Questions and Answers #4 Negative Numbers & Coordinate Plane

Questions and Answers Sheet 4 Negative Numbers & Coordinate Plane Question #1 What is the sign of 𝑎7 2 ∗ (−3/7) when a is a negative number? Answer: Any real number raised to an even power is going to be positive. ∴ 𝑎72 is positive. Since a positive times a negative is negative, 3 𝑎72 ⋅ (− ) is negative. 7 Question #2 For how many integers a is it true that 𝑎2 − 8 is a negative number? Answer: Well for 𝑎2 − 8 to be a negative number, it tantamounts to solve the inequality 𝑎2 − 8 < 0 Now this is nothing but (𝑎 − 2√2)(𝑎 + 2√2) < 0 1. for 𝑎 < −2√2, both 𝑎 − 2√2 and 𝑎 + 2√2 are negative and hence 𝑎2 − 8 > 0 2. for 𝑎 > 2√2, both 𝑎 − 2√2 and 𝑎 + 2√2 are positive and hence 𝑎2 − 8 > 0 3. But for −2√2 < 𝑎 < 2√2 , while 𝑎 − 2√2 is negative, 𝑎 + 2√2 is positive and hence 𝑎2 − 8 > 0 Hence possible integer values of a, for which 𝑎2 − 8 > 0 are -2,-1,0,1 and 2. Question #3 If A is a negative number, and 𝐵 = 0 − 𝐴2 . Is B a negative number or a positive number? Answer: if A is a negative imaginary number, with no real part, e.g., A=-i , then B will be positive note that 𝐴 = √−𝐵 and check section "Square roots of negative numbers" in the wikipage above for the fallacies that one must pay attention to when dealing with square roots of negative numbers Question #4 Natural log of a negative number ln(−1) = ln(𝑒 𝑖𝜋 ) = 𝑖 𝜋 Answer: Context is important here. In the context of real numbers, negative numbers have no logarithms (and neither does 0) because log(𝑥) is a number y such that 𝑒 𝑦 = 𝑥 and 𝑒 𝑦 is always greater than 0 . On the other hand, in the context of complex numbers, every complex number other than 0 has logarithms. In fact, any such complex number has infinitely many logarithms! You are right when you claim that 𝑖𝜋 is a logarithm of -1 . However, every complex number of the form 𝜋𝑖 + 2𝜋 𝑖𝑛 (with 𝑛 ∈ 𝑍𝑍) is also a logarithm of -1 , since 𝑒 𝜋𝑖+2𝑛𝜋𝑖 = 𝑒 𝜋𝑖 𝑒 2𝜋𝑖𝑛 = (−1) × 1 = −1. Question #5 Is the 𝑎𝑟𝑐𝑡𝑎𝑛 of a negative number always negative and 𝑎𝑟𝑐𝑡𝑎𝑛 of a positve number always positive? Answer: Hint: arctan(0) = 0, 𝑑 𝑑𝑥 arctan(𝑥) = 1 1+𝑥 2 > 0. Question #6 We know that negative number to the power of any integers or some fractions will always have a solution. 1 Is it possible for us to solve (−2)3 or (−2)𝑒 , by modifying/extending our current number system? Answer: Consider the value of the "number" (−2)𝑒 ; this is equivalent to considering 𝑥 = (−2)𝑒 ln 𝑥 = 𝑒 ln(−2) ln 𝑥 𝑒 𝑒 1 2 𝑒 = ln(−2) ln 𝑥 𝑒 𝑒 = −2 ln 𝑥 𝑒 ln 𝑥 𝑒 = −1 + ln 1 2 = −1 Given the equation 𝑒 𝑎+𝑏𝑖 = 𝑒 𝑎 (cos 𝑏 + 𝑖 sin 𝑏), we only need to find x such that ln 𝑥 𝑒 = 𝑖𝜋 + 2𝑛𝑖𝜋 − ln 1 2 or ln 𝑥 = 𝑒𝑖𝜋 + 2𝑛𝑒𝑖𝜋 + 𝑒 ln 2 which is 𝑥 = 𝑒 𝑒𝑖𝜋+2𝑛𝑒𝑖𝜋+𝑒 ln 2 = 2𝑒 (cos(𝑒𝜋 + 2𝑛𝑒𝜋) + 𝑖 sin(𝑒𝜋 + 2𝑛𝑒𝜋)) which is an infinite set of unique complex values in a circle of radius 2𝑒 Question #7 I am overthinking this but (−4)(−5/2) is not a valid equation, am I correct? Working through the problem gives me 1/(−45/2 ) which then works out to 1/√−45 which leaves a negative number in the square root, which is not valid. Answer: It is perfectly valid, the result is simply not a Real Number, it is what is called an Imaginary Number (or a Complex Number in the case in which you had a Real Numbered component as well), which you may not be familiar with. The result you gave simplifies down to − 1 32 𝑖, where i is called the Imaginary Unit, and is equal to √−1 Question #8 Why 134 mod -26 is equal 22 ? I found an explanation, which says: 134 = (−26 ∗ −6) − 22 However, 134 = (−26 ∗ −5) + 4, so why 134 mod -26 is not equal 4 if 134 = (26 ∗ 5) + 4 ⇒ 134 𝑚𝑜𝑑 26 = 4 All i could find, is about a situation, if i take a modulo of negative number (-134 mod 26) , but that is not what i want. Answer: The first sentence in the post is wrong. -136 in not congruent to 22 modulo -26. It is congruent to -22. In the second sentence you had an error (now fixed). You mean −26 ∗ −5. 4 and -22 are the same answer modulo -26 since they differ by a multiple of -26. Arithmetic modulo m is essentially the same as arithmetic modulo -m. Sometimes you want to require an answer that lies between 0 and the positive value |m|-1, sometimes any equivalent answer will do. If you are working with or writing a computer program to do modular arithmetic the program will make a choice - usually the representative between 0 and |m|-1.