary numbers could use the code as reference. Mediaeval Algebra in Western Europe was first learnt from the works of al-Khowarizmi, Abu Kamil and Fibonacci. it is a complete quadratic if b 0. Since the discriminant b 2 - 4 ac is 0, the equation has one root. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. 2 2 +7 +2 ≥ 0 32. Algebra-help. radical (symbol, expression). The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. Real part Imaginary part. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. When this occurs, the equation has no roots (zeros) in the set of real numbers. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. Quadratic Function Graph • max/min • vertex • axis of symmetry • y intercept • domain/range 7. Solve 3 – 4i = x + yi Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. Solved Name Unit 4 Solving Quadratic Equations Date B. 5 Solving Quadratic Equations – Factoring. 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of imaginary quadratic fields. Is Zero Considered a Pure Imaginary Number (as 0i)? [12/02/2003] In the complex plane, zero (0 + 0i) is on both the real and pure imaginary axes. Normally it is mentioned in chapter related to complex numbers where the reader is made aware of the power of complex numbers in solving polynomial equations. If and is not equal to 0, the complex number is called a pure imaginary number. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Ncert Exemplar Class 11 Maths Solutions Chapter 5 Free Pdf. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. Quadratic Equations solving quadratic equations by completing the square the quadratic formula long division of a polynomial by a. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. These are sometimes called pure imaginary numbers. Its solution may be presented as x = √a. Newton did not include imaginary quantities within the notion of number, and that G. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex. the effect that changing. In the 17th century, René Descartes (1596–1650) referred to them as imaginary numbers. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. If the real part of a complex number is 0, then it is called a pure imaginary number. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. 2 Power Functions with Modeling. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Day 10 I can find complex solutions of quadratic equations. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. (ii) Determine the other root of the equation, giving your answer in the form p + iq. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. is the imaginary part of the complex number. Zero Factor Property – basis for solving quadratic equations. Well, this time, I would like to write about quadratic equation. i i i is "a" solution to the quadratic equation x 2 = A pure imaginary number is a complex number having its real part zero. College Algebra (11th Edition) answers to Chapter 1 - Section 1. These are all quadratic equations in disguise:. If one complex number is known, the conjugate can be obtained immediately by changing the sign of the imaginary part. Complex Number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1, which is called an imaginary number because there is no real number that satisfies this equation. Imaginary numbers are based on the mathematical number $$i$$. Imaginary Unit i, Complex Number, Standard Form of a complex number, Imaginary Number, Pure Imaginary. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Therefore a complex number is the sum of a real number and a pure imaginary one. 4) Using a quadratic equation solver, we wind up with this: x = (2. The standard form of The solution set of equation 25x2 — I = 0 is: The quadratic. quadratic equations. Consider the pure quadratic equation: x 2 = a , where a – a known value. Solving Quadratic Equations Pure Imaginary Numbers. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. 3 10 4 3 9. • Use the discriminant to find the number of x-intercepts/real solutions/zeros/roots. number, pure imaginary number 3. We use the function func:scipy. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Solve quadratic equations by inspection (e. Improper Fraction. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. -5x2 + 12x - 8 = 0 4. Solving Quadratics with Imaginary Solutions Name_____ Date_____ Period____ ©M M2O0M1_6k GK_ultYaQ hSqoTfftTwwalrmed qLULvCm. , the square root of -1. The diagram shows how different types of complex numbers are related. Quadratic inequality in two variables: Quadratic inequality in one variable: Linear inequality in two variables: Solve the equation using any method. Answer by math_helper(1904) ( Show Source ):. These are all quadratic equations in disguise:. Real part + bi Imaginary part Sec. Perform operations with pure imaginary numbers and complex numbers Use complex conjugates to write quotients of complex numbers in standard form Graph quadratic functions Solve quadratic equations - Set - Element - Subset - Universal Set - Complement - Union - Intersection - Empty Set - Imaginary Unit - Complex Number. Galerkin (HDG) method for solving the Helmholtz equation with impedance boundary condition: (1. By using this website, you agree to our Cookie Policy. Imaginary numbers. Which statement about the solutions x = 5 and x = –20 is true? asked by T on June 2, 2016; Algebra 2 help :) Any number in the form of a+-bi , where a and b are real numbers and b not equal 0 is considered a pure imaginary number. Therefore, the rules for some imaginary numbers are:. Imaginary Part. Numbers like —2 — i and - 2 + i that include a real term and an imaginary term are called complex numbers. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. The concept was discussed in a recent thread, where we pointed out that the definition used for real radicands doesn't apply here, as there are no "positive" complex numbers; in cases like yours, in fact, both roots (2 - i and -2 + i) have a negative sign somewhere, so. The aim of this paper is to study t k and the value of N k /ℚ ( η k ). Horizontal Line Equation. The difference is that the root is not real. Upon completing this goal the student will be able to: * solve quadratic equations by graphing, factoring, and completing the square. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. Solve quadratic equations with complex number solutions. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. I will even skip a match if it means swiping the largest number out of the corner. They will also analyze situations involving quadratic functions and formulate quadratic equations to solve problems. From this starting point evolves a rich and exciting world of the number system that encapsulates everything we have known before: integers, rational, and real numbers. Simplify the expression: 16. Imaginary Numbers. They are factoring, using the square roots, completing the square and using the quadratic formula. x2 + 4x + 5 = 0 c. The Imaginary Unit i Not all quadratic equations have real-number solutions. Imaginary numbers are applied to square roots of negative numbers, allowing them to be simplified in terms of i. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Solve equations Quadratic in Form by substitution: Step 1: Determine the appropriate substitution and write the equation in the form au2 + bu + c = 0 Step 2: Solve the equation (using any method). " Imaginary numbers allow for complex analysis, which allows engineers to solve practical problems working in the plane. -2-Create your own worksheets like this one with Infinite Algebra 2. Complex Solution to Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, we can use complex numbers and the imaginary root to express the solutions. It also provides solutions to the problematic quadratic equations and all other polynomial equations In the form p(x). Write and graph an equation of a parabola with its vertex at (h,k) and an equation of a circle, ellipse, or hyperbola with its center at (h,k) Classify a conic using its equation : Quadratic Systems : Solve systems of quadratic equations by finding points of intersection Solve systems of quadratic equations using substitution. Quadratic equation usually used to find the unknown number(s) of x in the equation. Solve quadratic equations by inspection (e. for solving quadratic 2a equations. Note that each of these numbers is pure imaginary with positive coefficient. 5 Relation of the Roots. 6 Complex and Imaginary Numbers Objectives What is an imaginary number? What is a complex number? Jan 3010:53 AM 1 Complex Numbers 2010 September 15, 2010 Warmup: Solve using the quadratic formula. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. Now, by applying algebra techniques we can solve the equation. • Solve quadratic equations by factoring. Unique Math Equation Stickers designed and sold by artists. Continuing coursework from the Algebra II A, this title covers the review of square roots, radicals, complex pure and imaginary numbers, solving and factoring, identifying and evaluating the discriminant of a quadratic equation, rewriting equations, solving problems with number lines, graphing parabola, circle parts and formulas, hyperbola. fsolve to solve it. a coefficient has on. Quadratic Equation Solver. 1 Unit Objectives 4. Myung-Hwan Kim, Introduction to Universal Positive Quadratic Forms over Real Number Fields, Proc. The Unit Imaginary Number, i, has an interesting property. (Substitute your values back into the original subst. 4c Calculate the discriminant of a quadratic equation to determine the number of real & complex solutions. All non-imaginary numbers are real. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. Quadratic Equation. Cauchy-Riemann equations, harmonic functions. imaginary quadratic base eld Groups of Special Units, University of Georgia 2009 Invited number theory seminar about my thesis research Conference Organization West Coast Number Theory Conference 2015 - present On organizational committee and grant commitee Selected Conferences and Scholarly Activities [email protected] 2016 -present. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. \)The trajectory of such a solution consists of one point, namely $$c\ ,$$ and such a point is called an equilibrium. doc Author: E0022430 Created Date: 2/9/2010 12:03:19 PM. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. Polynomials with Complex Solutions. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Solve quadratic equations by factoring. If 𝑏=0, then the number 𝑎+𝑏𝑖=𝑎 is a real number. xx2 12 35 0 2. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. When the radicand in the quadratic formula (the discriminant Delta) is negative it means that you cannot find pure Real solutions to your equation.  Find the value of the discriminant. You want the square root of a number less than zero? That’s absurd!. Complete quadratic equation: If the equation having x and x2 terms such an. LOVE IT!! Reply Delete. Impossible Event. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Horizontal Shrink. xx2 10 25 64 4. The solution set is The Quadratic Formula If we start with the equation ax2 + bx + c = 0, for a > 0, and complete the square to solve for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. State the number of complex roots of the equation x 3 2x2 3x 0. web; books; video; audio; software; images; Toggle navigation. 156 pure imaginary number, p. complex numbers are required to be covered. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. pure imaginary number. Negative 4, if I take a square root, I'm going to get an imaginary number. fsolve to do that. Steacie Memorial Fellowship U of T’s team of students place 4 th in the 2017 Putnam Competition! Three faculty, R. Complex Numbers H2 Maths Tuition Tips. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Quadratic Formula 9. There are various methods through which a quadratic equation can be solved. 10 points for the best working. The number has a non-zero real part and pure imaginary part. An imaginary number is an even root of a negative number. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. • Find square roots and perform operations with pure imaginary numbers. Imaginary numbers are used to help us work with numbers that involve taking the square root of a negative number. The problem was with certain cubic equations, for example x3 −6x+2 = 0. LOVE IT!! Reply Delete. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. = −1, and every complex number has the form a + biwith a and b real. I been trying to figure out how to set up this equation to add two complex numbers for Java. In a similar way, we can find the square root of any negative number. We often use the notation z= a+ib, where aand bare real. • Estimate solutions of quadratic equations by graphing. Students apply these. Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Identity (Equation) Identity Matrix. Example 2A: Solving a Quadratic Equation with Imaginary Solutions Take square roots. 3  Notes  Solving Quadratics with Imaginary Numbers. Khan Academy Video: Quadratic Formula 1; Need more problem types? Try MathPapa Algebra. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Powered by Cognero Page 1. To ensure that every quadratic equation has a solution, we need a new set of numbers that includes the real numbers. Solving quadratic equations can sometimes be quite difficult. Math is the basic building blocks that deals with all sort of calculations such as Addition, subtraction, multiplication, division and much more. 2 Problem 101E. Well, this time, I would like to write about quadratic equation. The imaginary unit i is the complex. • Complete the square to solve quadratic equations or to convert from standard to vertex form. Students work extensively with factoring quadratics using various factoring techniques. Finding the values or real and imaginary numbers in standard form. This book has been requested by many readers. Journal Canadien de Math\'ematiques 44. Complex Solutions of Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, you often obtain a result such as which you know is not a real number. In this paper, we present a new method for solving standard quaternion equations. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. a unique quadratic function. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. x2 + 9 = 0 b. Imaginary numbers are called so because they lie in the imaginary plane, they arise from taking square roots of negative numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. the square, or using. To find complex number solutions of quadratic equations. (Substitute your values back into the original subst. By using this website, you agree to our Cookie Policy. ExampleUse the formula for solving a quadratic equation to solve x2 − 2x+10=0. x2 + 4x + 5 = 0 c. The others are right - pick a corner and keep the largest numbers closest to that corner. But this pure oscillation would be the B equals 0 with undamped. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. Lastly, Allen defined a complex number as one which is not real (p. pure imaginary number. Lesson 2: Solving Square Root Equations Lesson 3: The Basic Exponent Properties Lesson 4: Fractional Exponents Revisited Lesson 5: More Exponent Practice Lesson 6: The Quadratic Formula Lesson 7: More Work with the Quadratic Formula Unit 9 Lesson 1: Imaginary Numbers Lesson 2: Complex Numbers Lesson 3: Solving Quadratics with Complex Solutions. Number Theory 85 (2000), 201-219. 1) 10x2 - 4x + 10 = 02) x2 - 6x + 12 = 0 3) 5x2 - 2x + 5 = 04) 4b2 - 3b + 2 = 0. But this is really two. (used with a sing. In quadratic planes, imaginary numbers show up in equations. Solve quadratic equations by inspection (e. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. 22 (1996), 425-434. Pure Mathematics 2 & 3. First published in 1975, this classic book gives a systematic account of transcendental number theory, that is those numbers which cannot be expressed as the roots of algebraic equations having rational coefficients. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. Example: 3i If a ≠0 and b ≠ 0, the complex number is a nonreal complex number. You want the square root of a number less than zero? That’s absurd!. Impossible Event. Solved Name Unit 4 Solving Quadratic Equations Date B. 7) 10n2 - n - 8 = 08) 8p2 - 12p + 7 = 0 9) 2r2 + 2r + 6 = 0 10) 11r2 - 5r - 12 = 7 11) -14 + a = -3a2 12) -5 = 11b2 - 2b 13) 3n2 + 10n = -12 - 8n2 + 10n14) r2 - 2r - 4 = 2r2 + 8 Find the discriminant of each quadratic equation then state the number and type of solutions. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. 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Equations such as +1 0 have no real solution, so mathematicians defined the imaginary numbers to represent their solu ions. x2 + 4x + 5 = 0 c. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. As humans have solved new problems, equations, they have needed to create more numbers. 146 Solving Quadratic. or Quadratic Equations That Can Be Solved by Factoring, Applications of the Pythagorean Theorem Pg. 3 Example 4 Solve: x 2 x 6 0 1 223 x 2 No real-number solutions To solve such equations, we must define the square root of a negative number. Students will solve quadratic equations using graphs, tables, and algebraic methods. nth roots of a complex number The technique is the same for finding nth roots of any complex number. 2 Basic I can use the quadratic formula to solve a quadratic equation. • finding an equation for the common perpendicular to two skew lines 4 Complex numbers 4. When you need guidance on algebra exam or concepts of mathematics, Algebra-help. 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Note: It is not necessary to find the roots. In this paper a new algorithm for solving algebraic Riccati equations (both continuous-time and discrete-time versions) is presented. Pure STEP 3 Questions 2012 S3 Q6 1Preparation The STEP question involves complex numbers and the Argand diagram. Solve the equation x2 +4x+5 = 0. Quadratic Formula 9. 8x2 - 4x + 5 = 0 3. Workshops in Pure Math. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. If 𝑏≠ 0, then the number 𝑎+𝑏𝑖 is called an imaginary number. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. And this happens when b squared is smaller than 4ac. Improper Fraction. To find complex number solutions of quadratic equations. This page will try to solve a quadratic equation by factoring it first. complex number standard form EXAMPLE 1 imaginary unit GOAL 1 Solve quadratic equations with complex solutions and perform operations with complex numbers. Any number that is a non-repeating decimal is irrational. 2x2 - 10x + 25 = 0 5. 1) u k2u= f in ; @u @n (1. Real and imaginary numbers; Addition, subtraction and multiplying complex numbers and simplifying powers of i; Complex conjugates; Division of a complex number by a complex number; Argand diagrams; Modulus and argument of a complex number; Solving problems with complex numbers; Square roots of a complex number; Solving quadratic equations with. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. Horizontal Reflection. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. The solution to this particular equation is called the imaginary number i: i2 = 1,1 and one way to de ne the set of complex numbers is as the set of all expressions of type x+ iywhere xand yare real. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. • Writing quadratic equations in different forms reveals different key features. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. 5th Class Maths Worksheets. When the real part is zero we often will call the complex number a purely imaginary number. f x Ax Bx C = + + = 0 Equation 1. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i^2 = -1. You can use the imaginary unit to write the square root of any negative number. 7 3 2i i i 11. To find complex number solutions of quadratic equations. Students work extensively with factoring quadratics using various factoring techniques. For example, as follows:. Journal of Pure and Applied Algebra, volume 215, number 6, pages 1371--1397, 2011. 2) + iku= g on ; where 2Rd;d= 1;2;3 is a convex polyhedral domain, := @, k˛1 is known as the wave number, i = p 1 denotes the imaginary unit, and ndenotes the unit outward normal to @. The name comes from "quad" meaning square, as the variable is squared (in other words x 2). Full text of "The theory of equations: with an introduction to the theory of binary algebraic forms" See other formats. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. The imaginary unit “i” is used to represent: i 1 and i2 1 Ex. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNINGUPDATE: This unit now contains a Google document with links to instructional videos to help with remote teaching during COVID-19 school closures. square roots denoted by s and ºs. Simplify the expression: 16. 3 - Complex Numbers - 1. (a = 0) So, a number is either real or imaginary, and some imaginary numbers are pure imaginary numbers. It "cycles" through 4 different values each time we multiply:. Excel in math and science. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. (a) x 2 −1=0 (b) x2 −x −6 =0 (c) x 2 −2x −2 =0 (d) x2 −2x +2 =0 You should have found (a), (b) and (c) straightforward to solve. in the complex number. Pure imaginary numbers – numbers in the form bi – where i= −1. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. An equivalent form is b2 — 4ac. Quadratic formula: A quadratic formula is the solution of a quadratic equation ax 2 + bx + c = 0, where a ≠ 0, given by. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. The imaginary part is zero. And you would be right. Pure quadratic equation The number of methods to solve a quadratic e uatlon Is: Which equation is called exponential equation? A solution of equation which does not satisfy the equation is called: An equation in which variable occurs under radical sign is called. Joel Kamnitzer awarded a 2018 E. 3 Solving Quadratic Equations 4. These are solutions where appear the imaginary unit i. Imaginary numbers. x2 =-1 *This section may be omitted without any loss of continuity. Horizontal Parabola. Well, this time, I would like to write about quadratic equation. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). It is also called an "Equation of Degree 2" (because of the "2" on the x) A "Standard" Quadratic Equation looks like this: The letters a, b and c are coefficients (you know those values). Unit 4 Solving Quadratic Equations Homework 2 Answer Key. We call athe real part and bthe imaginary part of z. Page 126 Solving Quadratic Equations Freyer Model. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. 9) The matrix of this system is A = 0 I F 0 , where F = −Λ+bK. The discriminant is the radicand in the quadratic formula. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). Introduction Fundamental theorem of algebra is one of the most famous results provided in higher secondary courses of mathematics. Solving Quadratic Equations by Finding Square Roots. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. Use the Quadratic Formula to solve the quadratic equation. In this paper, we present a new method for solving standard quaternion equations. Many answers. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots.


# A square matrix has the same minimal polynomial over its base field as it has over an extension field

I think I have heard that the following is true before, but I don't know how to prove it:

Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.

Is this true? Would anyone be willing to provide a proof?

Attempt at a proof:

Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.

Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.

Does this look to be correct?

More generally, one might prove the following

Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$.

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There's the saying, "Look before you leap". I think I've managed to prove this. Please confirm if my answer is correct. –  iroiroaru Sep 21 '11 at 18:43
I think you already posted before under a different account (the "above" instead of "over"); I also remember your user name. Have you considered registering, so that all your activity is under the same user name? –  Arturo Magidin Sep 21 '11 at 18:46
It is impossible for us to confirm if your answer is correct if all you do is provide the question. If you want us to "confirm if [your] answer is correct", why not post your proof ? –  Arturo Magidin Sep 21 '11 at 18:49
I am in the process of writing it! –  iroiroaru Sep 21 '11 at 18:51
Hi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn't be necessary. e- I'm done writing my proof. –  iroiroaru Sep 21 '11 at 18:53

Written before/while the OP was adding his/her own proof, which is essentially the same as what follows.

Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\mathbb{C}$.

Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $\mu_{\mathbb{C}}(x)$ divides $\mu_{\mathbb{R}}(x)$.

I claim that $\mu_{\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\cdots + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\cdots + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu_{\mathbb{C}}(x)$ are real numbers.

Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED

So, yes, your argument is correct.

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Another way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:

Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$.

Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v_i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\lambda_i = \sum_j \alpha_{ij} e_j$, with $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 0$. $\square$

Now consider a field extension $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \deg \mu_F$. On the other hand it is clear that $\mu_F$ divides $\mu_K$. So $\mu_F = \mu_K$.

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As Andrea explained, the statement in the question results immediately from the following one.

Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$.

But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.

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This looks correct.

Another way to see it is that you can find the minimal polynomial of the matrix by computing the invariant factors of the matrix $A-XId$ over $\mathbb{R}$. Since the same process (with same operations) may be done over $\mathbb{C}$, their minimal polynomial is the same.

sorry, i don't know the english word for the "invariant factors", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$.

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Don't apologise, I'm having trouble with English as well! Since Arturo posted what seems like a more straightforward proof (well, it's the one I thought of...), I've accepted his answer, but thank you for your input and I will consider your idea. –  iroiroaru Sep 21 '11 at 19:07


# How many non empty subsets of {1, 2, …, n} satisfy that the sum of their elements is even?

The question I am working on is the case for $n$ = 9. How many non-empty subsets of $\{1,2,...,9\}$ have that the sum of their elements is even?

My solution is that the sum of elements is even if and only if the subset contains an even number of odd numbers. Since this is precisely half of all of the subsets the answer is $\frac{2^{9}}{2}=2^8$. Then the question specifies non-empty so final answer is $2^8-1$. Is this correct? In general I guess the solutions is $2^{n}-1$. My problem is why do exactly half of the total amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number of odd numbers and even number of odd numbers?

Let $S$ be a subset of $\{0,1,2,\dots,9\}$, possibly empty. Note that $1+2+\cdots +9=45$. So the sum of the elements of $S$ is even if and only if the sum of the elements of the complement of $S$ is odd.

Divide the subsets of $\{1,2,\dots,9\}$ into complementary pairs. There are $2^8$ such pairs, and exactly one element of each pair has even sum. Thus there are $2^8$ subsets with even sum, and $2^8-1$ if we exclude the empty set.

Remark: Suppose that $1+2+\cdots+n$ is odd. This is the case when $n\equiv 1\pmod{4}$ and when $n\equiv 2\pmod{4}$. Then the same argument shows that there are $2^{n-1}$ subsets with even sum.

We can use another argument for the general case. Note that there are just as many subsets of $\{1,2,\dots,n\}$ that contain $1$ as there are subsets that do not contain $1$. And for any subset of $A$ of $\{2,3,\dots,n\}$, we have that $A$ has even sum if and only if $A\cup\{1\}$ has odd sum, and $A$ has odd sum if and only if $A\cup\{1\}$ has even sum. Thus in general there are $2^{n-1}$ subsets with even sum.

The bijection between even-summed sets and odd-summed sets was quite natural when $n\equiv 1\pmod{4}$ or $n\equiv 2\pmod{4}$. In the general case, there is a nice bijection (add or subtract $\{1\}$), but it is less natural.

Let's first count all subsets of $\{1,\ldots,n\}$ with even sum. Removing the empty sets then makes us have to subtract one from this result.

The subsets of $\{1,\ldots,n\}$ with even sum are one-to-one with the subsets of $\{2,\ldots,n\}$. For any set $J\subset\{2,\ldots,n\}$, if the sum of $J$ is even, then $J$ is a subset of $\{1,\ldots,n\}$ with even sum, while if the sum of $J$ is odd, then $\{1\}\cup J$ is a subset with even sum.

Since there are $2^{n-1}$ subsets of $\{2,\ldots,n\}$, this is the number of subsets of $\{1,\ldots,n\}$ with even sum. Remove the empty set, and you get $2^{n-1}-1$.

What you did is fine, we can get an alternative proof if we recall how we prove that there are $2^{n-1}$ subsets of $\{1,2\dots n\}$ of even cardinality.

Let $E$ be the set of subsets of even cardinity and let $O$ be the set of subsets of odd cardinality, pick an arbitrary element $a\in\{1,2,3\dots n\}$.

Then $f:E\rightarrow O$ defined as $X\mapsto \{a\}\Delta X$ is a bijection right?

Well, if $E'$ is the set of subsets with even sum and $O'$ is the set of subsets with odd sum and $a\in\{1,2,3\dots n\}$ is odd.

$f:E'\rightarrow O'$ defined as $X\mapsto \{a\}\Delta X$ is also a bijection.

So in any finite subset $A$ of positive integers, exactly half of the subsets have even sum, unless all of the elements of $A$ are even, in which case all the subsets clearly have even sum.

$\Delta$ is just the symmetric difference of sets, so $\{a\}\Delta X$ is $\{a\}\cup X$ if $a$ was not in $X$ and is $X\setminus\{a\}$ if $a$ was in $X$.


# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$.

I believe I have been able to construct in two ways, using the field axioms, that if $x \neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.

My question, then, is which of these is 'more' standard in a first-year analysis course?

Proof 1: Assuming I can use arithmetic .

Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*}

Example 2: Without assuming arithmetic, and mirroring Rudin.

\begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \neq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ & = \frac{1}{x} \left(x \cdot y\right) & & \text{Associativity of multiplication} \\ & = \frac{1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \cdot x\right) z & & \text{Associativity of multiplication} \\ & = 1z & & \text{Inverse properties} \\ & = z \end{align*} Thanks in advance.

• If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20
• In your "mirroring Rudin" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. – James Aug 17 '18 at 18:28
• Also, I think your worry about "using arithmetic" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. – James Aug 17 '18 at 18:31
• One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37

The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial.
More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{align}
On the other hand, the other proof seems shorter \begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}(xz) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{-1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && \text{(multiplicative identity)} \end{align} and less “rabbit out of a top hat”.


Does this question have two answers correct?

A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $$18$$ seconds from the start. The period of the simple pendulum is

(A) $$0.9$$ sec

(B) $$1.8$$ sec

(C) $$2.7$$ sec

(D) $$3.6$$ sec

I was given a formula for such questions:

$$T = \frac {T_1 T_2} {T_1-T_2} \qquad (T_1>T_2)$$

where $$T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again.

I took $$T_1$$ as the seconds pendulum, i.e., $$T_1=2$$ seconds.

Using the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are:

$$1.8\ \ 3.6\ \ 5.4\ \ 7.2\ \ 9.0\ \ 10.8\ \ 12.6\ \ 14.4\ \ 16.8\ \ 18.0$$

seconds for simple pendulum, and $$2, 4, 6, 8, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds.

However, I also tried option A, and got the timestamps as:

$$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7.2\ \ 8.1\ \ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.4\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$

seconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase after intervals of $$18$$ seconds.

According to the answer key, the answer is only B. Is A also correct, or am I missing something?

The key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $$\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\frac{9}{11}$$ of the way through a period (try dividing $$\frac{18}{11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase.

I'd highlight that this means care is needed to derive the $$\frac{T_1 T_2}{T_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum are in phase after $$18$$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$t/T_2$$ is an integer.

For the explicit derivation: the pendulums are in phase at time $$t$$ if and only if $$t/T_1 - t/T_2 = n$$ for some integer $$n$$. Solving for $$t$$ yields

$$t = n \frac{T_1 T_2}{T_1 - T_2},$$

and hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\frac{T_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of generality). In particular, the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\frac{T_1 T_2}{T_1 - T_2}$$ as claimed.

• Yes, this is the key. The formula gives the time for the first time they are in phase. Any period which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time.
– nasu
Apr 3 at 20:41

Good work, but your issue is that you're taking too narrow a view of what "swinging in phase" means. To be in phase, the two pendulums simply need to be at the same point in their cycle -- meaning at the same angle and swinging in same direction. What you're doing with your timestamps approach is to identify only those instants when the two pendulums have returned exactly to their starting position at the same time.

And they will definitely be in phase when they are both back at their starting positions at the same time (since the problem specified that they started off in phase), but the trick is that they could also be in phase at points before that as well.

Imagine a pendulum with a period of 10 seconds, and one with a period of only 1 second:

• After 1 second, P2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So P2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with P2 basically doing all the work).
• Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% through its 1-second cycle, while P1 will only be 11% through its cycle
• Another 0.01 seconds after that (1.11s total), P2 will be 11% through its cycle, while P1 is 11.1% through its cycle.
• You can see where this is headed -- P2 will "catch" P1 for the first time at 1.111111...s (aka 10/9 seconds).

You can validate that from your formula: T2*T1/(T2-T1) = 10*1/(10-1) = 10/9 = 1.11111...

So these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through the cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see if they're in phase.

Your method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first time, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase at the end of a complete cycle) is 10s (since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = 9) that they've been back in phase with each other.

In your question, with P1 of 2s and P2 of 0.9s, the "beat frequency" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their starting points and in phase, at the end of the 9th complete cycle for P1, the 20th complete cycle for P2 -- and it's the 11th time (18/1.63) that they've returned to phase with each other.

With P1 of 2s and P2 of 1.8s, the "beat frequency" is 2*1.8/(2-1.8) = 18 seconds, AND the least common integer multiple of the two periods also happens to be 18s. At 18s, P1 will have completed 9 cycles, P2 will have completed 10 cycles, and it will be the first time they're back in phase together.

You could argue the question is slightly ambiguous by saying "They again swing in phase after an interval of 18 seconds from the start" -- that is, it doesn't specify that "They, for the first time since the start, swing again in phase after an interval of 18 seconds from the start", but I think the "first time" part is pretty heavily implied.

In your analysis with the time stamps, you deduced that either answers (a) and (b) are possibly correct. See also helloworld's answer on how you can deduce further which of these is correct by again looking closer at the phase relationship.

But how you arrive at the correct answer lies in the not-so-obvious wording of the question. First, do you know what "a seconds pendulum" is? Your question reads:

A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum

Note the words and a seconds pendulum. A seconds pendulum is a pendulum that has a period of precisely $$2$$ seconds. That is, one second per swing in one direction, or two seconds to complete a swing in both directions (one full period is two seconds)$$^1$$.

This means with $$T_1=2\ \text{sec}$$ and with $$T=18\ \text{sec}$$ to retain an in-phase relationship, gives the only possible correct answer of $$T_2=1.8\ \text{sec}$$. The answer cannot possibly be 0.9 sec. Nor can it be the other two possibilities (c) and (d).

$$^1$$From the wording of the question alone, you can deduce that the only possible answers are (a) and (b), since "the length is less than that of a second's pendulum" or less than 2 seconds, since we know that the period of a simple pendulum is proportional to $$l^{\frac 12}$$. And one could have even deduced the correct answer from this information alone. The equation you quoted is a transpose of the equation $$\frac 1T=\frac 1T_1-\frac 1T_2$$ and by adding $$\frac{1}{18}$$ and $$\frac 12$$ $$(=\frac{9}{18})$$ to each other and then just flipping the fraction to get 1.8 seconds.

• I get why 1.8 seconds is an answer; I'm asking whether 0.9 seconds is also an answer or not, and if not, why not? Apr 3 at 9:52
• It can't be the correct answer because you are already told that one pendulum has a period of two seconds, then you are asked to calculate the period of the other pendulum. It's not an answer, it's the only answer. Apr 3 at 9:54
• Could you please justify that without directly using the formula? On trying it manually, $0.9$ seconds seems to work for me Apr 3 at 10:09
• I agree this answer doesn’t explain why answer a. is invalid. Another answer clarifies succinctly. Apr 3 at 21:22
• But why can there be only one possible answer? An assertion is not the same as an explanation. Apr 3 at 23:33

Let's look at the equations

$$x_1(t)=\sin\left(\frac{2\pi}{T_1}\,t\right)\\ x_2(t)=\sin\left(\frac{2\pi}{T_2}\,t\right)$$

for $$~t=T~$$ is $$~x_1(T)=0~$$ only if

$$~\frac {T}{T_1}=1,2,\ldots n~\quad$$

and $$~x_2(T)=0~$$ only if

$$~\frac {T}{a\,T_2}=1,2,\ldots n~$$

with $$~T=\frac{T_1\,T_2}{T_1-T_2}=\frac{2*1.8}{2-1.8}=18~$$ thus $$a=1~,T_2\mapsto 1.8\quad ,n=\frac{18}{1.8}=10~\surd\\ a=\frac 12~,T_2\mapsto \frac 12*1.8=0.9\quad ,n= \frac{18*2}{0.9}=40~\surd\\ a=2~,T_2\mapsto 2*1.8=3.6\quad ,n=\frac{18}{3.6}=5~\surd$$

I don't think that you need the formula $$~T=\frac{T_1\,T_2}{T_1-T_2}~$$ .

for a given period $$~T,~\frac {T}{T_1}~$$ must be integer.

the period $$~T_2=\frac{T}{n}~$$ and for $$~T_2 \le T_1=2\quad \Rightarrow n\gt \frac T2$$

Example

$$T=18~,n\gt 9\\ T_2=\left[\frac 95,{\frac {18}{11}},\frac 32,{\frac {18}{13}},{\frac {9}{7}},\frac 65,{\frac { 9}{8}}\,\ldots\right]$$

My goodness the other answers are woefully overcomplicating this.

The question says

A simple pendulum (whose length is less than that of a second's pendulum)

The length of the pendulum is less than that of a second's pendulum. Therefore it will swing faster, therefore it will have a shorter period than $$1s$$. Only one answer has a period less than $$1s$$.

More generally, equate $$18$$ swings of the 'second's pendulum' with some integer number of swings of the other pendulum: $$18 s = n T$$ It is generally assumed, by the wording, that they were in sync only after $$18s$$ and no sooner than that. In this case, it must be that $$\gcd(18,n) = 1.$$ You're also told that the simple pendulum is shorter than the second's pendulum. Therefore it swings faster, with shorter period, and $$n>18$$.

Therefore, $$n>18$$ with $$\gcd(18,n) = 1$$. The smallest valid $$n$$ is $$19$$, which implies a period of $$T = \frac{18s}{n} = \frac{18s}{19} \approx 0.947s \approx 0.9s$$ In principle the answer could also be $$n=23, 25, 29, \dots$$ . However, the question only offers one choice for $$n>18$$, i.e. $$T<1s$$, so that's the only choice.

You might say that the 'second's pendulum' has a period of $$2s$$. Unless the question specifically states that a second's pendulum has a period of $$2s$$, or if you were taught this specific fact, I think it is a coincidence that the term 'second's pendulum' refers to a pendulum with a specific period of $$2s$$; common parlance would assume it would have a period of $$1s$$.

If you insist on the $$2s$$, then we have $$18 \times 2s = nT$$ with $$\gcd(18,n) = 1$$ and $$n>18$$. The answers are the same, except multiplied by $$2$$.

For $$n=19$$ we have $$T \approx 1.89s \not \approx 1.8s$$ For $$n=23$$ we have $$T \approx 1.57s$$ ...and subsequent choices will have a lower period $$T$$. There is one correct choice, which is $$n=41$$ which gives $$T \approx 0.88s \approx 0.9s$$ so, the answer is again A.

The only correct answer is A.

• No. The answer is subtle. A "seconds pendulum" has a period of two seconds, meaning either (a) or (b) are correct. And the correct answer is (b). See my answer above for a further explanation. Cheers. Apr 5 at 2:33
• @josephh I addressed this. Apr 5 at 2:35
• Did you look at this link where the seconds pendulum is explained? "Unless the question specifically states that a second's pendulum has a period of 2s" It does by mentioning it is. BTW 2 seconds means time for two full swings. Cheers. Apr 5 at 2:38
• @josephh I addressed this. Read my answer before commenting please. Using $2s$ gives the same answer $A$ anyway. Apr 5 at 2:42
• Under the interpretation that the seconds pendulum has a 1s period, observe that a T1=0.9s pendulum and T2=1s pendulum are already in phase at 9s, so that isn't the right answer. Under the interpretation that it has a 2s period, a T1=0.9s pendulum and T2=2s pendulum are already in phase at 18/11s, so that's also not correct. Apr 8 at 1:22


# Documentation/Calc Functions/MOD

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MOD

Mathematical

## Summary:

Calculates the remainder when one number (the dividend or numerator) is divided by another number (the divisor or denominator). This is known as the modulo operation.

Often the dividend and divisor will be integer values (Euclidean division). However, MOD accepts and processes real numbers with non-zero fractional parts.

## Syntax:

MOD(Dividend; Divisor)

## Returns:

Returns a real number that is the remainder when one number is divided by another number. The value returned has the same sign as the divisor.

## Arguments:

Dividend is a real number, or a reference to a cell containing that number, that is the dividend of the divide operation.

Divisor is a real number, or a reference to a cell containing that number, that is the divisor of the divide operation.

• If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! error.
• If Divisor is equal to 0, then MOD reports a #DIV/0! error.

• For real x and y (y <>0), MOD implements the following formula:

$\displaystyle{ \text{MOD}(x,y)~=~x-\left(y\times \text{INT} \left(\frac{x}{y}\right)\right) }$

The INT function always rounds down (toward -∞) and returns the largest integer less than or equal to a given number. This means that when the dividend and divisor have different signs, MOD may produce results that appear counterintuitive. For example, the formula =MOD(7, -3) returns -2; this is because the fraction $\displaystyle{ \left(\frac{7}{-3}\right) }$ is rounded to -3 by the INT function.
• For more general information about the modulo operation, visit Wikipedia’s Modulo operation page.

## Examples:

Formula Description Returns
=MOD(11; 3) Here the function returns the remainder when 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. 2
=MOD(-11; 3) Here the function returns the remainder when -11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. 1
=MOD(11; -3) Here the function returns the remainder when 11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1
=MOD(-11; -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. -2
=MOD(D1; D2) where cells D1 and D2 contain the numbers 11.25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in this example. 1.25

MOD