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The Deadly Sins of Algebra There are some algebraic misconceptions that are so damaging to your quantitative and formal reasoning ability, you might as well be said not to have any such reasoning ability. If they go uncorrected, they will all but rule out success in any STEM course of study, and even if you somehow manage to earn passing grades, you will never be able to reliably work with mathematics in your professional career. Trains might collide, bridges collapse and rockets explode because of you, though more likely, an employer will simply let you go when they discover that those mathematics credits donβt translate into reliable practical problem solving ability. This handout is intended to help you to un-learn these misconceptions. Misconception #1 The square root of a sum is the sum of the square roots: βπ₯ + π¦ = βπ₯ + βπ¦. Pick x to be 9 and y to be 16. Then βπ₯ + π¦ = β9 + 16 = β25 = 5 but βπ₯ + βπ¦ = β9 + β16 = 3 + 4 = 7. Five is not equal to seven. Five is in fact less than seven. Here is a way to look at this misconception from a geometric point of view. Flying from Phoenix to Houston (1015 miles east), and then from Houston to Chicago (943 miles north) takes more miles than flying from Phoenix to Chicago directly, which only takes 1453 miles, rather than 1015+943 = 1958 miles. These three cities form approximately a right triangle. According to the Pythagorean theorem, in a right triangle, we have π2 + π 2 = π 2 , with the usual notations. So the hypotenuse is π = βπ2 + π 2 . If you could break up the square root of a sum as the sum of the square roots, then π would be equal to π + π. That is an absurdity, because c is clearly shorter. If βπ₯ + π¦ = βπ₯ + βπ¦ was true, then there would be no shortcuts in the universe. But there are, and the square root of the sum of two positive quantities is always less than the sum of the roots. Correction of Misconception #1 The square root of a sum of positive numbers x,y is always less than the sum of the square roots: βπ₯ + π¦ < βπ₯ + βπ¦ © R. Boerner, ASU School of Mathematical and Statistical Sciences 2014 Misconception #2 π β1 (π₯) = 1 π(π₯) . Specifically, the inverse (arc) trigonometric functions are just reciprocals of the associated trig functions: arcsin π₯ = sinβ1 π₯ = 1 sin(π₯) = sec π₯, etc. If this was the case, you should ask yourself why we would have all these different names for the same function. Why would one and the same function be called βarcsinβ as well as βsecantβ? It wouldnβt be. This is your common sense clue that arcsin and sec are two very different functions, and that the apparent power of -1 in the notation " sinβ1 π₯ " means something else entirely. 1 While it is true that if π₯ is a number, π₯ β1 means βx raised to the power of -1β, or , for a function, the notation π β1 refers to the inverse function, not the reciprocal function. π₯ The inverse function of π is the function that reverses the role of input and output for π. It takes an output of π and recovers the input that lead to that output. It is found by switching the roles of x and y in the definition. For example, if π(π₯) = π₯ + 1 then π is the function βadd oneβ. It adds one to every input. Its inverse function will undo the operation by subtracting one from every input: π β1 (π₯) = π₯ β 1 Compare this to the reciprocal function : (π(π₯))β1 = 1 1 = π(π₯) π₯ + 1 These are two very different functions. This means that " sinβ1 π₯ " is not the function β1 divided by sin xβ. It is the function βfind the angle (or π π arc) in the range [β , ] whose sine value is xβ. Thatβs why sinβ1 is also referred to as the arcsin. 2 2 Correction of Misconception #2 π β1 (π₯) and 1 π(π₯) are generally totally different functions. sinβ1 π₯ = arcsin π₯ is a totally different function than 1 sin(π₯) = sec π₯. © R. Boerner, ASU School of Mathematical and Statistical Sciences 2014 Misconception #3: βCancelling Everything in Sightβ This is the idea that if you find the same quantity in two different places in an algebraic expression, once in a numerator, once in a denominator (possibly belonging to different fractions), then they cancel: π₯+2 =π₯ 2 3 1 1 1 + = + π₯ 3+2 π₯ 2 βCancelling Everything in Sightβ1 is ultimately ignorance of the algebraic order of operations. For a quantity to multiplicatively cancel itself, it needs to meet itself directly in a division, according to the order of operations. Whether this is the case becomes apparent when you write the expression with a slash for division and parentheses. In the expression π₯+2 2 , 2 is not divided by 2. Before the division happens, there is an addition. So the expression is really (π₯ + 2)/2. 2 does not cancel against the (different) quantity (π₯ + 2). For that to 2 happen, the expression would have to be π₯ + 2/2 which is π₯ + . 2 Another way to think about this situation correctly is to realize that division is really just multiplication, namely, multiplication by the reciprocal, so the distributive law applies: π₯+2 1 1 1 π₯ = (π₯ + 2) β = π₯ β + 2 β = + 1 2 2 2 2 2 Correction of Misconception #3 The only situation where a quantity can be multiplicatively canceled is the case of a common factor in a quotient: ππ + ππ π + π = ππ + ππ π + π 1 Credit for term goes to Thomas Scofield of Calvin College. © R. Boerner, ASU School of Mathematical and Statistical Sciences 2014 Misconception #4 You solve a quadratic equation by moving the constant term to the right side and then factoring out the x on the left side: π₯ 2 + π₯ β 6 = 0 is solved by rewriting the equation as π₯ 2 + π₯ = 6 and then as π₯(π₯ + 1) = 6 So π₯ = 6 or π₯ + 1 = 6, meaning π₯ = 6 or π₯ = 5. This incorrect solution is based on the idea that a product is 6 when one of the factors is 6, which is contrary to the basic laws of arithmetic and to common sense. You know that the value of a product depends on both factors. When a product is 6, there are many ways in which the two factors can conspire to make that happen: 1 β 6, 2 β 3, 11 β 6 6 , etc. In fact, there 11 are infinitely many ways: for any nonzero x, π₯ β = 6. Therefore, the mere fact that a product of two π₯ factors is 6 implies no specific values of the factors involved. π₯(π₯ + 1) = 6 is a dead end. It cannot be used to determine what x might be, except by guessing. If that still doesnβt convince you, think about the value of the expression π₯(π₯ + 1) when π₯ = 6. Itβs not 6, itβs 42. And when π₯ = 5, π₯(π₯ + 1) is not 6, but 30. There is a grain of truth hidden in this misconception. There is one special number that forces a product to be equal to that number, regardless of other factors in the product: zero. A product is zero when one of the factors is zero. This means that a factorization is only useful when it is equal to zero. No other number will do. In the example given above, we factor π₯ 2 + π₯ β 6 = 0 as (π₯ β 2)(π₯ + 3) = 0 and conclude that π₯ = 2 or π₯ = β3. Correction of Misconception #4 We solve a quadratic equation by moving all terms to the left side, thereby creating a zero on the right side, and then factoring the left side. If no factorization of the left side is apparent, we use the quadratic formula. © R. Boerner, ASU School of Mathematical and Statistical Sciences 2014 Misconception #5 Forgetting the negative root of π₯ 2 = π, where π is a positive number. This one is not so much a misconception as it is a common oversight. Correction of Misconception #5 The indicated quadratic equation has two solutions: π₯ = ±βπ. Example: π₯ 2 = 9 has two solutions, π₯ = 3 and π₯ = β3. © R. Boerner, ASU School of Mathematical and Statistical Sciences 2014