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2.1 A Preview of Calculus

Learning objectives.

• 2.1.1 Describe the tangent problem and how it led to the idea of a derivative.
• 2.1.2 Explain how the idea of a limit is involved in solving the tangent problem.
• 2.1.3 Recognize a tangent to a curve at a point as the limit of secant lines.
• 2.1.4 Identify instantaneous velocity as the limit of average velocity over a small time interval.
• 2.1.5 Describe the area problem and how it was solved by the integral.
• 2.1.6 Explain how the idea of a limit is involved in solving the area problem.
• 2.1.7 Recognize how the ideas of limit, derivative, and integral led to the studies of infinite series and multivariable calculus.

As we embark on our study of calculus, we shall see how its development arose from common solutions to practical problems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two key problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent to a curve at a point; and (2) the area problem, or how to determine the area under a curve.

The Tangent Problem and Differential Calculus

Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at the graphs of the three lines f ( x ) = −2 x − 3 , g ( x ) = 1 2 x + 1 , f ( x ) = −2 x − 3 , g ( x ) = 1 2 x + 1 , and h ( x ) = 2 , h ( x ) = 2 , shown in Figure 2.2 .

As we move from left to right along the graph of f ( x ) = −2 x − 3 , f ( x ) = −2 x − 3 , we see that the graph decreases at a constant rate. For every 1 unit we move to the right along the x -axis, the y -coordinate decreases by 2 units. This rate of change is determined by the slope (−2) of the line. Similarly, the slope of 1/2 in the function g ( x ) g ( x ) tells us that for every change in x of 1 unit there is a corresponding change in y of 1/2 unit. The function h ( x ) = 2 h ( x ) = 2 has a slope of zero, indicating that the values of the function remain constant. We see that the slope of each linear function indicates the rate of change of the function.

Compare the graphs of these three functions with the graph of k ( x ) = x 2 k ( x ) = x 2 ( Figure 2.3 ). The graph of k ( x ) = x 2 k ( x ) = x 2 starts from the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowly at first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single number represents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinear function?

We can approximate the rate of change of a function f ( x ) f ( x ) at a point ( a , f ( a ) ) ( a , f ( a ) ) on its graph by taking another point ( x , f ( x ) ) ( x , f ( x ) ) on the graph of f ( x ) , f ( x ) , drawing a line through the two points, and calculating the slope of the resulting line. Such a line is called a secant line. Figure 2.4 shows a secant line to a function f ( x ) f ( x ) at a point ( a , f ( a ) ) . ( a , f ( a ) ) .

We formally define a secant line as follows:

The secant to the function f ( x ) f ( x ) through the points ( a , f ( a ) ) ( a , f ( a ) ) and ( x , f ( x ) ) ( x , f ( x ) ) is the line passing through these points. Its slope is given by

The accuracy of approximating the rate of change of the function with a secant line depends on how close x is to a . As we see in Figure 2.5 , if x is closer to a , the slope of the secant line is a better measure of the rate of change of f ( x ) f ( x ) at a .

The secant lines themselves approach a line that is called the tangent to the function f ( x ) f ( x ) at a ( Figure 2.6 ). The slope of the tangent line to the graph at a measures the rate of change of the function at a . This value also represents the derivative of the function f ( x ) f ( x ) at a , or the rate of change of the function at a . This derivative is denoted by f ′ ( a ) . f ′ ( a ) . Differential calculus is the field of calculus concerned with the study of derivatives and their applications.

For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit this applet ( Note: this site requires a Java browser plugin): Math Insight .

Example 2.1 illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or, equivalently, the rate of change of the function at the point at which the slopes are calculated.

Example 2.1

Finding slopes of secant lines.

Estimate the slope of the tangent line (rate of change) to f ( x ) = x 2 f ( x ) = x 2 at x = 1 x = 1 by finding slopes of secant lines through ( 1 , 1 ) ( 1 , 1 ) and each of the following points on the graph of f ( x ) = x 2 . f ( x ) = x 2 .

• ( 2 , 4 ) ( 2 , 4 )
• ( 3 2 , 9 4 ) ( 3 2 , 9 4 )

Use the formula for the slope of a secant line from the definition.

• m sec = 4 − 1 2 − 1 = 3 m sec = 4 − 1 2 − 1 = 3
• m sec = 9 4 − 1 3 2 − 1 = 5 2 = 2.5 m sec = 9 4 − 1 3 2 − 1 = 5 2 = 2.5

The point in part b. is closer to the point ( 1 , 1 ) , ( 1 , 1 ) , so the slope of 2.5 is closer to the slope of the tangent line. A good estimate for the slope of the tangent would be in the range of 2 to 2.5 ( Figure 2.7 ).

Checkpoint 2.1

Estimate the slope of the tangent line (rate of change) to f ( x ) = x 2 f ( x ) = x 2 at x = 1 x = 1 by finding the slope of the secant line through ( 1 , 1 ) ( 1 , 1 ) and the point ( 5 4 , 25 16 ) ( 5 4 , 25 16 ) on the graph of f ( x ) = x 2 . f ( x ) = x 2 .

We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rate of change of position, suppose that we have a function, s ( t ) , s ( t ) , that gives the position of an object along a coordinate axis at any given time t . Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time t = a ? t = a ? We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define the average velocity of an object over a time period to be the change in its position divided by the length of the time period.

Let s ( t ) s ( t ) be the position of an object moving along a coordinate axis at time t . The average velocity of the object over a time interval [ a , t ] [ a , t ] where a < t a < t (or [ t , a ] [ t , a ] if t < a ) t < a ) is

As t is chosen closer to a , the average velocity becomes closer to the instantaneous velocity. Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point a , we let the x -values approach a in the slope of the secant line. Similarly, to find the instantaneous velocity at time a , we let the t -values approach a in the average velocity. This process of letting x or t approach a in an expression is called taking a limit . Thus, we may define the instantaneous velocity as follows.

For a position function s ( t ) , s ( t ) , the instantaneous velocity at a time t = a t = a is the value that the average velocities approach on intervals of the form [ a , t ] [ a , t ] and [ t , a ] [ t , a ] as the values of t become closer to a , provided such a value exists.

Example 2.2 illustrates this concept of limits and average velocity.

Example 2.2

Finding average velocity.

A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground t seconds later (for 0 ≤ t ≤ 2 ) 0 ≤ t ≤ 2 ) is given by s ( t ) = −16 t 2 + 64 . s ( t ) = −16 t 2 + 64 . Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time t = 0.5 . t = 0.5 .

• [ 0.49 , 0.5 ] [ 0.49 , 0.5 ]
• [ 0.5 , 0.51 ] [ 0.5 , 0.51 ]

Substitute the data into the formula for the definition of average velocity.

• v ave = s ( 0.5 ) − s ( 0.49 ) 0.5 − 0.49 = −15.84 v ave = s ( 0.5 ) − s ( 0.49 ) 0.5 − 0.49 = −15.84
• v ave = s ( 0.51 ) − s ( 0.5 ) 0.51 − 0.5 = −16.16 v ave = s ( 0.51 ) − s ( 0.5 ) 0.51 − 0.5 = −16.16

The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.

Checkpoint 2.2

An object moves along a coordinate axis so that its position at time t is given by s ( t ) = t 3 . s ( t ) = t 3 . Estimate its instantaneous velocity at time t = 2 t = 2 by computing its average velocity over the time interval [ 2 , 2.001 ] . [ 2 , 2.001 ] .

The Area Problem and Integral Calculus

We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities of work—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the x -axis over an interval ( Figure 2.8 )?

As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area by dividing up the interval [ a , b ] [ a , b ] into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles ( Figure 2.9 ).

As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of f ( x ) f ( x ) and the x -axis over the interval [ a , b ] . [ a , b ] . Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and their applications.

Example 2.3

Estimation using rectangles.

Estimate the area between the x -axis and the graph of f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 over the interval [ 0 , 3 ] [ 0 , 3 ] by using the three rectangles shown in Figure 2.10 .

The areas of the three rectangles are 1 unit 2 , 2 unit 2 , and 5 unit 2 . Using these rectangles, our area estimate is 8 unit 2 .

Checkpoint 2.3

Estimate the area between the x -axis and the graph of f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 over the interval [ 0 , 3 ] [ 0 , 3 ] by using the three rectangles shown here:

Other Aspects of Calculus

So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two dimensions; however, there is no good reason to restrict our investigation to two dimensions. Suppose, for example, that instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a rock fired from a catapult at a given time, or of an airplane moving in three dimensions. We might want to graph real-value functions of two variables or determine volumes of solids of the type shown in Figure 2.11 . These are only a few of the types of questions that can be asked and answered using multivariable calculus . Informally, multivariable calculus can be characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.

Section 2.1 Exercises

For the following exercises, points P ( 1 , 2 ) P ( 1 , 2 ) and Q ( x , y ) Q ( x , y ) are on the graph of the function f ( x ) = x 2 + 1 . f ( x ) = x 2 + 1 .

[T] Complete the following table with the appropriate values: y -coordinate of Q , the point Q ( x , y ) , Q ( x , y ) , and the slope of the secant line passing through points P and Q . Round your answer to eight significant digits.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at x = 1 . x = 1 .

Use the value in the preceding exercise to find the equation of the tangent line at point P . Graph f ( x ) f ( x ) and the tangent line.

For the following exercises, points P ( 1 , 1 ) P ( 1 , 1 ) and Q ( x , y ) Q ( x , y ) are on the graph of the function f ( x ) = x 3 . f ( x ) = x 3 .

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at x = 1 . x = 1 .

For the following exercises, points P ( 4 , 2 ) P ( 4 , 2 ) and Q ( x , y ) Q ( x , y ) are on the graph of the function f ( x ) = x . f ( x ) = x .

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at x = 4 . x = 4 .

Use the value in the preceding exercise to find the equation of the tangent line at point P .

For the following exercises, points P ( 1.5 , 0 ) P ( 1.5 , 0 ) and Q ( ϕ , y ) Q ( ϕ , y ) are on the graph of the function f ( ϕ ) = cos ( π ϕ ) . f ( ϕ ) = cos ( π ϕ ) .

[T] Complete the following table with the appropriate values: y -coordinate of Q , the point Q ( φ , y ) , Q ( φ , y ) , and the slope of the secant line passing through points P and Q . Round your answer to eight significant digits.

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at φ = 1.5 . φ = 1.5 .

For the following exercises, points P ( −1 , −1 ) P ( −1 , −1 ) and Q ( x , y ) Q ( x , y ) are on the graph of the function f ( x ) = 1 x . f ( x ) = 1 x .

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at x = −1 . x = −1 .

For the following exercises, the position function of a ball dropped from the top of a 200-meter tall building is given by s ( t ) = 200 − 4.9 t 2 , s ( t ) = 200 − 4.9 t 2 , where position s is measured in meters and time t is measured in seconds. Round your answer to eight significant digits.

[T] Compute the average velocity of the ball over the given time intervals.

• [ 4.99 , 5 ] [ 4.99 , 5 ]
• [ 5 , 5.01 ] [ 5 , 5.01 ]
• [ 4.999 , 5 ] [ 4.999 , 5 ]
• [ 5 , 5.001 ] [ 5 , 5.001 ]

Use the preceding exercise to guess the instantaneous velocity of the ball at t = 5 t = 5 sec.

For the following exercises, consider a stone tossed into the air from ground level with an initial velocity of 15 m/sec. Its height in meters at time t seconds is h ( t ) = 15 t − 4.9 t 2 . h ( t ) = 15 t − 4.9 t 2 .

[T] Compute the average velocity of the stone over the given time intervals.

• [ 1 , 1.05 ] [ 1 , 1.05 ]
• [ 1 , 1.01 ] [ 1 , 1.01 ]
• [ 1 , 1.005 ] [ 1 , 1.005 ]
• [ 1 , 1.001 ] [ 1 , 1.001 ]

Use the preceding exercise to guess the instantaneous velocity of the stone at t = 1 t = 1 sec.

For the following exercises, consider a rocket shot into the air that then returns to Earth. The height of the rocket in meters is given by h ( t ) = 600 + 78.4 t − 4.9 t 2 , h ( t ) = 600 + 78.4 t − 4.9 t 2 , where t is measured in seconds.

[T] Compute the average velocity of the rocket over the given time intervals.

• [ 9 , 9.01 ] [ 9 , 9.01 ]
• [ 8.99 , 9 ] [ 8.99 , 9 ]
• [ 9 , 9.001 ] [ 9 , 9.001 ]
• [ 8.999 , 9 ] [ 8.999 , 9 ]

Use the preceding exercise to guess the instantaneous velocity of the rocket at t = 9 t = 9 sec.

For the following exercises, consider an athlete running a 40-m dash. The position of the athlete is given by d ( t ) = t 3 6 + 4 t , d ( t ) = t 3 6 + 4 t , where d is the position in meters and t is the time elapsed, measured in seconds.

[T] Compute the average velocity of the runner over the given time intervals.

• [ 1.95 , 2.05 ] [ 1.95 , 2.05 ]
• [ 1.995 , 2.005 ] [ 1.995 , 2.005 ]
• [ 1.9995 , 2.0005 ] [ 1.9995 , 2.0005 ]
• [ 2 , 2.00001 ] [ 2 , 2.00001 ]

Use the preceding exercise to guess the instantaneous velocity of the runner at t = 2 t = 2 sec.

For the following exercises, consider the function f ( x ) = | x | . f ( x ) = | x | .

Sketch the graph of f over the interval [ −1 , 2 ] [ −1 , 2 ] and shade the region above the x -axis.

Use the preceding exercise to find the aproximate value of the area between the x -axis and the graph of f over the interval [ −1 , 2 ] [ −1 , 2 ] using rectangles. For the rectangles, use the square units, and approximate both above and below the lines. Use geometry to find the exact answer.

For the following exercises, consider the function f ( x ) = 1 − x 2 . f ( x ) = 1 − x 2 . ( Hint : This is the upper half of a circle of radius 1 positioned at ( 0 , 0 ) .) ( 0 , 0 ) .)

Sketch the graph of f over the interval [ −1 , 1 ] . [ −1 , 1 ] .

Use the preceding exercise to find the aproximate area between the x -axis and the graph of f over the interval [ −1 , 1 ] [ −1 , 1 ] using rectangles. For the rectangles, use squares 0.4 by 0.4 units, and approximate both above and below the lines. Use geometry to find the exact answer.

For the following exercises, consider the function f ( x ) = − x 2 + 1 . f ( x ) = − x 2 + 1 .

Approximate the area of the region between the x -axis and the graph of f over the interval [ −1 , 1 ] . [ −1 , 1 ] .

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What Is 6÷2(1+2) = ? Mathematician Explains The Correct Answer

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What is 6÷2(1+2) = ? What is 6/2(1+2) = ?

The problem often generates debate and has millions of comments on Facebook, Twitter, YouTube and other social media sites.

I posted a video with the correct answer.

Keep reading for a text explanation.

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The expression can be simplified by the order of operations, often remembered by the acronyms PEMDAS/BODMAS.

First evaluate P arentheses/ B rackets, then evaluate E xponents/ O rders, then evaluate M ultiplication- D ivision, and finally evaluate A ddition- S ubtraction.

Everyone is in agreement about the first step: simplify the addition inside of the parentheses.

6÷2(1+2) = 6÷2(3)

This is where the debate starts.

The answer is 9

If you type 6÷2(3) into a calculator, Google or WolframAlpha , the input has to be parsed and then computed. All of these will first convert the parentheses into an implied multiplication. The expression becomes the following.

6÷2(3) = 6÷2×3

According to the order of operations, division and multiplication have the same precedence, so the correct order is to evaluate from left to right. First take 6 and divide it by 2, and then multiply by 3.

6÷2×3 = 3×3 = 9

This gets to the correct answer of 9.

This is without argument the correct answer of how to evaluate this expression according to current usage.

Some people have a different interpretation. And while it’s not the correct answer today, it would have been regarded as the correct answer 100 years ago.

The other result of 1

Suppose it was 1917 and you saw 6÷2(3) in a textbook. What would you think the author was trying to write?

Historically the symbol ÷ was used to mean you should divide by the entire product on the right of the symbol (see longer explanation below).

Under that interpretation:

6÷2(3) = 6÷(2(3)) (Important: this is outdated usage!)

From this stage, the rest of the calculation works by the order of operations. First we evaluate the multiplication inside the parentheses. So we multiply 2 by 3 to get 6. And then we divide 6 by 6.

6÷(2(3)) = 6÷6 = 1

This gives the result of 1. This is not the correct answer; rather it is what someone might have interpreted the expression according to old usage.

The symbol ÷ historical use

Textbooks often used ÷ to denote the divisor was the whole expression to the right of the symbol. For example, a textbook would have written:

9 a 2 ÷3 a = 3 a (Important: this is outdated usage!)

This indicates that the divisor is the entire product on the right of the symbol. In other words, the problem is evaluated:

9 a 2 ÷3 a = 9 a 2 ÷(3 a ) (Important: this is outdated usage!)

I suspect the custom was out of practical considerations. The in-line expression would have been easier to typeset, and it takes up less space compared to writing a fraction as a numerator over a denominator:

The in-line expression also omits the parentheses of the divisor. This is like how trigonometry books commonly write sin 2θ to mean sin (2θ) because the argument of the function is understood, and writing parentheses every time would be cumbersome.

However, that practice of the division symbol was confusing, and it went against the order of operations. It was something of a well-accepted exception to the rule.

Today this practice is discouraged, and I have never seen a mathematician write an ambiguous expression using the division symbol. Textbooks always have proper parentheses, or they explain what is to be divided. Because mathematical typesetting is much easier today, we almost never see ÷ as a symbol, and instead fractions are written with the numerator vertically above the denominator.

*Note: I get many, many emails arguing with me about these order of operations problems, and most of the time people have misunderstood my point, not read the post fully, or not read the sources. If you send an email on this problem, I may not have time to reply.

A Related Problem

In 2019, the problem 8 ÷ 2(2 + 2) = went viral. I elaborated about binary expression trees and addressed common misconceptions like “Isn’t the answer ambiguous?” and “What about the distributive property?” For details please see my article 8 ÷ 2(2 + 2) = ? The Correct Answer Explained .

1. In 2013, Slate explained this problem and provided a bit about the history of the division symbol.

http://www.slate.com/articles/health_and_science/science/2013/03/facebook_math_problem_why_pemdas_doesn_t_always_give_a_clear_answer.html

2. The historical usage of ÷ is documented the following journal article from 1917. Read the second and third pages of the article (pages 94 and 95) for the usage of ÷ in evaluating expressions. Notice the author points out this was something of an “exception” to the order of operations.

Lennes, N. J. “Discussions: Relating to the Order of Operations in Algebra.” The American Mathematical Monthly 24.2 (1917): 93-95. Web. http://www.jstor.org/stable/2972726?seq=1#page_scan_tab_contents

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Presh talwalkar.

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13 thoughts on “What Is 6÷2(1+2) = ? Mathematician Explains The Correct Answer”

Hi, thanks for this good article. So what’s 6/2(1+2) then, 9 or 1? Do you interpret ÷ different than / if it is written inline? Kind regards

@Manuel: Hi, the expression 6/2(1+2) equals 9. In modern usage, ÷ and / have the same interpretation. Some people might write x/2y to mean x/(2y)–we try to avoid the confusion by putting parenthesis or indicating in the text “x divided by 2y.”

Can you write 6÷2(1+2) as (6/1)*(1/(2*(1+2))=(6/1)*(1/6)=6/6=1 with use of Commutative, Associative and Distributive Laws ?

If you give google 6/(2*(1+2)) for example you get 1

Question: according to your explanation Pemdas:= Parenthesis, Exponents, multiplication, division, addition ,substraction. Bodmas:= Brackets, (?), division, multiplication, addition, substraction. i.e. the “O” is nothing

From my old school teachings “O” = Of which is written as , to use this example, 2(3). That takes precedence to the divide.

More importantly this example clearly demonstrates that the order in which multiplication and division is applied significantly affects the answer which in maths one would deduce that there must be an order or precedence. I did Maths at university inn the early 90’s and at no point was it taught that Multiplication and division had the same precedence.

Please explain.

Another thought: You are saying I would be wrong to do the following: 6 / 2(1+2) — sorry no devide by symbol =6 / (2*1+2*2) = 6/(2+4) =6/(6) =1

@anzio You are correct. His argument of order of operations is wrong, because he resolves the inside of the paranteses instead of resolving the parenteses itself, as his PEMDAS/BODMAS system quite clearly states. The addition is the last thing that is supposed to be done.

I’m sorry, but I’m so upset with this article… So, correct answer is 9, so be it, but there is something, I just want to say. About this: “This is without argument the correct answer of how to evaluate this expression according to current usage.” And this: “Some people might write x/2y to mean x/(2y)”

So “some people” are a whole lot of mathematicians, physicists and other academics, professors, doctors, students alike, who have a nerve to insist, that multiplication denoted by juxtaposition should has higher precedence than division. And that rule used not only in some personal notes, but in scientific journals and academic literature too. I can understand if you don’t like it for some reason, but just denying it as something outdated used by bunch of freaks, sorry, but it’s quite insulting.

On the topic itself, I fail to understand, why this simple rule is still so unwelcome sometimes. So, some people might write x/2(y+z) to mean (x/2)(y+z). Why on Earth not to put it simply: x(y+z)/2? Quite different from x/2(y+z) without any new parenthesis. I see no ambiguity here. Instead people suggest to use parenthesis in every single denominator like x/(2y), which is a damn nightmare in mathematical analysis, and all this just to avoid following simple rule.

I’m sorry for my horrible English and if I sound rude – I definitely don’t want to insult anyone. I just find your attitude highly upsetting.

And finally, and I’m sorry one last time, but I’m pretty sure, that history of ÷ symbol has nothing to do with “wrong” answer. The rule about multiplication precedence over division is simply much more widespread, than you think, even outside high school, I suppose. Eh.

The “modern” version is incorrect for a simple reason: ALGEBRA. Algebra is the reason for PEMDAS, so it stands to reason that we should follow the procedure exactly the way we would if the numbers were variables.

If we were to substitute the first 2 with x, we get 6/x(1+2) 6/(x+2x) 6/3x 2/x since x=2 the answer is 1. That is unarguably the correct answer. Sorry.

I find it fascinating that this has elicited an emotional response. I also initially got an answer of 1, following what I was taught back in the 70s in primary school as ‘Please My Dear Aunt Sally’, I suppose which now corresponds to PEMDAS. I reached the answer of 1 because I took the MD part of PEMDAS to mean that multiplication would precede division. I’m willing to accept the convention that while P always precedes E and E always precedes M, that M and D should actually not precede one another but instead be evaluated from left to right. I think we have to accept Presh’s conclusion that the answer must be 9, and that this is the correct convention, because this has been codified widely, demonstrated by the Google and WolframAlpha points. If a convention so basic as this is NOT agreed upon, then we should expect satellites and airplanes to fall out of the sky and all kinds of mayhem (“Scotty, I need warp 6/2(1+2) now!” “No dammit, I said warp 9, not warp 1!”).

But again, it’s the emotion I find interesting. In the discussion of the problem we use adjectives such as unarguable and “not arguable” which I think reduces the quality of the discussion and leads to the emotional response. The analysis is clearly arguable, because it was argued. Unfortunately, one side in this case has to be wrong, else we would have mayhem. I accept that my own initial answer to the math question was wrong, but I have learned something, and am not upset by this. We just need to accept the convention and move on.

If I have 6 apples and I want to give this 6 apples on 2 groups of one boy and two girl the answer in one apple for each other. The answer can’t be 9 apple.

This guy that made this video is just a person seeking argument and ‘likes’ or ‘hits’. I teach aeronautical engineering and I Guarantee you that if the author of this post built aircraft based on his misguided interpretation of basic mathematics planes would crash and people would die. Thank the gods he dosent build planes. I think the crutch of his failings is thy he has added ,if there are two operations of the same level then work from left to right. This is just wrong. Left and right is irrelevant in mathematics. The quoted rule is simple multiplication first then divison as it is written. I find it annoying that I’m sitting here arguing with one that does not understand the basics of mathematics.

Some calculators follow a different order of precedence and give different answers for a/bc and a/b*c depending on entry. Engineers need to know how their calculators work and should proceed accordingly. That doesn’t change how PEMDAS should be interpreted, which by modern convention is (PE)(MD)(AS), in order of left to right. That unambiguously leads to 9 as the answer to the puzzle question if we apply that algorithm correctly. If you apply another algorithm, such as the algorithm used to program certain older TI calculators, and believe that is correct, you may come up with a different answer.

If you need more proof that 1 is the correct answer, let’s try some reverse algebra. We’ll set the problem equal to one of the suspected answers and then substitute a variable for one or more of the numbers, then solve for the variable.

First, we’ll let x/2(1+2)=9. The answer comes out to be 54. Hmm. Next, we’ll let x/2(1+2)=1. The answer comes out to be 6. Next, we’ll let 6/x(1+2)=9. The answer comes out to be 2/9. Not looking good for the 9 answer. Next, we’ll let 6/x(1+2)=1. The answer comes out to be 2. Next, we’ll let 6/2x=9. The answer comes out to be 1/3, which is not 1+2. Finally, we’ll let 6/2x=1. The answer comes out to be 3, which is 1+2.

I hope that read this, Presh Talwalker. You really need to be shown the correct way to do this.

Comments are closed.

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Home > CC3 > Chapter 2 > Lesson 2.1.1

Lesson 2.1.1, lesson 2.1.2, lesson 2.1.3, lesson 2.1.4, lesson 2.1.5, lesson 2.1.6, lesson 2.1.7, lesson 2.1.8, lesson 2.1.9.

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